Euclidean Jordan algebras and some conditions over the spectra of a strongly regular graph

Luís Vieira

doi:10.1051/fopen/2019017

Open Access

Review

Issue		4open Volume 2, 2019 Mathematical Models


Article Number		21
Number of page(s)		19
Section		Mathematics - Applied Mathematics
DOI		https://doi.org/10.1051/fopen/2019017
Published online		02 July 2019

4open 2019, 2, 21

Review Article

Euclidean Jordan algebras and some conditions over the spectra of a strongly regular graph

Luís Vieira^*

Department of Cívil Engineering of University of Porto, Street Roberto Frias, 0351-4200-465 Porto, Portugal

^* Corresponding author: This email address is being protected from spambots. You need JavaScript enabled to view it.

Received: 24 January 2019
Accepted: 13 May 2019

Abstract

Let G be a primitive strongly regular graph G such that the regularity is less than half of the order of G and A its matrix of adjacency, and let 𝒜 be the real Euclidean Jordan algebra of real symmetric matrices of order n spanned by the identity matrix of order n and the natural powers of A with the usual Jordan product of two symmetric matrices of order n and with the inner product of two matrices being the trace of their Jordan product. Next the spectra of two Hadamard series of 𝒜 associated to A² is analysed to establish some conditions over the spectra and over the parameters of G.

Key words: Euclidean Jordan algebras / Strongly regular graphs / Admissibility conditions

© L. Vieira, Published by EDP Sciences, 2019

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Introduction

The Euclidean Jordan algebras have many applications in several areas of mathematics. Some authors applied the theory of Euclidean Jordan algebras to interior-point methods [1–10], others applied this theory to combinatorics [11–14], and to statistics [15–17]. More actually, some authors extended the properties of the real symmetric matrices to the elements of simple real Euclidean Jordan algebras, see [18–24].

A good exposition about Jordan algebras can be founded in the beautiful work of K. McCrimmon, A taste of Jordan algebras, see [25], or for a more abstract survey one must cite the work of N. Jacobson, Structure and Representations of Jordan Algebras, see [26] and the PhD thesis of Michael Baes, Spectral Functions and Smoothing Techniques on Jordan Algebras, see [27].

For a well based understanding of the results of Euclidean Jordan algebras we must cite the works of Faraut and Korányi, Analysis on Symmetric Cones, see [28], the PhD thesis Jordan algebraic approach to symmetric optimization of Manuel Vieira, see [29], and the PhD thesis A Gershgorin type theorem, special inequalities and simultaneous stability in Euclidean Jordan algebras of Melanie Moldovan, see [30].

But for a very readable text on Euclidean Jordan algebras we couldn’t avoid of indicating, the chapter written by F. Alizadeh and S. H. Schmieta, Symmetric Cones, Potential Reduction Methods and Word-By-Word Extensions of the book Handbook of semi-definite programming, Theory, Algorithms and Applications, see [31], and the chapter, written by F. Alizadeh, An Introduction to Formally Real Jordan Algebras and Their Applications in Optimization of the book Handbook on Semidefinite, Conic and Polynomial Optimization, see [32].

In this paper we establish some admissibility conditions in an algebraic asymptotically way over the parameters and over the spectra of a primitive strongly regular graph.

This paper is organized as follows. In the second section we expose the most important concepts and results about Jordan algebras and Euclidean Jordan algebras, without presenting any proof of these results. Nevertheless some bibliography is present on the subject. In the following section we present some concepts about simple graphs and namely strongly regular graphs needed for a clear understanding of this work. In the last section we consider a three dimensional real Euclidean Jordan algebra $A$ $Mathematical equation: $ \mathcal{A}$$ associated to the adjacency matrix of a primitive strongly regular graph and we establish some admissibility conditions over, in an algebraic asymptotic way, the spectra and over the parameters of a strongly regular graph.

Principal results on Euclidean Jordan algebras

Herein, we describe the principal definitions, results and the more relevant theorems of the theory of Euclidean Jordan algebras without presenting the proof of them.

To make this exposition about Euclidean Jordan algebras we have recurred to the the monograph, Analysis on Symmetric Cones of Faraut and Kóranyi [28], and to the book A taste of Jordan algebra of Kevin McCrimmon [25]. But for general Jordan algebras very readable expositions can be found in the book Statistical Applications of Jordan algebras of James D. Malley [17].

Now, we will present only the main results about Euclidean Jordan algebras needed for this paper.

A Jordan algebra $A$ $Mathematical equation: $ \mathcal{A}$$ over a field $K$ $Mathematical equation: $ \mathbb{K}$$ with characteristic ≠ 2 is a vector space over the field $K$ $Mathematical equation: $ \mathbb{K}$$ with a operation of multiplication ⋄ such that for any x and y in $A$ $Mathematical equation: $ \mathcal{A}$$ , x ⋄ y = y ⋄ x and x ⋄ (x ^2⋄ ⋄ y) = x ^2⋄ ⋄ (x ⋄ y), where x ^2⋄ = x ⋄ x. We will suppose throughout this paper that if $A$ $Mathematical equation: $ \mathcal{A}$$ is a Jordan algebra then $A$ $Mathematical equation: $ \mathcal{A}$$ has a unit element that we will denote it by e.

When the field $K$ $Mathematical equation: $ \mathbb{K}$$ is the field of the reals numbers we call the Jordan algebra a real Jordan algebra. Since we are only interested in finite dimensional real Jordan algebras with unit element, we only consider Jordan algebras that are real finite dimensional Jordan algebras and that have an unit element e and that are equipped with an operation of multiplication that we denote by ⋄.

The real vector space of real symmetric matrices, $A = Sym (n, R)$ $Mathematical equation: $ \mathcal{A}=\enspace \mathrm{Sym}\enspace (n,\mathbb{R})$$ , of order n, with the operation $x ⋄ y = \frac{xy + yx}{2}$ $Mathematical equation: $ x\diamond y=\frac{{xy}+{yx}}{2}$$ is a real Jordan algebra.

We must note, that we define the powers of an element x in $A$ $Mathematical equation: $ \mathcal{A}$$ in the usual way $x^{0 ⋄} = e, x^{1 ⋄} = x, x^{2 ⋄} = x ⋄ x$ $Mathematical equation: $ {x}^{0\diamond }=\mathbf{e},\hspace{0.5em}{x}^{1\diamond }=x,\hspace{0.5em}{x}^{2\diamond }=x\diamond x$$ and $x^{k ⋄} = x ⋄ x^{(k - 1) ⋄}$ $Mathematical equation: $ {x}^{k\diamond }=x\diamond {x}^{(k-1)\diamond }$$ for any natural number k. Hence, we have $x^{2 ⋄} = x ⋄ x = \frac{xx + xx}{2} = \frac{2 x^{2}}{2} = x^{2}$ $Mathematical equation: $ {x}^{2\diamond }=x\diamond x=\frac{{xx}+{xx}}{2}=\frac{2{x}^2}{2}={x}^2$$ and therefore by induction over $N$ $Mathematical equation: $ \mathbb{N}$$ we conclude that $x^{k ⋄} = x^{k}$ $Mathematical equation: $ {x}^{k\diamond }={x}^k$$ for any natural number k, where x ^k represents the usual power of order k of a squared symmetric matrix. $A$ $Mathematical equation: $ \mathcal{A}$$ is a Jordan algebra since for x and y in $A$ $Mathematical equation: $ \mathcal{A}$$ we have $x ⋄ y = \frac{xy + yx}{2} = \frac{yx + xy}{2} = y ⋄ x$ $Mathematical equation: $ x\diamond y=\frac{{xy}+{yx}}{2}=\frac{{yx}+{xy}}{2}=y\diamond x$$ and

$\begin{matrix} x ⋄ (x^{2 ⋄} ⋄ y) & = x ⋄ (x^{2} ⋄ y) = x ⋄ (\frac{x^{2} y + y x^{2}}{2}) \\ \begin{array}{l} = \frac{x (\frac{x^{2} y + y x^{2}}{2}) + (\frac{x^{2} y + y x^{2}}{2}) x}{2} = \frac{\frac{x^{3} y + xy x^{2} + x^{2} yx + y x^{3}}{2}}{2} \\ = \frac{\frac{x^{3} y + x^{2} yx + xy x^{2} + y x^{3}}{2}}{2} = \frac{\frac{x^{2} (xy + yx) + (xy + yx) x^{2}}{2}}{2} \\ = \frac{x^{2} \frac{xy + yx}{2} + \frac{xy + yx}{2} x^{2}}{2} = \frac{x^{2} (x ⋄ y) + (x ⋄ y) x^{2}}{2} \\ = \frac{x^{2 ⋄} (x ⋄ y) + (x ⋄ y) x^{2 ⋄}}{2} = x^{2 ⋄} ⋄ (x ⋄ y) . \end{array} \end{matrix}$ $Mathematical equation: $$ \begin{array}{cc}x\diamond \left({x}^{2\diamond }\diamond y\right)& =x\diamond \left({x}^2\diamond y\right)=x\diamond \left(\frac{{x}^2y+y{x}^2}{2}\right)\\ & \begin{array}{l}\\ =\frac{x\left(\frac{{x}^2y+y{x}^2}{2}\right)+\left(\frac{{x}^2y+y{x}^2}{2}\right)x}{2}=\frac{\frac{{x}^3y+{xy}{x}^2+{x}^2{yx}+y{x}^3}{2}}{2}\\ =\frac{\frac{{x}^3y+{x}^2{yx}+{xy}{x}^2+y{x}^3}{2}}{2}=\frac{\frac{{x}^2\left({xy}+{yx}\right)+\left({xy}+{yx}\right){x}^2}{2}}{2}\\ =\frac{{x}^2\frac{{xy}+{yx}}{2}+\frac{{xy}+{yx}}{2}{x}^2}{2}=\frac{{x}^2\left(x\diamond y\right)+\left(x\diamond y\right){x}^2}{2}\\ =\frac{{x}^{2\diamond }\left(x\diamond y\right)+\left(x\diamond y\right){x}^{2\diamond }}{2}={x}^{2\diamond }\diamond \left(x\diamond y\right).\end{array}\end{array} $$$

Let’s consider another example of Euclidean Jordan algebra. Let consider the real vector space $A_{n + 1} = R^{n + 1}$ $Mathematical equation: $ {\mathcal{A}}_{n+1}={\mathbb{R}}^{n+1}$$ equipped with the product ⋄ such that

$z ⋄ w = [\begin{array}{l} z | w \\ z_{1} \overset{̅}{w} + w_{1} \overset{̅}{x} \end{array}],$ $Mathematical equation: $$ z\diamond w=\left[\begin{array}{l}z|w\\ {z}_1\bar{w}+{w}_1\bar{x}\end{array}\right], $$$ where $\begin{matrix} z = [\begin{array}{l} z_{1} \\ z_{2} \\ ⋮ \\ z_{n} \\ z_{n + 1} \end{array}], & w = [\begin{array}{l} w_{1} \\ w_{2} \\ ⋮ \\ w_{n} \\ w_{n + 1} \end{array}], & \overset{̅}{z} = [\begin{array}{l} z_{2} \\ z_{3} \\ ⋮ \\ z_{n - 1} \\ z_{n} \\ z_{n + 1} \end{array}] \end{matrix}$ $Mathematical equation: $ \begin{array}{ccc}z=\left[\begin{array}{l}{z}_1\\ {z}_2\\ \vdots \\ {z}_n\\ {z}_{n+1}\end{array}\right],& w=\left[\begin{array}{l}{w}_1\\ {w}_2\\ \vdots \\ {w}_n\\ {w}_{n+1}\end{array}\right],& \bar{z}=\left[\begin{array}{l}{z}_2\\ {z}_3\\ \vdots \\ {z}_{n-1}\\ {z}_n\\ {z}_{n+1}\end{array}\right]\end{array}\hspace{0.5em}$$ and $\overset{̅}{w} = [\begin{array}{l} w_{2} \\ w_{3} \\ ⋮ \\ w_{n - 1} \\ w_{n} \\ w_{n + 1} \end{array}] .$ $Mathematical equation: $ \bar{w}=\left[\begin{array}{l}{w}_2\\ {w}_3\\ \vdots \\ {w}_{n-1}\\ {w}_n\\ {w}_{n+1}\end{array}\right].$$ Now we will show that $z ⋄ w = w ⋄ z$ $Mathematical equation: $ z\diamond w=w\diamond z$$ and that $z ⋄ (z^{2 ⋄} ⋄ w) = z^{2 ⋄} (z ⋄ w) .$ $Mathematical equation: $ z\diamond \left({z}^{2\diamond }\diamond w\right)={z}^{2\diamond }\left(z\diamond w\right).$$ We have

$z ⋄ w = [\begin{array}{l} z | w \\ z_{1} \overset{̅}{w} + w_{1} \overset{̅}{z} \end{array}] = [\begin{array}{l} w | z \\ w_{1} \overset{̅}{z} + z_{1} \overset{̅}{w} \end{array}] = w ⋄ z .$ $Mathematical equation: $$ z\diamond w=\left[\begin{array}{l}z|w\\ {z}_1\bar{w}+{w}_1\bar{z}\end{array}\right]=\left[\begin{array}{l}w|z\\ {w}_1\bar{z}+{z}_1\bar{w}\end{array}\right]=w\diamond z. $$$ Herein, we must say that the element $e = [\begin{array}{l} 1 \\ \overset{̅}{0} \end{array}]$ $Mathematical equation: $ \mathbf{e}=\left[\begin{array}{l}1\\ \bar{0}\end{array}\right]$$ is the unit of the Euclidean Jordan algebra $A_{n + 1} .$ $Mathematical equation: $ {\mathcal{A}}_{n+1}.$$ Indeed, we have

$e ⋄ [\begin{array}{l} x_{1} \\ \overset{̅}{x} \end{array}] = [\begin{array}{l} 1 \\ \overset{̅}{0} \end{array}] ⋄ [\begin{array}{l} x_{1} \\ \overset{̅}{x} \end{array}] = [\begin{array}{l} x_{1} + \overset{̅}{x} | \overset{̅}{0} \\ x_{1} \overset{̅}{0} + 1 \overset{̅}{x} \end{array}] = [\begin{array}{l} x_{1} \\ \overset{̅}{x} \end{array}] .$ $Mathematical equation: $$ \mathbf{e}\diamond \left[\begin{array}{l}{x}_1\\ \bar{x}\end{array}\right]=\left[\begin{array}{l}1\\ \bar{0}\end{array}\right]\diamond \left[\begin{array}{l}{x}_1\\ \bar{x}\end{array}\right]=\left[\begin{array}{l}{x}_1+\bar{x}|\bar{0}\\ {x}_1\bar{0}+1\bar{x}\end{array}\right]=\left[\begin{array}{l}{x}_1\\ \bar{x}\end{array}\right]. $$$

Now since the operation ⋄ is commutative, we showed that

$e ⋄ [\begin{array}{l} x_{1} \\ \overset{̅}{x} \end{array}] = [\begin{array}{l} x_{1} \\ \overset{̅}{x} \end{array}] ⋄ e = [\begin{array}{l} x_{1} \\ \overset{̅}{x} \end{array}] .$ $Mathematical equation: $$ \mathbf{e}\diamond \left[\begin{array}{l}{x}_1\\ \bar{x}\end{array}\right]=\left[\begin{array}{l}{x}_1\\ \bar{x}\end{array}\right]\diamond \mathbf{e}=\left[\begin{array}{l}{x}_1\\ \bar{x}\end{array}\right]. $$$

Hence, e is a unit of the Jordan algebra $A_{n + 1} .$ $Mathematical equation: $ {\mathcal{A}}_{n+1}.$$ Now, we have

$\begin{array}{l} z ⋄ (z^{2 ⋄} ⋄ w) \\ = z ⋄ ((z ⋄ z) ⋄ w) \\ = z ⋄ ([\begin{array}{l} | | z | |^{2} \\ 2 z_{1} \overset{̅}{z} \end{array}] ⋄ [\begin{array}{l} w_{1} \\ \overset{̅}{w} \end{array}]) = [\begin{array}{l} z_{1} \\ \overset{̅}{z} \end{array}] ⋄ [\begin{array}{l} | | z | |^{2} w_{1} + 2 z_{1} \overset{̅}{z} | \overset{̅}{w} \\ | | z | |^{2} \overset{̅}{w} + 2 z_{1} w_{1} \overset{̅}{z} \end{array}] \\ = [\begin{array}{l} z_{1} w_{1} (| | z | |^{2} + 2 | | \overset{̅}{z} | |^{2}) + (| | z | |^{2} + 2 z_{1}^{2}) \overset{̅}{z} | \overset{̅}{w} \\ z_{1} | | z | |^{2} \overset{̅}{w} + 2 z_{1}^{2} w_{1} \overset{̅}{z} + (| | z | |^{2} w_{1} + 2 z_{1} \overset{̅}{z} | \overset{̅}{w}) \overset{̅}{z} \end{array}] \\ = [\begin{array}{l} z_{1} w_{1} (| | z | |^{2} + 2 | | \overset{̅}{z} | |^{2}) + (| | z | |^{2} + 2 z_{1}^{2}) \overset{̅}{z} | \overset{̅}{w} \\ z_{1} | | z | |^{2} \overset{̅}{w} + (2 z_{1}^{2} w_{1} + | | z | |^{2} w_{1} + 2 z_{1} \overset{̅}{z} | \overset{̅}{w}) \overset{̅}{z} \end{array}] \end{array}$ $Mathematical equation: $$ \begin{array}{l}z\diamond \left({z}^{2\diamond }\diamond w\right)\\ =z\diamond \left(\left(z\diamond z\right)\diamond w\right)\\ =z\diamond \left(\left[\begin{array}{l}||z|{|}^2\\ 2{z}_1\bar{z}\\ \end{array}\right]\diamond \left[\begin{array}{l}{w}_1\\ \bar{w}\end{array}\right]\right)=\left[\begin{array}{l}{z}_1\\ \bar{z}\end{array}\right]\diamond \left[\begin{array}{l}\left|\left|z\right|{|}^2{w}_1+2{z}_1\bar{z}\right|\bar{w}\\ ||z|{|}^2\bar{w}+2{z}_1{w}_1\bar{z}\end{array}\right]\\ =\left[\begin{array}{l}{z}_1{w}_1\left(\left|\left|z\right|{|}^2+2\right|\left|\bar{z}\right|{|}^2\right)+(||z|{|}^2+2{z}_1^2)\bar{z}|\bar{w}\\ {z}_1||z|{|}^2\bar{w}+2{z}_1^2{w}_1\bar{z}+\left(\left|\left|z\right|{|}^2{w}_1+2{z}_1\bar{z}\right|\bar{w}\right)\bar{z}\end{array}\right]\\ =\left[\begin{array}{l}{z}_1{w}_1\left(\left|\left|z\right|{|}^2+2\right|\left|\bar{z}\right|{|}^2\right)+(||z|{|}^2+2{z}_1^2)\bar{z}|\bar{w}\\ {z}_1||z|{|}^2\bar{w}+\left(2{z}_1^2{w}_1+\left|\left|z\right|{|}^2{w}_1+2{z}_1\bar{z}\right|\bar{w}\right)\bar{z}\end{array}\right]\end{array} $$$ and

$\begin{array}{l} z^{2 ⋄} ⋄ (z ⋄ w) \\ = ([\begin{array}{l} z_{1} \\ \overset{̅}{z} \end{array}] ⋄ [\begin{array}{l} z_{1} \\ \overset{̅}{z} \end{array}]) ⋄ ([\begin{array}{l} z_{1} \\ \overset{̅}{z} \end{array}] ⋄ [\begin{array}{l} w_{1} \\ \overset{̅}{w} \end{array}]) = ([\begin{array}{l} | | z | |^{2} \\ 2 z_{1} \overset{̅}{z} \end{array}]) ⋄ ([\begin{array}{l} z_{1} w_{1} + \overset{̅}{z} | \overset{̅}{w} \\ z_{1} \overset{̅}{w} + w_{1} \overset{̅}{z} \end{array}]) \\ = [\begin{array}{l} z_{1} w_{1} (| | z | |^{2} + 2 | | \overset{̅}{z} | |^{2}) + (2 z_{1}^{2} + | | \overset{̅}{z} | |^{2}) \overset{̅}{z} | \overset{̅}{w} \\ z_{1} | | z | |^{2} \overset{̅}{w} + (w_{1} | | z | |^{2} + 2 z_{1}^{2} w_{1} + 2 z_{1} \overset{̅}{z} | \overset{̅}{w}) \overset{̅}{z} . \end{array}] \end{array}$ $Mathematical equation: $$ \begin{array}{l}{z}^{2\diamond }\diamond \left(z\diamond w\right)\\ =\left(\left[\begin{array}{l}{z}_1\\ \bar{z}\end{array}\right]\diamond \left[\begin{array}{l}{z}_1\\ \bar{z}\end{array}\right]\right)\diamond \left(\left[\begin{array}{l}{z}_1\\ \bar{z}\end{array}\right]\diamond \left[\begin{array}{l}{w}_1\\ \bar{w}\end{array}\right]\right)=\left(\left[\begin{array}{l}||z|{|}^2\\ 2{z}_1\bar{z}\end{array}\right]\right)\diamond \left(\left[\begin{array}{l}{z}_1{w}_1+\bar{z}|\bar{w}\\ {z}_1\bar{w}+{w}_1\bar{z}\end{array}\right]\right)\\ =\left[\begin{array}{l}{z}_1{w}_1(||z|{|}^2+2||\bar{z}|{|}^2)+(2{z}_1^2+||\bar{z}|{|}^2)\bar{z}|\bar{w}\\ {z}_1||z|{|}^2\bar{w}+({w}_1||z|{|}^2+2{z}_1^2{w}_1+2{z}_1\bar{z}|\bar{w})\bar{z}.\end{array}\right]\end{array} $$$ Therefore, since for any z and w in $A_{n + 1}$ $Mathematical equation: $ {\mathcal{A}}_{n+1}$$ we have z ⋄ w = w ⋄ z and $z ⋄ (z^{2 ⋄} ⋄ w) = z^{2} ⋄ (z ⋄ w)$ $Mathematical equation: $ z\diamond ({z}^{2\diamond }\diamond w)={z}^2\diamond (z\diamond w)$$ . Then we conclude that $A_{n + 1}$ $Mathematical equation: $ {\mathcal{A}}_{n+1}$$ is a Jordan algebra.

Now we define a Jordan subalgebra of the Euclidean Jordan algebra $A = Sym (n, R)$ $Mathematical equation: $ \mathcal{A}=\enspace \mathrm{Sym}\enspace (n,\mathbb{R})$$ . Let’s consider the set ${B_{1}, B_{2}, \dots, B_{l}}$ $Mathematical equation: $ \{{B}_1,{B}_2,\cdots,{B}_l\}$$ of symmetric matrices of $M_{n} (R)$ $Mathematical equation: $ {M}_n(\mathbb{R})$$ such that

B ₁ = I _n,
$(B_{i})_{jk} \in {0,1}$ $Mathematical equation: $ ({B}_i{)}_{{jk}}\in \{\mathrm{0,1}\}$$ , for $j, k \in {1, \dots, l}$ $Mathematical equation: $ j,k\in \{1,\cdots,l\}$$ and $B_{i} = B_{i}^{T}$ $Mathematical equation: $ {B}_i={B}_i^T$$ for i = 1,⋯, l,
B ₁ + B ₂+⋯+B _l = J _n,
$\forall i, j \in {1, \dots, l}, \forall k \in {1, \dots, l} \exists α_{ij}^{k} \in R : B_{i} B_{j} = \sum_{k = 1}^{l} α_{ij}^{k} B_{k}$ $Mathematical equation: $ \forall i,j\in \left\{1,\cdots,l\right\},\forall k\in \left\{1,\cdots,l\right\}\exists {\alpha }_{{ij}}^k\in \mathbb{R}:{B}_i{B}_j={\sum }_{k=1}^l {\alpha }_{{ij}}^k{B}_k$$ .

Then the real vector space $B$ $Mathematical equation: $ \mathcal{B}$$ spanned by the set ${B_{1}, B_{2}, \dots, B_{l}}$ $Mathematical equation: $ \{{B}_1,{B}_2,\cdots,{B}_l\}$$ with the Jordan product ⋄ is a Jordan algebra. Firstly, we must say that $B$ $Mathematical equation: $ \mathcal{B}$$ is closed for the Jordan product. Indeed, for C and D in $B$ $Mathematical equation: $ \mathcal{B}$$ we have $C = α_{1} B_{1} + \dots + α_{l} B_{l}$ $Mathematical equation: $ C={\alpha }_1{B}_1+\cdots +{\alpha }_l{B}_l$$ and $D = β_{1} B_{1} + \dots + β_{l} B_{l} .$ $Mathematical equation: $ D={\beta }_1{B}_1+\cdots +{\beta }_l{B}_l.$$ Therefore

$\begin{array}{l} C^{T} = (α_{1} B_{1} + \dots + α_{l} B_{l})^{T} = α_{1} B_{1}^{T} + \dots + α_{l} B_{l}^{T} \\ = α_{1} B_{1} + \dots + α_{l} B_{l} = C \end{array}$ $Mathematical equation: $$ \begin{array}{l}{C}^T=({\alpha }_1{B}_1+\cdots +{\alpha }_l{B}_l{)}^T={\alpha }_1{B}_1^T+\cdots +{\alpha }_l{B}_l^T\\ ={\alpha }_1{B}_1+\cdots +{\alpha }_l{B}_l=C\end{array} $$$ and

$\begin{array}{l} D^{T} = (β_{1} B_{1} + \dots + β_{l} B_{l})^{T} = β_{1} B_{1}^{T} + \dots + β_{l} B_{l}^{T} \\ = β_{1} B_{1} + \dots + β_{l} B_{l} = D . \end{array}$ $Mathematical equation: $$ \begin{array}{l}{D}^T=({\beta }_1{B}_1+\cdots +{\beta }_l{B}_l{)}^T={\beta }_1{B}_1^T+\cdots +{\beta }_l{B}_l^T\\ ={\beta }_1{B}_1+\cdots +{\beta }_l{B}_l=D.\end{array} $$$

Since $C ⋄ D = \frac{CD + DC}{2}$ $Mathematical equation: $ C\diamond D=\frac{{CD}+{DC}}{2}$$ then

$\begin{array}{l} (C ⋄ D)^{T} = \frac{(CD + DC)^{T}}{2} = \frac{(CD)^{T} + (DC)^{T}}{2} = \frac{D^{T} C^{T} + C^{T} D^{T}}{2} \\ = \frac{DC + CD}{2} = \frac{CD + DC}{2} = C ⋄ D . \end{array}$ $Mathematical equation: $$ \begin{array}{l}(C\diamond D{)}^T=\frac{({CD}+{DC}{)}^T}{2}=\frac{({CD}{)}^T+({DC}{)}^T}{2}=\frac{{D}^T{C}^T+{C}^T{D}^T}{2}\\ =\frac{{DC}+{CD}}{2}=\frac{{CD}+{DC}}{2}=C\diamond D.\end{array} $$$

So $C ⋄ D$ $Mathematical equation: $ C\diamond D$$ is a symmetric matrix of $M_{n} (R)$ $Mathematical equation: $ {M}_n(\mathbb{R})$$ , but from property iv) we conclude that $C ⋄ D = \sum_{k = 1}^{l = 1} γ_{k} B_{k},$ $Mathematical equation: $ C\diamond D={\sum }_{k=1}^{l=1} {\gamma }_k{B}_k,$$ therefore $C ⋄ D$ $Mathematical equation: $ C\diamond D$$ in $B .$ $Mathematical equation: $ \mathcal{B}.$$ Now, we must say, that for any matrices X and $Y \in B$ $Mathematical equation: $ Y\in \mathcal{B}$$ we have X ⋄ Y = Y ⋄ X and $X ⋄ (X^{2} ⋄ Y) = X^{2} ⋄ (X ⋄ Y)$ $Mathematical equation: $ X\diamond ({X}^2\diamond Y)={X}^2\diamond (X\diamond Y)$$ since $B$ $Mathematical equation: $ \mathcal{B}$$ is a subalgebra of $A = Sym (n, R)$ $Mathematical equation: $ \mathcal{A}=\enspace \mathrm{Sym}\enspace (n,\mathbb{R})$$ .

Let $A$ $Mathematical equation: $ \mathcal{A}$$ be a n-dimensional Jordan algebra. Then $A$ $Mathematical equation: $ \mathcal{A}$$ is power associative, this is, is an algebra such that for any x in $A$ $Mathematical equation: $ \mathcal{A}$$ the algebra spanned by x and e is associative. For x in $A$ $Mathematical equation: $ \mathcal{A}$$ we define rank (x) as being the least natural number k such that ${e, x^{1 ⋄}, \dots, x^{k ⋄}}$ $Mathematical equation: $ \{\mathbf{e},{x}^{1\diamond },\dots,{x}^{k\diamond }\}$$ is a linearly dependent set and we write rank (x) = k. Now since $\dim (A) = n$ $Mathematical equation: $ \mathrm{dim}(\mathcal{A})=n$$ then $rank (x) \leq n .$ $Mathematical equation: $ \mathrm{rank}\enspace (x)\le n.$$ The rank of A is defined as being the natural number $r = rank (A) = \max {rank (x) : x \in A} .$ $Mathematical equation: $ r=\enspace \mathrm{rank}\enspace (A)=\enspace \mathrm{max}\enspace \{\enspace \mathrm{rank}\enspace (x):x\in A\}.$$ An element x in $A$ $Mathematical equation: $ \mathcal{A}$$ is regular if $rank (x) = r .$ $Mathematical equation: $ \mathrm{rank}\enspace (x)=r.$$ Now, we must observe that the set of regular elements of $A$ $Mathematical equation: $ \mathcal{A}$$ is a dense set in $A$ $Mathematical equation: $ \mathcal{A}$$ . Let’s consider a regular element x of $A$ $Mathematical equation: $ \mathcal{A}$$ and r = rank (x).

Then, there exist real numbers $α_{1} (x), α_{2} (x), \dots, α_{r - 1} (x)$ $Mathematical equation: $ {\alpha }_1(x),{\alpha }_2(x),\dots,{\alpha }_{r-1}(x)$$ and $α_{r} (x)$ $Mathematical equation: $ {\alpha }_r(x)$$ such that

$x^{r ⋄} - α_{1} (x) x^{(r - 1) ⋄} + \dots + (- 1)^{r} α_{r} (x) e = 0,$ $Mathematical equation: $$ {x}^{r\diamond }-{\alpha }_1(x){x}^{(r-1)\diamond }+\cdots +(-1{)}^r{\alpha }_r(x)\mathbf{e}=0, $$$ (1)where 0 is the zero vector of $A$ $Mathematical equation: $ \mathcal{A}$$ . Taking into account (1) we conclude that the polynomial p(x, −) define by the equality (2).

$p (x, λ) = λ^{r} - α_{1} (x) λ^{r - 1} + \dots + (- 1)^{r} α_{r} (x),$ $Mathematical equation: $$ p(x,\lambda )={\lambda }^r-{\alpha }_1(x){\lambda }^{r-1}+\cdots +(-1{)}^r{\alpha }_r(x), $$$ (2)is the minimal polynomial of x. When x is a non regular element of $A$ $Mathematical equation: $ \mathcal{A}$$ the minimal polynomial of x has a degree less than r. The polynomial p(x, −) is called the characteristic polynomial of x. Now, we must say that the coefficients $α_{i} (x)$ $Mathematical equation: $ {\alpha }_i(x)$$ are homogeneous polynomials of degree i on the coordinates of x on a fixed basis of $A$ $Mathematical equation: $ \mathcal{A}$$ . Since the set of regular elements of $A$ $Mathematical equation: $ \mathcal{A}$$ is a dense set in $A$ $Mathematical equation: $ \mathcal{A}$$ then we extend the definition of characteristic polynomial to non regular elements of $A$ $Mathematical equation: $ \mathcal{A}$$ by continuity. The roots of the characteristic polynomial p(x, −) of x are called the eigenvalues of x. The coefficient $α_{1} (x)$ $Mathematical equation: $ {\alpha }_1(x)$$ of the characteristic polynomial p(x, −) is called the trace of x and we denote it by Trace (x) and we call the coefficient $α_{r} (x)$ $Mathematical equation: $ {\alpha }_r(x)$$ the determinant of x and we denote it by Det (x).

Let $A$ $Mathematical equation: $ \mathcal{A}$$ be a real finite dimensional associative algebra with the bilinear map $(x, y) \mapsto x ⋄ y$ $Mathematical equation: $ (x,y)\mapsto x\diamond y$$ . We introduce on $A$ $Mathematical equation: $ \mathcal{A}$$ a structure of Jordan algebra by considering a new product • defined by $x \cdot y = (x ⋄ y + y ⋄ x) / 2$ $Mathematical equation: $ x\middot y=(x\diamond y+y\diamond x)/2$$ for all x and y in $A$ $Mathematical equation: $ \mathcal{A}$$ . The product · is called the Jordan product of x by y. Let $A$ $Mathematical equation: $ \mathcal{A}$$ be a real Jordan algebra and x be a regular element of $A .$ $Mathematical equation: $ \mathcal{A}.$$ Then we have $rank (x) = r = rank (A) .$ $Mathematical equation: $ \mathrm{rank}\enspace (x)=r=\enspace \mathrm{rank}\enspace (\mathcal{A}).$$ We define the linear operator $L_{⋄} (x)$ $Mathematical equation: $ {L}_{\diamond }(x)$$ such that $L_{⋄} (x) z = x ⋄ z, \forall z \in A .$ $Mathematical equation: $ {L}_{\diamond }(x)z=x\diamond z,\forall z\in \mathcal{A}.$$ We define the real vector space $R [x]$ $Mathematical equation: $ \mathbb{R}[x]$$ by $R [x] = {z \in A : \exists γ_{0}, γ_{1}, \dots, γ_{r - 1} \in R : z = γ_{0} e + γ_{1} x^{1 ⋄} + \dots + γ_{r - 1} x^{(r - 1) ⋄}} .$ $Mathematical equation: $ \mathbb{R}[x]=\{z\in \mathcal{A}:\exists {\gamma }_0,{\gamma }_1,\cdots,{\gamma }_{r-1}\in \mathbb{R}:z={\gamma }_0\mathbf{e}+{\gamma }_1{x}^{1\diamond }+\cdots +{\gamma }_{r-1}{x}^{(r-1)\diamond }\}.$$ The restriction of the linear operator $L_{⋄} (x)$ $Mathematical equation: $ {L}_{\diamond }(x)$$ to $R [x]$ $Mathematical equation: $ \mathbb{R}[x]$$ we call $L_{⋄}^{0} (x) .$ $Mathematical equation: $ {L}_{\diamond }^0(x).$$ We must note now that $trace (L_{⋄}^{0} (x)) = a_{1} (x) = Trace (x)$ $Mathematical equation: $ \mathrm{trace}\enspace ({L}_{\diamond }^0(x))={a}_1(x)=\enspace \mathrm{Trace}\enspace (x)$$ and $\det (L_{⋄}^{0} (x)) = a_{r} (x) = Det (x) .$ $Mathematical equation: $ \mathrm{det}\enspace ({L}_{\diamond }^0(x))={a}_r(x)=\enspace \mathrm{Det}\enspace (x).$$

A Jordan algebra is simple if and only if does not contain any nontrivial ideal. An Euclidean Jordan algebra $A$ $Mathematical equation: $ \mathcal{A}$$ is a Jordan algebra with an inner product ·|· such that $L_{⋄} (x) y | z = y | L_{⋄} (x) z$ $Mathematical equation: $ {L}_{\diamond }(x)y|z=y|{L}_{\diamond }(x)z$$ , for all x, y and z in $A .$ $Mathematical equation: $ \mathcal{A}.$$ Herein, we must say that an Euclidean Jordan algebra is simple if and only if it can’t be written as a direct sum of two Euclidean Jordan algebras. But it is already proved that every Euclidean Jordan algebra is a direct orthogonal sum of simple Euclidean Jordan algebras.

The Jordan algebra $A = Sym (n, R)$ $Mathematical equation: $ \mathcal{A}=\enspace \mathrm{Sym}\enspace (n,\mathbb{R})$$ equipped with the Jordan product $x ⋄ y = \frac{xy + yx}{2}$ $Mathematical equation: $ x\diamond y=\frac{{xy}+{yx}}{2}$$ with xy and yx the usual products of matrices of order n, x by y and of y by x and with the inner product $x | y = trace (L_{⋄} (x) y)$ $Mathematical equation: $ x|y=\enspace \mathrm{trace}\enspace ({L}_{\diamond }(x)y)$$ for x and y in $A$ $Mathematical equation: $ \mathcal{A}$$ is an Euclidean Jordan algebra. Indeed, we have

$\begin{array}{l} L_{⋄} (x) y | z = x ⋄ y | z \\ = trace ((x ⋄ y) ⋄ z) \\ = trace ((\frac{xy + yx}{2}) ⋄ z) \\ = trace (\frac{(\frac{xy + yx}{2}) z + z (\frac{xy + yx}{2})}{2}) \\ = trace (\frac{(xy) z}{4} + \frac{(yx) z}{4} + \frac{z (xy)}{4} + \frac{z (yx)}{4}) \\ = trace (\frac{(xy) z}{4}) + trace (\frac{(yx) z}{4}) + trace (\frac{z (xy)}{4}) + trace (\frac{z (yx)}{4}) \\ = trace (\frac{(yx) z}{4}) + trace (\frac{z (xy)}{4}) + trace (\frac{(xy) z}{4}) + trace (\frac{z (yx)}{4}) \\ = trace (\frac{y (xz)}{4}) + trace (\frac{(zx) y}{4}) + trace (\frac{(xy) z}{4}) + trace (\frac{(zy) x}{4}) \\ = trace (\frac{y (xz)}{4}) + trace (\frac{y (zx)}{4}) + trace (\frac{z (xy)}{4}) + trace (\frac{x (zy)}{4}) \\ = trace (\frac{y (xz)}{4}) + trace (\frac{y (zx)}{4}) + trace (\frac{(zx) y}{4}) + trace (\frac{(xz) y}{4}) \\ = trace (\frac{y (\frac{xz + zx}{2}) + (\frac{xz + zx}{2}) y}{2}) \\ = trace (\frac{y (x ⋄ z) + (x ⋄ z) y}{2}) \\ = trace (y ⋄ (x ⋄ z)) \\ = trace (L_{⋄} (y) (x ⋄ z)) \\ = y | x ⋄ z \\ = y | L_{⋄} (x) z . \end{array}$ $Mathematical equation: $$ \begin{array}{l}{L}_{\diamond }(x)y|z=x\diamond y|z\\ =\enspace \mathrm{trace}\enspace ((x\diamond y)\diamond z)\\ =\enspace \mathrm{trace}\enspace \left(\left(\frac{{xy}+{yx}}{2}\right)\diamond z\right)\\ =\enspace \mathrm{trace}\enspace \left(\frac{\left(\frac{{xy}+{yx}}{2}\right)z+z\left(\frac{{xy}+{yx}}{2}\right)}{2}\right)\\ =\enspace \mathrm{trace}\enspace \left(\frac{({xy})z}{4}+\frac{({yx})z}{4}+\frac{z({xy})}{4}+\frac{z({yx})}{4}\right)\\ =\enspace \mathrm{trace}\enspace \left(\frac{({xy})z}{4}\right)+\enspace \mathrm{trace}\enspace \left(\frac{({yx})z}{4}\right)+\enspace \mathrm{trace}\enspace \left(\frac{z({xy})}{4}\right)+\mathrm{trace}\enspace \left(\frac{z({yx})}{4}\right)\\ =\enspace \mathrm{trace}\enspace \left(\frac{({yx})z}{4}\right)+\enspace \mathrm{trace}\enspace \left(\frac{z({xy})}{4}\right)+\mathrm{trace}\enspace \left(\frac{({xy})z}{4}\right)+\mathrm{trace}\left(\frac{z({yx})}{4}\right)\\ =\enspace \mathrm{trace}\enspace \left(\frac{y({xz})}{4}\right)+\enspace \mathrm{trace}\enspace \left(\frac{({zx})y}{4}\right)+\enspace \mathrm{trace}\enspace \left(\frac{({xy})z}{4}\right)+\enspace \mathrm{trace}\enspace \left(\frac{({zy})x}{4}\right)\\ =\enspace \mathrm{trace}\enspace \left(\frac{y({xz})}{4}\right)+\enspace \mathrm{trace}\left(\frac{y({zx})}{4}\right)+\enspace \mathrm{trace}\enspace \left(\frac{z({xy})}{4}\right)+\enspace \mathrm{trace}\enspace \left(\frac{x({zy})}{4}\right)\\ =\enspace \mathrm{trace}\enspace \left(\frac{y({xz})}{4}\right)+\enspace \mathrm{trace}\enspace \left(\frac{y({zx})}{4}\right)+\enspace \mathrm{trace}\enspace \left(\frac{({zx})y}{4}\right)+\enspace \mathrm{trace}\enspace \left(\frac{({xz})y}{4}\right)\\ =\enspace \mathrm{trace}\enspace \left(\frac{y\left(\frac{{xz}+{zx}}{2}\right)+\left(\frac{{xz}+{zx}}{2}\right)y}{2}\right)\\ =\enspace \mathrm{trace}\enspace \left(\frac{y(x\diamond z)+(x\diamond z)y}{2}\right)\\ =\enspace \mathrm{trace}\enspace \left(y\diamond (x\diamond z)\right)\\ =\enspace \mathrm{trace}\enspace ({L}_{\diamond }(y)(x\diamond z))\\ =y|x\diamond z\\ =y|{L}_{\diamond }(x)z.\end{array} $$$

Now, we will show that the $A_{n + 1} = R^{n + 1}$ $Mathematical equation: $ {\mathcal{A}}_{n+1}={\mathbb{R}}^{n+1}$$ is an Euclidean Jordan algebra relatively to the inner product $[\begin{array}{l} x_{1} \\ \overset{̅}{x} \end{array}] | [\begin{array}{l} y_{1} \\ \overset{̅}{y} \end{array}] = x_{1} y_{1} + \overset{̅}{x} | \overset{̅}{y} .$ $Mathematical equation: $ \left[\begin{array}{l}{x}_1\\ \bar{x}\end{array}\right]|\left[\begin{array}{l}{y}_1\\ \bar{y}\end{array}\right]={x}_1{y}_1+\bar{x}|\bar{y}.$$ We have the following calculations

$\begin{array}{l} L_{⋄} (x) y | z = (x ⋄ y) | z \\ = ([\begin{array}{l} x_{1} \\ \overset{̅}{x} \end{array}] ⋄ [\begin{array}{l} y_{1} \\ \overset{̅}{y} \end{array}]) | [\begin{array}{l} z_{1} \\ \overset{̅}{z} \end{array}] \\ = [\begin{array}{l} x_{1} y_{1} + \overset{̅}{x} | \overset{̅}{y} \\ x_{1} \overset{̅}{y} + y_{1} \overset{̅}{x} \end{array}] | [\begin{array}{l} z_{1} \\ \overset{̅}{z} \end{array}] \\ = x_{1} y_{1} z_{1} + z_{1} \overset{̅}{x} | \overset{̅}{y} + x_{1} \overset{̅}{y} | \overset{̅}{z} + y_{1} \overset{̅}{x} | \overset{̅}{z} \end{array}$ $Mathematical equation: $$ \begin{array}{l}{L}_{\diamond }(x)y|z=(x\diamond y)|z\\ =\left(\left[\begin{array}{l}{x}_1\\ \bar{x}\end{array}\right]\diamond \left[\begin{array}{l}{y}_1\\ \bar{y}\end{array}\right]\right)|\left[\begin{array}{l}{z}_1\\ \bar{z}\end{array}\right]\\ =\left[\begin{array}{l}{x}_1{y}_1+\bar{x}|\bar{y}\\ {x}_1\bar{y}+{y}_1\bar{x}\\ \end{array}\right]|\left[\begin{array}{l}{z}_1\\ \bar{z}\end{array}\right]\\ ={x}_1{y}_1{z}_1+{z}_1\bar{x}|\bar{y}+{x}_1\bar{y}|\bar{z}+{y}_1\bar{x}|\bar{z}\end{array} $$$ and

$\begin{array}{l} y | (L_{⋄} (x) z) = y | (x ⋄ z) = [\begin{array}{l} y_{1} \\ \overset{̅}{y} \end{array}] | ([\begin{array}{l} x_{1} \\ \overset{̅}{x} \end{array}] ⋄ [\begin{array}{l} z_{1} \\ \overset{̅}{z} \end{array}]) \\ [\begin{array}{l} y_{1} \\ \overset{̅}{y} \end{array}] | [\begin{array}{l} x_{1} z_{1} + \overset{̅}{x} | \overset{̅}{z} \\ x_{1} \overset{̅}{z} + z_{1} \overset{̅}{x} \end{array}] \\ = x_{1} y_{1} z_{1} + y_{1} \overset{̅}{x} | \overset{̅}{z} + x_{1} \overset{̅}{y} | \overset{̅}{z} + z_{1} \overset{̅}{y} | \overset{̅}{x} \\ = x_{1} y_{1} z_{1} + z_{1} \overset{̅}{y} | \overset{̅}{x} + x_{1} \overset{̅}{y} | \overset{̅}{z} + y_{1} \overset{̅}{x} | \overset{̅}{z} . \end{array}$ $Mathematical equation: $$ \begin{array}{l}y|({L}_{\diamond }(x)z)=y|(x\diamond z)=\left[\begin{array}{l}{y}_1\\ \bar{y}\\ \end{array}\right]|\left(\left[\begin{array}{l}{x}_1\\ \bar{x}\end{array}\right]\diamond \left[\begin{array}{l}{z}_1\\ \bar{z}\end{array}\right]\right)\\ \left[\begin{array}{l}{y}_1\\ \bar{y}\\ \end{array}\right]|\left[\begin{array}{l}{x}_1{z}_1+\bar{x}|\bar{z}\\ {x}_1\bar{z}+{z}_1\bar{x}\\ \end{array}\right]\\ ={x}_1{y}_1{z}_1+{y}_1\bar{x}|\bar{z}+{x}_1\bar{y}|\bar{z}+{z}_1\bar{y}|\bar{x}\\ ={x}_1{y}_1{z}_1+{z}_1\bar{y}|\bar{x}+{x}_1\bar{y}|\bar{z}+{y}_1\bar{x}|\bar{z}.\end{array} $$$

Hence, we conclude that $L_{⋄} (x) y | z = y | L_{⋄} (x) z .$ $Mathematical equation: $ {L}_{\diamond }(x)y|z=y|{L}_{\diamond }(x)z.$$ And therefore $A_{n + 1}$ $Mathematical equation: $ {\mathcal{A}}_{n+1}$$ is an Euclidean Jordan algebra.

Now, we will analyse the rank of the Euclidean Jordan algebra $A = Sym (n, R)$ $Mathematical equation: $ \mathcal{A}=\enspace \mathrm{Sym}\enspace (n,\mathbb{R})$$ . Let consider the element x of $A$ $Mathematical equation: $ \mathcal{A}$$ with n distinct eigenvalues λ ₁, λ ₂,…, λ _n−1 and λ _n, and $B = {v_{1}, v_{2}, \dots, v_{n}}$ $Mathematical equation: $ \mathcal{B}=\{{v}_1,{v}_2,\cdots,{v}_n\}$$ an orthonormal basis of $R^{n}$ $Mathematical equation: $ {\mathbb{R}}^n$$ of eigenvectors of x such that $x v_{i} = λ_{i} v_{i}$ $Mathematical equation: $ x{v}_i={\lambda }_i{v}_i$$ for i=1,⋯, n. Considering the notation e = I _n, then we have:

$\begin{matrix} e = v_{1} v_{1}^{T} + v_{2} v_{2}^{T} + \dots + v_{n} v_{n}^{T} \\ x^{1 ⋄} = λ_{1} v_{1} v_{1}^{T} + λ_{2} v_{2} v_{2}^{T} + \dots + λ_{n} v_{n} v_{n}^{T} \\ \begin{matrix} x^{2 ⋄} = λ_{1}^{2} v_{1} v_{1}^{T} + λ_{2}^{2} v_{2}^{T} + \dots + λ_{n}^{2} v_{n} v_{n}^{T} \\ ⋮ = ⋮ \\ x^{n - 1 ⋄} = λ^{n - 1} v_{1} v_{1}^{T} + λ_{2}^{n - 1} v_{2} v_{2}^{T} + \dots + λ_{n}^{n - 1} v_{n} v_{n}^{T} . \end{matrix} \end{matrix}$ $Mathematical equation: $$ \begin{array}{c}\mathbf{e}={v}_1{v}_1^T+{v}_2{v}_2^T+\cdots +{v}_n{v}_n^T\\ {x}^{1\diamond }={\lambda }_1{v}_1{v}_1^T+{\lambda }_2{v}_2{v}_2^T+\cdots +{\lambda }_n{v}_n{v}_n^T\\ \begin{array}{c}{x}^{2\diamond }={\lambda }_1^2{v}_1{v}_1^T+{\lambda }_2^2{v}_2^T+\cdots +{\lambda }_n^2{v}_n{v}_n^T\\ \vdots =\vdots \\ {x}^{n-1\diamond }={\lambda }^{n-1}{v}_1{v}_1^T+{\lambda }_2^{n-1}{v}_2{v}_2^T+\cdots +{\lambda }_n^{n-1}{v}_n{v}_n^T.\end{array}\end{array} $$$

Therefore, the set $X_{n - 1} = {e, x^{1 ⋄}, x^{2 ⋄}, \dots, x^{n - 1 ⋄}}$ $Mathematical equation: $ {\mathcal{X}}_{n-1}=\{\mathbf{e},{x}^{1\diamond },{x}^{2\diamond },\cdots,{x}^{n-1\diamond }\}$$ is a linearly independent set of $A$ $Mathematical equation: $ \mathcal{A}$$ if and only if the set

$S_{n - 1} = {(1,1, \dots, 1), (λ_{1}, λ_{2}, \dots, λ_{n}), (λ_{1}^{2}, λ_{2}^{2}, \dots, λ_{n}^{2}) \dots, (λ_{1}^{n - 1}, λ_{2}^{n - 1}, \dots, λ_{n}^{n - 1})}$ $Mathematical equation: $$ {S}_{n-1}=\left\{(\mathrm{1,1},\cdots,1),({\lambda }_1,{\lambda }_2,\cdots,{\lambda }_n),({\lambda }_1^2,{\lambda }_2^2,\cdots,{\lambda }_n^2)\cdots,({\lambda }_1^{n-1},{\lambda }_2^{n-1},\cdots,{\lambda }_n^{n-1})\right\} $$$ is a linearly independent set of $R^{n} .$ $Mathematical equation: $ {\mathbf{R}}^n.$$ But, since

$| \begin{array}{l} 1 & 1 & \dots & 1 \\ λ_{1} & λ_{2} & \dots & λ_{n} \\ λ_{1}^{2} & λ_{2}^{2} & \dots & λ_{n}^{2} \\ ⋮ & ⋮ & \dots & ⋮ \\ λ_{1}^{n - 1} & λ_{2}^{n - 1} & \dots & λ_{n}^{n - 1} \end{array} | = \prod_{1 \leq i < j \leq n} (λ_{j} - λ_{i}) \neq 0$ $Mathematical equation: $$ \left|\begin{array}{llll}1& 1& \cdots & 1\\ {\lambda }_1& {\lambda }_2& \cdots & {\lambda }_n\\ {\lambda }_1^2& {\lambda }_2^2& \cdots & {\lambda }_n^2\\ \vdots & \vdots & \cdots & \vdots \\ {\lambda }_1^{n-1}& {\lambda }_2^{n-1}& \cdots & {\lambda }_n^{n-1}\end{array}\right|=\prod_{1\le i < j\le n} ({\lambda }_j-{\lambda }_i)\ne 0 $$$ then the set $S_{n - 1}$ $Mathematical equation: $ {S}_{n-1}$$ is a linearly independent set of $R^{n}$ $Mathematical equation: $ {\mathbb{R}}^n$$ and therefore the set

$X_{n - 1} = {e, x^{1 ⋄}, x^{2 ⋄}, \dots, x^{n - 1 ⋄}}$ $Mathematical equation: $$ {\mathcal{X}}_{n-1}=\{\mathbf{e},{x}^{1\diamond },{x}^{2\diamond },\cdots,{x}^{n-1\diamond }\} $$$ is a linearly independent set of $A .$ $Mathematical equation: $ \mathcal{A}.$$ The set $X_{n} = {e, x^{1 ⋄}, x^{2 ⋄}, \dots, x^{n ⋄}}$ $Mathematical equation: $ {\mathcal{X}}_n=\{\mathbf{e},{x}^{1\diamond },{x}^{2\diamond },\cdots,{x}^{n\diamond }\}$$ is a linear dependent set of $A$ $Mathematical equation: $ \mathcal{A}$$ since the set

$S_{n} = {(1,1, \dots, 1), (λ_{1}, λ_{2}, \dots, λ_{n}), \dots, (λ_{1}^{n - 1}, λ_{2}^{n - 1}, \dots, λ_{n}^{n - 1}), (λ_{1}^{n}, λ_{2}^{n}, \dots, λ_{n}^{n})}$ $Mathematical equation: $$ {S}_n=\{(\mathrm{1,1},\cdots,1),({\lambda }_1,{\lambda }_2,\cdots,{\lambda }_n),\cdots,({\lambda }_1^{n-1},{\lambda }_2^{n-1},\cdots,{\lambda }_n^{n-1}),({\lambda }_1^n,{\lambda }_2^n,\cdots,{\lambda }_n^n)\} $$$ is a linearly dependent set of $R^{n}$ $Mathematical equation: $ {\mathbb{R}}^n$$ because the dimension of $R^{n}$ $Mathematical equation: $ {\mathbb{R}}^n$$ is n. Therefore, we conclude that rank (x) = n.

Let x be an element of $A$ $Mathematical equation: $ \mathcal{A}$$ with k distinct non null eigenvalues λ_is, and let $v_{i_{1}}, v_{i_{2}}, \dots, v_{i_{l - 1}}$ $Mathematical equation: $ {v}_{{i}_1},{v}_{{i}_2},\cdots,{v}_{{i}_{l-1}}$$ and $v_{i_{l}}$ $Mathematical equation: $ {v}_{{i}_l}$$ be an orthonormal basis of eigenvectors of x of the eigenvector space of x associated λ _i, this is $x v_{i_{j}} = λ_{i} v_{i_{j}}, j = 1, \dots, l_{i} .$ $Mathematical equation: $ x{v}_{{i}_j}={\lambda }_i{v}_{{i}_j},j=1,\cdots,{l}_i.$$ Now, we consider the elements $u_{i} = \sum_{j = 1}^{l_{i}} v_{i_{j}}$ $Mathematical equation: $ {u}_i={\sum }_{j=1}^{{l}_i} {v}_{{i}_j}$$ and we have $x = λ_{1} u_{1} u_{1}^{T} + λ_{2} u_{2} u_{2}^{T} + \dots + λ_{k} u_{k} u_{k}^{T} .$ $Mathematical equation: $ x={\lambda }_1{u}_1{u}_1^T+{\lambda }_2{u}_2{u}_2^T+\cdots +{\lambda }_k{u}_k{u}_k^T.$$ Therefore, we have

$\begin{matrix} e = u_{1} u_{1}^{T} + u_{2} u_{2}^{T} + \dots + u_{k} u_{k}^{T} \\ x^{1 ⋄} = λ_{1} u_{1} u_{1}^{T} + λ_{2} u_{2} u_{2}^{T} + \dots + λ_{k} u_{k} u_{k}^{T} \\ \begin{matrix} ⋮ = ⋮ \\ x^{k - 1 ⋄} = λ_{1}^{k - 1} u_{1} u_{1}^{T} + λ_{2}^{k - 1} u_{2} u_{2}^{T} + \dots + λ_{k}^{k - 1} u_{k} u_{k}^{T} \end{matrix} \end{matrix}$ $Mathematical equation: $$ \begin{array}{c}\mathbf{e}={u}_1{u}_1^T+{u}_2{u}_2^T+\cdots +{u}_k{u}_k^T\\ {x}^{1\diamond }={\lambda }_1{u}_1{u}_1^T+{\lambda }_2{u}_2{u}_2^T+\cdots +{\lambda }_k{u}_k{u}_k^T\\ \begin{array}{c}\vdots =\vdots \\ {x}^{k-1\diamond }={\lambda }_1^{k-1}{u}_1{u}_1^T+{\lambda }_2^{k-1}{u}_2{u}_2^T+\cdots +{\lambda }_k^{k-1}{u}_k{u}_k^T\end{array}\end{array} $$$ and, since

$| \begin{array}{l} 1 & 1 & \dots & 1 \\ λ_{1} & λ_{2} & \dots & λ_{k} \\ λ_{1}^{2} & λ_{2}^{2} & \dots & λ_{k}^{2} \\ ⋮ & ⋮ & \dots & ⋮ \\ λ_{1}^{k - 1} & λ_{2}^{k - 1} & \dots & λ_{k}^{k - 1} \end{array} | = \prod_{1 \leq i < j \leq k} (λ_{j} - λ_{i}) \neq 0$ $Mathematical equation: $$ \left|\begin{array}{llll}1& 1& \cdots & 1\\ {\lambda }_1& {\lambda }_2& \cdots & {\lambda }_k\\ {\lambda }_1^2& {\lambda }_2^2& \cdots & {\lambda }_k^2\\ \vdots & \vdots & \cdots & \vdots \\ {\lambda }_1^{k-1}& {\lambda }_2^{k-1}& \cdots & {\lambda }_k^{k-1}\end{array}\right|=\prod_{1\le i < j\le k} ({\lambda }_j-{\lambda }_i)\ne 0 $$$ then the set $X_{k - 1} = {e, x^{1 ⋄}, x^{2 ⋄}, \dots, x^{k - 1 ⋄}}$ $Mathematical equation: $ {\mathcal{X}}_{k-1}=\{\mathbf{e},{x}^{1\diamond },{x}^{2\diamond },\cdots,{x}^{k-1\diamond }\}$$ is a linearly independent set of $A .$ $Mathematical equation: $ \mathcal{A}.$$ And, the set $X_{k} = {e, x^{1 ⋄}, x^{2 ⋄}, \dots, x^{k - 1 ⋄}, x^{k ⋄}}$ $Mathematical equation: $ {X}_k=\{\mathbf{e},{x}^{1\diamond },{x}^{2\diamond },\cdots,{x}^{k-1\diamond },{x}^{k\diamond }\}$$ is a linearly dependent set of $A$ $Mathematical equation: $ \mathcal{A}$$ since $\dim (R^{k}) = k$ $Mathematical equation: $ \mathrm{dim}({\mathbb{R}}^k)=k$$ and therefore $rank (x) = k .$ $Mathematical equation: $ \mathrm{rank}\enspace (x)=k.$$ If x has k distinct eigenvalues where k−1 eigenvalues are non null and one is null then one proves one a similar way that $rank (x) = k$ $Mathematical equation: $ \mathrm{rank}\enspace (x)=k$$ .

Therefore we conclude that $rank (A) = n$ $Mathematical equation: $ \mathrm{rank}\enspace (\mathcal{A})=n$$ and the regular elements of $A$ $Mathematical equation: $ \mathcal{A}$$ are the elements x of $A$ $Mathematical equation: $ \mathcal{A}$$ with n distinct eigenvalues.

The characteristic polynomial of a regular element of $A$ $Mathematical equation: $ \mathcal{A}$$ is a monic polynomial of minimal degree $n = rank (A) .$ $Mathematical equation: $ n=\enspace \mathrm{rank}\enspace (A).$$ Now, let x be an element of $A$ $Mathematical equation: $ \mathcal{A}$$ with n distinct eigenvalues λ ₁, λ ₂,⋯, λ _n−1 and λ _n then by the Theorem of Cayley-Hamilton we conclude that the polynomial p such that

$p (λ) = (λ - λ_{1}) (λ - λ_{2}) \dots (λ - λ_{n})$ $Mathematical equation: $$ p(\lambda )=(\lambda -{\lambda }_1)(\lambda -{\lambda }_2)\cdots (\lambda -{\lambda }_n) $$$ is the monic polynomial of minimal degree of x. Therefore since the monic polynomial of minimal degree of element x is unique we conclude that the characteristic polynomial of x, p(x, −) is such that $p (x, λ) = p (λ) .$ $Mathematical equation: $ p(x,\lambda )=p(\lambda ).$$ This is $p (x, λ) = (λ - λ_{1}) (λ - λ_{2}) \dots (λ - λ_{n}) .$ $Mathematical equation: $ p(x,\lambda )=(\lambda -{\lambda }_1)(\lambda -{\lambda }_2)\cdots (\lambda -{\lambda }_n).$$ So, we have $p (x, λ) = λ^{n} - (λ_{1} + λ_{2} + \dots + λ_{n}) λ^{n - 1} + \dots + (- 1)^{n} λ_{1} λ_{2} \dots λ_{n} .$ $Mathematical equation: $ p(x,\lambda )={\lambda }^n-({\lambda }_1+{\lambda }_2+\cdots +{\lambda }_n){\lambda }^{n-1}+\cdots +(-1{)}^n{\lambda }_1{\lambda }_2\cdots {\lambda }_n.$$ Therefore, we have $Trace (x) = λ_{1} + λ_{2} + \dots + λ_{n}$ $Mathematical equation: $ \mathrm{Trace}\enspace (x)={\lambda }_1+{\lambda }_2+\cdots +{\lambda }_n$$ and $Det (x) = λ_{1} λ_{2} \dots λ_{n}$ $Mathematical equation: $ \mathrm{Det}\enspace (x)={\lambda }_1{\lambda }_2\cdots {\lambda }_n$$ .

Now, we will show that $rank (A_{n + 1}) = 2$ $Mathematical equation: $ \mathrm{rank}\enspace ({\mathcal{A}}_{n+1})=2$$ To came to this conclusion, we will firstly show that for $\overset{̅}{x} \neq \overset{̅}{0}, rank ([\begin{array}{l} x_{1} \\ \overset{̅}{x} \end{array}]) = 2$ $Mathematical equation: $ \bar{x}\ne \bar{0},\hspace{0.5em}\mathrm{rank}\enspace \left(\left[\begin{array}{l}{x}_1\\ \bar{x}\end{array}\right]\right)=2$$ and that for $x_{1} \neq 0, rank ([\begin{array}{l} x_{1} \\ \overset{̅}{0} \end{array}]) = 1 .$ $Mathematical equation: $ {x}_1\ne 0,\hspace{0.5em}\mathrm{rank}\enspace \left(\left[\begin{array}{l}{x}_1\\ \bar{0}\end{array}\right]\right)=1.$$ So, let suppose $\overset{̅}{x} \neq \overset{̅}{0}$ $Mathematical equation: $ \bar{x}\ne \bar{0}$$ then, we have

$α [\begin{array}{l} x_{1} \\ \overset{̅}{x} \end{array}] + β [\begin{array}{l} 1 \\ \overset{̅}{0} \end{array}] = [\begin{array}{l} 0 \\ \overset{̅}{0} \end{array}] \Leftrightarrow (α = 0) \land (β = 0) .$ $Mathematical equation: $$ \alpha \left[\begin{array}{l}{x}_1\\ \bar{x}\end{array}\right]+\beta \left[\begin{array}{l}1\\ \bar{0}\end{array}\right]=\left[\begin{array}{l}0\\ \bar{0}\end{array}\right]\iff (\alpha =0)\wedge (\beta =0). $$$ Therefore, the set ${[\begin{array}{l} 1 \\ \overset{̅}{0} \end{array}], [\begin{array}{l} x_{1} \\ \overset{̅}{x} \end{array}]}$ $Mathematical equation: $ \left\{\left[\begin{array}{l}1\\ \bar{0}\end{array}\right],\left[\begin{array}{l}{x}_1\\ \bar{x}\end{array}\right]\right\}$$ is a linearly independent set of $A_{n + 1} .$ $Mathematical equation: $ {\mathcal{A}}_{n+1}.$$ Now, we have ${[\begin{array}{l} x_{1} \\ \overset{̅}{x} \end{array}]}^{2 ⋄} = [\begin{array}{l} x_{1}^{2} + | | \overset{̅}{x} | |^{2} \\ 2 x_{1} \overset{̅}{x} \end{array}]$ $Mathematical equation: $ {\left[\begin{array}{l}{x}_1\\ \bar{x}\end{array}\right]}^{2\diamond }=\left[\begin{array}{l}{x}_1^2+||\bar{x}|{|}^2\\ 2{x}_1\bar{x}\end{array}\right]$$ and since

$[\begin{array}{l} x_{1}^{2} + | | \overset{̅}{x} | |^{2} \\ 2 x_{1} \overset{̅}{x} \end{array}] = α [\begin{array}{l} x_{1} \\ \overset{̅}{x} \end{array}] + β [\begin{array}{l} 1 \\ \overset{̅}{o} \end{array}] \Leftrightarrow (α = 2 x_{1}) \land (β = | | \overset{̅}{x} | | - x_{1}^{2})$ $Mathematical equation: $$ \left[\begin{array}{l}{x}_1^2+||\bar{x}|{|}^2\\ 2{x}_1\bar{x}\end{array}\right]=\alpha \left[\begin{array}{l}{x}_1\\ \bar{x}\end{array}\right]+\beta \left[\begin{array}{l}1\\ \bar{o}\end{array}\right]\iff \left(\alpha =2{x}_1\right)\wedge \left(\beta =\left|\left|\bar{x}\right|\right|-{x}_1^2\right) $$$ we conclude that $rank ([\begin{array}{l} x_{1} \\ \overset{̅}{x} \end{array}]) = 2 .$ $Mathematical equation: $ \mathrm{rank}\enspace \left(\left[\begin{array}{l}{x}_1\\ \bar{x}\end{array}\right]\right)=2.$$ If $x_{1} \neq 0,$ $Mathematical equation: $ {x}_1\ne 0,$$ we have $[\begin{array}{l} x_{1} \\ \overset{̅}{0} \end{array}] = x_{1} [\begin{array}{l} 1 \\ \overset{̅}{0} \end{array}] .$ $Mathematical equation: $ \left[\begin{array}{l}{x}_1\\ \bar{0}\end{array}\right]={x}_1\left[\begin{array}{l}1\\ \bar{0}\end{array}\right].$$ Then the set ${[\begin{array}{l} 1 \\ \overset{̅}{0} \end{array}], [\begin{array}{l} x_{1} \\ \overset{̅}{x} \end{array}]}$ $Mathematical equation: $ \left\{\left[\begin{array}{l}1\\ \bar{0}\end{array}\right],\left[\begin{array}{l}{x}_1\\ \bar{x}\end{array}\right]\right\}$$ is a linearly dependent set of $A_{n + 1}$ $Mathematical equation: $ {\mathcal{A}}_{n+1}$$ then $rank ([\begin{array}{l} x_{1} \\ \overset{̅}{0} \end{array}]) = 1 .$ $Mathematical equation: $ \mathrm{rank}\enspace \left(\left[\begin{array}{l}{x}_1\\ \bar{0}\end{array}\right]\right)=1.$$ And, therefore $rank (A_{n + 1}) = 2 .$ $Mathematical equation: $ \mathrm{rank}\enspace ({\mathcal{A}}_{n+1})=2.$$ And the regular elements of $A_{n + 1}$ $Mathematical equation: $ {\mathcal{A}}_{n+1}$$ are the elements of $A_{n + 1}$ $Mathematical equation: $ {\mathcal{A}}_{n+1}$$ such that $\overset{̅}{x} \neq \overset{̅}{0} .$ $Mathematical equation: $ \bar{x}\ne \bar{0}.$$

Since, when $\overset{̅}{x} \neq \overset{̅}{0}$ $Mathematical equation: $ \bar{x}\ne \bar{0}$$ we have

${[\begin{array}{l} x_{1} \\ \overset{̅}{x} \end{array}]}^{2 ⋄} - 2 x_{1} [\begin{array}{l} x_{1} \\ \overset{̅}{x} \end{array}] + (x_{1}^{2} - | | \overset{̅}{x} | |^{2}) [\begin{array}{l} 1 \\ \overset{̅}{0} \end{array}] = [\begin{array}{l} 0 \\ \overset{̅}{0} \end{array}] .$ $Mathematical equation: $$ {\left[\begin{array}{l}{x}_1\\ \bar{x}\end{array}\right]}^{2\diamond }-2{x}_1\left[\begin{array}{l}{x}_1\\ \bar{x}\end{array}\right]+({x}_1^2-||\bar{x}|{|}^2)\left[\begin{array}{l}1\\ \bar{0}\end{array}\right]=\left[\begin{array}{l}0\\ \bar{0}\end{array}\right]. $$$ Then, supposing $\overset{̅}{x} \neq \overset{̅}{0}$ $Mathematical equation: $ \bar{x}\ne \bar{0}$$ and considering the notation $x = [\begin{array}{l} x_{1} \\ \overset{̅}{x} \end{array}],$ $Mathematical equation: $ x=\left[\begin{array}{l}{x}_1\\ \bar{x}\end{array}\right],$$ we conclude that the caractheristic polynomial of x is $p (x, λ) = λ^{2} - 2 x_{1} λ + (x_{1}^{2} - | | \overset{̅}{x} | |^{2}) .$ $Mathematical equation: $ p(x,\lambda )={\lambda }^2-2{x}_1\lambda +({x}_1^2-||\bar{x}|{|}^2).$$ Therefore, the eigenvalues of x are $λ_{1} (x) = x_{1} - | | \overset{̅}{x} | |$ $Mathematical equation: $ {\lambda }_1(x)={x}_1-||\bar{x}||$$ and $λ_{2} (x) = x_{1} + | | \overset{̅}{x} | |, Trace (x) = 2 x_{1}$ $Mathematical equation: $ {\lambda }_2(x)={x}_1+||\bar{x}||,\enspace \mathrm{Trace}\enspace (x)=2{x}_1$$ and $Det (x) = x_{1}^{2} - | | x | |^{2} .$ $Mathematical equation: $ \mathrm{Det}\enspace (x)={x}_1^2-||x|{|}^2.$$

Let $A$ $Mathematical equation: $ \mathcal{A}$$ be a real Euclidean Jordan algebra with unit element e. An element f in $A$ $Mathematical equation: $ \mathcal{A}$$ is an idempotent of $A$ $Mathematical equation: $ \mathcal{A}$$ if $f^{2 ⋄} = f$ $Mathematical equation: $ {f}^{2\diamond }=f$$ . Two idempotents f and g of $A$ $Mathematical equation: $ \mathcal{A}$$ are orthogonal if $f ⋄ g = 0 .$ $Mathematical equation: $ f\diamond g=0.$$ A set ${g_{1}, g_{2}, \dots, g_{k}}$ $Mathematical equation: $ \{{g}_1,{g}_2,\dots,{g}_k\}$$ of nonzero idempotents is a complete system of orthogonal idempotents of $A$ $Mathematical equation: $ \mathcal{A}$$ if $g_{i}^{2 ⋄} = g_{i}$ $Mathematical equation: $ {g}_i^{2\diamond }={g}_i$$ , for i = 1, …, k, $g_{i} ⋄ g_{j} = 0$ $Mathematical equation: $ {g}_i\diamond {g}_j=0$$ , for $i = j$ Mathematical equation: $ i=j$ , and $\sum_{i = 1}^{k} g_{i} = e$ $Mathematical equation: $ {\sum }_{i=1}^k {g}_i=\mathbf{e}$$ . An element g of $A$ $Mathematical equation: $ \mathcal{A}$$ is a primitive idempotent if it is a non null idempotent of $A$ $Mathematical equation: $ \mathcal{A}$$ and if cannot be written as a sum of two orthogonal nonzero idempotents of $A$ $Mathematical equation: $ \mathcal{A}$$ . We say that ${g_{1}, g_{2}, \dots, g_{l}}$ $Mathematical equation: $ \{{g}_1,{g}_2,\dots,{g}_l\}$$ is a Jordan frame of $A$ $Mathematical equation: $ \mathcal{A}$$ if ${g_{1}, g_{2}, \dots, g_{l}}$ $Mathematical equation: $ \{{g}_1,{g}_2,\dots,{g}_l\}$$ is a complete system of orthogonal idempotents such that each idempotent is primitive.

Let consider the matrices E _jj of the Euclidean Jordan algebra $A = Sym (n, R)$ $Mathematical equation: $ \mathcal{A}=\enspace \mathrm{Sym}\enspace (n,\mathbb{R})$$ with j = 1,⋯, n where E _jj is the square matrix of order n such that (E _jj)_jj = 1 and (E _jj)_ik = 0 if i ≠ j or k ≠ j.

Let k be a natural number such that 1 < k < n. Then $S = {E_{11} + E_{22} + \dots + E_{kk}, E_{k + 1 k + 1}, \dots, E_{nn}}$ $Mathematical equation: $ S=\{{E}_{11}+{E}_{22}+\cdots +{E}_{{kk}},{E}_{k+1\enspace k+1},\cdots,{E}_{{nn}}\}$$ is a complete system of orthogonal idempotents of $A$ $Mathematical equation: $ \mathcal{A}$$ and $S' = {E_{11}, E_{22}, \dots, E_{nn}}$ $Mathematical equation: $ S\mathrm{\prime}=\{{E}_{11},{E}_{22},\cdots,{E}_{{nn}}\}$$ is a Jordan frame of $A$ $Mathematical equation: $ \mathcal{A}$$ .

Let consider the Euclidean Jordan algebra $A_{n + 1}$ $Mathematical equation: $ {\mathcal{A}}_{n+1}$$ and x non zero element of $A_{n + 1}$ $Mathematical equation: $ {\mathcal{A}}_{n+1}$$ . Then the set $S = {g_{1}, g_{2}} = {\frac{1}{2} [\begin{matrix} 1 \\ - \frac{\overset{̅}{x}}{| | \overset{̅}{x} | |} \end{matrix}], \frac{1}{2} [\begin{matrix} 1 \\ \frac{\overset{̅}{x}}{| | \overset{̅}{x} | |} \end{matrix}]}$ $Mathematical equation: $ S=\{{g}_1,{g}_2\}=\left\{\frac{1}{2}\left[\begin{array}{c}1\\ -\frac{\bar{x}}{||\bar{x}||}\end{array}\right],\frac{1}{2}\left[\begin{array}{c}1\\ \frac{\bar{x}}{||\bar{x}||}\end{array}\right]\right\}$$ is a Jordan frame of the Euclidean Jordan algebra $A_{n + 1} .$ $Mathematical equation: $ {\mathcal{A}}_{n+1}.$$ Indeed, we have: $g_{1}^{2 ⋄} = g_{1}$ $Mathematical equation: $ {g}_1^{2\diamond }={g}_1$$ and $g_{2}^{2 ⋄} = g_{2}$ $Mathematical equation: $ {g}_2^{2\diamond }={g}_2$$

$\begin{array}{l} g_{1}^{2 ⋄} = g_{1} ⋄ g_{1} = \frac{1}{2} [\begin{matrix} 1 \\ - \frac{\overset{̅}{x}}{| | \overset{̅}{x} | |} \end{matrix}] ⋄ \frac{1}{2} [\begin{matrix} \begin{matrix} 1 \\ - \frac{\overset{̅}{x}}{| | \overset{̅}{x} | |} \end{matrix} \end{matrix}] \\ = [\begin{matrix} \frac{1}{4} + \frac{1}{4} \frac{\overset{̅}{x} | \overset{̅}{x}}{| | \overset{̅}{x} | |^{2}} \\ - \frac{1}{4} \frac{\overset{̅}{x}}{| | \overset{̅}{x} | |} - \frac{1}{4} \frac{\overset{̅}{x}}{| | \overset{̅}{x} | |} \end{matrix}] = \frac{1}{2} [\begin{matrix} 1 \\ - \frac{\overset{̅}{x}}{| | \overset{̅}{x} | |} \end{matrix}] = g_{1} \end{array}$ $Mathematical equation: $$ \begin{array}{l}{g}_1^{2\diamond }={g}_1\diamond {g}_1=\frac{1}{2}\left[\begin{array}{c}1\\ -\frac{\bar{x}}{||\bar{x}||}\end{array}\right]\diamond \frac{1}{2}\left[\begin{array}{c}\\ \begin{array}{c}1\\ -\frac{\bar{x}}{||\bar{x}||}\end{array}\end{array}\right]\\ =\left[\begin{array}{c}\frac{1}{4}+\frac{1}{4}\frac{\bar{x}|\bar{x}}{||\bar{x}|{|}^2}\\ -\frac{1}{4}\frac{\bar{x}}{||\bar{x}||}-\frac{1}{4}\frac{\overline{\mathrm{x}}}{||\bar{x}||}\end{array}\right]=\frac{1}{2}\left[\begin{array}{c}1\\ -\frac{\bar{x}}{||\bar{x}||}\end{array}\right]={g}_1\end{array} $$$ and, we have

$\begin{array}{l} g_{2}^{2 ⋄} = g_{2} ⋄ g_{2} = \frac{1}{2} [\begin{matrix} 1 \\ \frac{\overset{̅}{x}}{| | \overset{̅}{x} | |} \end{matrix}] ⋄ \frac{1}{2} [\begin{matrix} 1 \\ \frac{\overset{̅}{x}}{| | \overset{̅}{x} | |} \end{matrix}] \\ = [\begin{matrix} \frac{1}{4} + \frac{1}{4} \frac{\overset{̅}{x} | \overset{̅}{x}}{| | \overset{̅}{x} | |^{2}} \\ \frac{1}{4} \frac{\overset{̅}{x}}{| | \overset{̅}{x} | |} + \frac{1}{4} \frac{\overset{̅}{x}}{| | \overset{̅}{x} | |} \end{matrix}] = \frac{1}{2} [\begin{matrix} 1 \\ \frac{\overset{̅}{x}}{| | \overset{̅}{x} | |} \end{matrix}] = g_{2} \end{array}$ $Mathematical equation: $$ \begin{array}{l}{g}_2^{2\diamond }={g}_2\diamond {g}_2=\frac{1}{2}\left[\begin{array}{c}1\\ \frac{\bar{x}}{||\bar{x}||}\end{array}\right]\diamond \frac{1}{2}\left[\begin{array}{c}1\\ \frac{\bar{x}}{||\bar{x}||}\end{array}\right]\\ =\left[\begin{array}{c}\frac{1}{4}+\frac{1}{4}\frac{\bar{x}|\bar{x}}{||\bar{x}|{|}^2}\\ \frac{1}{4}\frac{\bar{x}}{||\bar{x}||}+\frac{1}{4}\frac{\bar{x}}{||\bar{x}||}\end{array}\right]=\frac{1}{2}\left[\begin{array}{c}1\\ \frac{\bar{x}}{||\bar{x}||}\end{array}\right]={g}_2\end{array} $$$

$\begin{array}{l} g_{1} + g_{2} = \frac{1}{2} [\begin{matrix} 1 \\ - \frac{\overset{̅}{x}}{| | \overset{̅}{x} | |} \end{matrix}] + \frac{1}{2} [\begin{matrix} 1 \\ \frac{\overset{̅}{x}}{| | \overset{̅}{x} | |} \end{matrix}] = [\begin{matrix} 1 \\ \overset{̅}{0} \end{matrix}] = e \end{array}$ $Mathematical equation: $$ \begin{array}{l}{g}_1+{g}_2=\frac{1}{2}\left[\begin{array}{c}1\\ -\frac{\bar{x}}{||\bar{x}||}\end{array}\right]+\frac{1}{2}\left[\begin{array}{c}1\\ \frac{\bar{x}}{||\bar{x}||}\end{array}\right]=\left[\begin{array}{c}1\\ \bar{0}\end{array}\right]=\mathbf{e}\end{array} $$$

$\begin{array}{l} g_{1} ⋄ g_{2} = [\begin{matrix} 1 \\ - \frac{\overset{̅}{x}}{2 | | \overset{̅}{x} | |} \end{matrix}] ⋄ [\begin{matrix} 1 \\ \frac{\overset{̅}{x}}{2 | | \overset{̅}{x} | |} \end{matrix}] = [\begin{matrix} 1 - \frac{| | \overset{̅}{x} | |^{2}}{| | \overset{̅}{x} | |^{2}} \\ \frac{\overset{̅}{x}}{| | \overset{̅}{x} | |} - \frac{\overset{̅}{x}}{| | \overset{̅}{x} | |} \end{matrix}] [\begin{matrix} 0 \\ \overset{̅}{0} \end{matrix}] . \end{array}$ $Mathematical equation: $$ \begin{array}{l}{g}_1\diamond {g}_2=\left[\begin{array}{c}1\\ -\frac{\bar{x}}{2\left|\left|\bar{x}\right|\right|}\end{array}\right]\diamond \left[\begin{array}{c}1\\ \frac{\bar{x}}{2\left|\left|\bar{x}\right|\right|}\end{array}\right]=\left[\begin{array}{c}1-\frac{||\bar{x}|{|}^2}{||\bar{x}|{|}^2}\\ \frac{\bar{x}}{\left|\left|\bar{x}\right|\right|}-\frac{\bar{x}}{\left|\left|\bar{x}\right|\right|}\\ \end{array}\right]\left[\begin{array}{c}0\\ \bar{0}\end{array}\right].\end{array} $$$

Therefore we conclude that ${g_{1}, g_{2}}$ $Mathematical equation: $ \{{g}_1,{g}_2\}$$ is a Jordan frame of $A_{n + 1}$ $Mathematical equation: $ {\mathcal{A}}_{n+1}$$ since $rank (A_{n + 1}) = 2 .$ $Mathematical equation: $ \mathrm{rank}\enspace ({\mathcal{A}}_{n+1})=2.$$

Theorem 1. ([28], p. 43). Let $A$ $Mathematical equation: $ \mathcal{A}$$ be a real Euclidean Jordan algebra. Then for x in $A$ $Mathematical equation: $ \mathcal{A}$$ there exist unique real numbers $β_{1} (x), β_{2} (x), \dots, β_{k} (x),$ $Mathematical equation: $ {\beta }_1(x),{\beta }_2(x),\dots,{\beta }_k(x),$$ all distinct, and a unique complete system of orthogonal idempotents ${g_{1}, g_{2}, \dots, g_{k}}$ $Mathematical equation: $ \{{g}_1,{g}_2,\dots,{g}_k\}$$ such that

$x = β_{1} (x) g_{1} + β_{2} (x) g_{2} + \dots + β_{k} (x) g_{k} .$ $Mathematical equation: $$ x={\beta }_1(x){g}_1+{\beta }_2(x){g}_2+\cdots +{\beta }_k(x){g}_k. $$$ (3)

The numbers $β_{j} (x)$ $Mathematical equation: $ {\beta }_j(x)$$ ’s of (3) are the eigenvalues of x and the decomposition (3) is the first spectral decomposition of x.

Theorem 2. ([28], p. 44). Let $A$ $Mathematical equation: $ \mathcal{A}$$ be a real Euclidean Jordan algebra with $rank (V) = r .$ $Mathematical equation: $ {rank}\enspace (\mathcal{V})=r.$$ Then for each x in $A$ $Mathematical equation: $ \mathcal{A}$$ there exists a Jordan frame ${g_{1}, g_{2}, \dots, g_{r}}$ $Mathematical equation: $ \{{g}_1,{g}_2,\cdots,{g}_r\}$$ and real numbers $β_{1} (x), \dots, β_{r - 1} (x)$ $Mathematical equation: $ {\beta }_1(x),\cdots,{\beta }_{r-1}(x)$$ and $β_{r} (x)$ $Mathematical equation: $ {\beta }_r(x)$$ such that

$x = β_{1} (x) g_{1} + β_{2} (x) g_{2} + \dots + β_{r} (x) g_{r} .$ $Mathematical equation: $$ x={\beta }_1(x){g}_1+{\beta }_2(x){g}_2+\cdots +{\beta }_r(x){g}_r. $$$ (4)

Remark 1. The decomposition (4) is called the second spectral decomposition of x. And we have $Det (x) = \prod_{j = 1}^{r} β_{j} (x), Trace (x) = \sum_{1}^{r} β_{j} (x)$ $Mathematical equation: $ \mathrm{Det}\enspace (x)={\prod }_{j=1}^r {\beta }_j(x),\enspace \mathrm{Trace}\enspace (x)=\sum_1^r{\beta }_j(x)$$ and $α_{k} (x) = \sum_{1 \leq i_{1} < \dots < i_{k} \leq r} β_{i_{1}} (x) \dots β_{i_{k}} (x) .$ $Mathematical equation: $ {\alpha }_k(x)=\sum_{1\le {i}_1<\cdots < {i}_k\le r}{\beta }_{{i}_1}(x)\dots {\beta }_{{i}_{\mathrm{k}}}(x).$$

Example 1. For $\overset{̅}{x} \neq 0$ $Mathematical equation: $ \bar{x}\ne 0$$ , the second spectral decomposition of $x = [\begin{array}{l} x_{1} \\ \overset{̅}{x} \end{array}]$ $Mathematical equation: $ x=\left[\begin{array}{l}{x}_1\\ \bar{x}\end{array}\right]$$ relatively to the Jordan frame $S = {g_{1}, g_{2}} = {\frac{1}{2} [\begin{matrix} 1 \\ - \frac{\overset{̅}{x}}{| | \overset{̅}{x} | |} \end{matrix}], \frac{1}{2} [\begin{matrix} 1 \\ \frac{\overset{̅}{x}}{| | \overset{̅}{x} | |} \end{matrix}]}$ $Mathematical equation: $ S=\{{g}_1,{g}_2\}=\left\{\frac{1}{2}\left[\begin{array}{c}1\\ -\frac{\bar{x}}{||\bar{x}||}\end{array}\right],\frac{1}{2}\left[\begin{array}{c}1\\ \frac{\bar{x}}{||\bar{x}||}\end{array}\right]\right\}$$ of $A_{n + 1}$ $Mathematical equation: $ {\mathcal{A}}_{n+1}$$ is

$x = (x_{1} - | | \overset{̅}{x} | |) g_{1} + (x_{1} + | | \overset{̅}{x} | |) g_{2} .$ $Mathematical equation: $$ x=({x}_1-||\bar{x}||){g}_1+({x}_1+||\bar{x}||){g}_2. $$$

An Euclidean Jordan algebra is called simple if and only if have only trivial ideals.

Any simple Euclidean Jordan algebra is isomorphic to one of the five Euclidean Jordan algebras that we describe below:

The spin Euclidean Jordan algebra $A_{n + 1}$ $Mathematical equation: $ {\mathcal{A}}_{n+1}$$ .
The Euclidean Jordan algebra $A = Sym (n, R)$ $Mathematical equation: $ \mathcal{A}=\enspace \mathrm{Sym}\enspace (n,\mathbb{R})$$ with the Jordan product of matrices and with an inner product of two symmetric matrices as being the trace of their Jordan product.
The Euclidean Jordan algebra $A = H_{n} (C)$ $Mathematical equation: $ \mathcal{A}={H}_n(\mathbb{C})$$ of hermitian matrices of complexes of order n equipped with the Jordan product of two hermitian matrices of complexes and with the scalar product of two Hermitian matrices of complexes as being the real part of the trace of their Jordan product.
The Euclidean Jordan algebra $A = H_{n} (H)$ $Mathematical equation: $ \mathcal{A}={H}_n(\mathbb{H})$$ , of hermitian matrices of quaternions of order n equipped with the Jordan product of hermitian matrices of quaternions and with the scalar product of two hermitian matrices of quaternions as being the real part of the trace of their the Jordan product.
The Euclidean Jordan algebra $A = H_{3} (O)$ $Mathematical equation: $ \mathcal{A}={H}_3(\mathbb{O})$$ of hermitian matrices of octonions of order n equipped with the Jordan product of two hermitian matrices of octonions and with the inner product of two hermitian matrices of octonions as being the real part of the trace of their Jordan product.

We now describe the Pierce decomposition of an Euclidean Jordan algebra relatively to one of its idempotents. But, first we must say that for any nonzero idempotent g of an Euclidean Jordan algebra $A$ $Mathematical equation: $ \mathcal{A}$$ the eigenvalues of the linear operator $L_{⋄} (g)$ $Mathematical equation: $ {L}_{\diamond }(g)$$ are $0, \frac{1}{2}$ $Mathematical equation: $ 0,\frac{1}{2}$$ and 1 and this fact permits us to say that, considering the eigenspaces $A (g, 0) = {x \in A : L_{⋄} (g) (x) = 0 x}, A (g, \frac{1}{2}) = {x \in A : L_{⋄} (g) (x) = \frac{1}{2} x}$ $Mathematical equation: $ \mathcal{A}(g,0)=\{x\in \mathcal{A}:{L}_{\diamond }(g)(x)=0x\},\mathcal{A}\left(g,\frac{1}{2}\right)=\left\{x\in \mathcal{A}:{L}_{\diamond }(g)(x)=\frac{1}{2}x\right\}$$ and $A (g, 1) = {x \in A : L_{⋄} (g) (x) = 1 x}$ $Mathematical equation: $ \mathcal{A}(g,1)=\{x\in \mathcal{A}:{L}_{\diamond }(g)(x)=1x\}$$ of L(g) associated to these eigenvalues we can decompose $A$ $Mathematical equation: $ \mathcal{A}$$ as an orthogonal direct sum $A = V (g, 0) + V (g, \frac{1}{2}) + V (g, 1) .$ $Mathematical equation: $ \mathcal{A}=V(g,0)+V\left(g,\frac{1}{2}\right)+V(g,1).$$ Now, we will describe the Pierce decomposition of the Euclidean Jordan algebra $A = Sym (n, R)$ $Mathematical equation: $ \mathcal{A}=\enspace \mathrm{Sym}\enspace (n,\mathbb{R})$$ relatively to an idempotent of the form $C = [\begin{array}{l} I_{k} & O_{k \times n - k} \\ O_{k \times n - k}^{T} & O_{n - k \times n - k} \end{array}]$ $Mathematical equation: $ C=\left[\begin{array}{ll}{I}_k& {O}_{k\times n-k}\\ {O}_{k\times n-k}^T& {O}_{n-k\times n-k}\end{array}\right]$$ .

Let $A = Sym (n, R)$ $Mathematical equation: $ \mathcal{A}=\enspace \mathrm{Sym}\enspace (n,\mathbb{R})$$ and $C = [\begin{array}{l} I_{k} & O_{k \times n - k} \\ O_{k \times n - k}^{T} & O_{n - k \times n - k} \end{array}] .$ $Mathematical equation: $ C=\left[\begin{array}{ll}{I}_k& {O}_{k\times n-k}\\ {\mathrm{O}}_{k\times n-k}^T& {O}_{n-k\times n-k}\end{array}\right].$$ Now, we will show that $A (C, 1) = {[\begin{array}{l} X_{k \times k} & O_{k \times n - k} \\ O_{k \times n - k}^{T} & O_{n - k \times n - k} \end{array}] : X_{k \times k} \in Sym (k, R)} .$ $Mathematical equation: $ \mathcal{A}(C,1)=\left\{\left[\begin{array}{ll}{X}_{k\times k}& {O}_{k\times n-k}\\ {O}_{k\times n-k}^T& {O}_{n-k\times n-k}\end{array}\right]:{X}_{k\times k}\in \enspace {Sym}\enspace (k,\mathbb{R})\right\}.$$ We have

$\begin{array}{l} [\begin{array}{l} I_{k} & O_{k \times n - k} \\ O_{n - k \times k}^{T} & O_{k \times k} \end{array}] ⋄ [\begin{array}{l} A_{k \times k} & A_{k \times n - k} \\ A_{k \times n - k}^{T} & A_{n - k \times n - k} \end{array}] = [\begin{array}{l} A_{k \times k} & A_{k \times n - k} \\ A_{k \times n - k}^{T} & A_{n - k \times n - k} \end{array}] \\ ⇕ \\ \frac{1}{2} ([\begin{array}{l} I_{k} & O_{k \times n - k} \\ O_{k \times n - k}^{T} & O_{n - k \times n - k} \end{array}] [\begin{array}{l} A_{k \times k} & A_{k \times n - k} \\ A_{k \times n - k}^{T} & A_{n - k \times n - k} \end{array}] \\ + [\begin{array}{l} A_{k \times k} & A_{k \times n - k} \\ A_{k \times n - k}^{T} & A_{n - k \times n - k} \end{array}] [\begin{array}{l} I_{k} & O_{k \times n - k} \\ O_{k \times n - k}^{T} & O_{n - k \times n - k} \end{array}]) = [\begin{array}{l} A_{k \times k} & A_{k \times n - k} \\ A_{k \times n - k}^{T} & A_{n - k \times n - k} \end{array}] \\ ⇕ \\ \frac{1}{2} [\begin{array}{l} A_{k \times k} & A_{k \times n - k} \\ O_{k \times n - k}^{T} & O_{n - k \times n - k} \end{array}] + \frac{1}{2} [\begin{array}{l} A_{k \times k} & O_{k \times n - k} \\ A_{k \times n - k}^{T} & O_{n - k \times n - k} \end{array}] = [\begin{array}{l} A_{k \times k} & A_{k \times n - k} \\ A_{k \times n - k}^{T} & A_{n - k \times n - k} \end{array}] \\ ⇕ \\ [\begin{array}{l} A_{k \times k} & \frac{1}{2} A_{k \times n - k} \\ \frac{1}{2} A_{k \times n - k}^{T} & O_{n - k \times n - k} \end{array}] = [\begin{array}{l} A_{k \times k} & A_{k \times n - k} \\ A_{k \times n - k}^{T} & A_{n - k \times n - k} \end{array}] \\ ⇕ \\ (A_{k \times n - k} = O_{k \times n - k}) \land (A_{n - k \times n - k} = O_{n - k \times n - k}) \end{array}$ $Mathematical equation: $$ \begin{array}{l}\left[\begin{array}{ll}{I}_k& {O}_{k\times n-k}\\ {O}_{n-k\times k}^T& {O}_{k\times k}\end{array}\right]\diamond \left[\begin{array}{ll}{A}_{k\times k}& {A}_{k\times n-k}\\ {A}_{k\times n-k}^T& {A}_{n-k\times n-k}\\ & \end{array}\right]=\left[\begin{array}{ll}{A}_{k\times k}& {A}_{k\times n-k}\\ {A}_{k\times n-k}^T& {A}_{n-k\times n-k}\\ & \end{array}\right]\\ \Updownarrow \\ \frac{1}{2}\left(\left[\begin{array}{ll}{I}_k& {O}_{k\times n-k}\\ {O}_{k\times n-k}^T& {O}_{n-k\times n-k}\end{array}\right]\left[\begin{array}{ll}{A}_{k\times k}& {A}_{k\times n-k}\\ {A}_{k\times n-k}^T& {A}_{n-k\times n-k}\end{array}\right]\right.\\ +\left.\left[\begin{array}{ll}{A}_{k\times k}& {A}_{k\times n-k}\\ {A}_{k\times n-k}^T& {A}_{n-k\times n-k}\end{array}\right]\left[\begin{array}{ll}{I}_k& {O}_{k\times n-k}\\ {O}_{k\times n-k}^T& {O}_{n-k\times n-k}\end{array}\right]\right)=\left[\begin{array}{ll}{A}_{k\times k}& {A}_{k\times n-k}\\ {A}_{k\times n-k}^T& {A}_{n-k\times n-k}\end{array}\right]\\ \Updownarrow \\ \frac{1}{2}\left[\begin{array}{ll}{A}_{k\times k}& {A}_{k\times n-k}\\ {O}_{k\times n-k}^T& {O}_{n-k\times n-k}\end{array}\right]+\frac{1}{2}\left[\begin{array}{ll}{A}_{k\times k}& {O}_{k\times n-k}\\ {A}_{k\times n-k}^T& {O}_{n-k\times n-k}\end{array}\right]=\left[\begin{array}{ll}{A}_{k\times k}& {A}_{k\times n-k}\\ {A}_{k\times n-k}^T& {A}_{n-k\times n-k}\end{array}\right]\\ \Updownarrow \\ \left[\begin{array}{ll}{A}_{k\times k}& \frac{1}{2}{A}_{k\times n-k}\\ \frac{1}{2}{A}_{k\times n-k}^T& {O}_{n-k\times \mathrm{n}-k}\end{array}\right]=\left[\begin{array}{ll}{A}_{k\times k}& {A}_{k\times n-k}\\ {A}_{k\times n-k}^T& {A}_{n-k\times n-k}\end{array}\right]\\ \Updownarrow \\ ({A}_{k\times n-k}={O}_{k\times n-k})\wedge ({A}_{n-k\times n-k}={O}_{n-k\times n-k})\end{array} $$$

Therefore, we have

$A (C, 1) = {[\begin{array}{l} A_{k \times k} & O_{k \times n - k} \\ O_{k \times n - k}^{T} & O_{n - k \times n - k} \end{array}] : A_{k \times k} \in M_{k} (R) \land (A_{k \times k} = A_{k \times k}^{T})}$ $Mathematical equation: $$ \mathcal{A}(C,1)=\left\{\left[\begin{array}{ll}{A}_{k\times k}& {O}_{k\times n-k}\\ {O}_{k\times n-k}^T& {O}_{n-k\times n-k}\end{array}\right]:{A}_{k\times k}\in {M}_k(\mathbb{R})\wedge ({A}_{k\times k}={A}_{k\times k}^T)\right\} $$$

Now, let calculate $A (C, \frac{1}{2}) .$ $Mathematical equation: $ \mathcal{A}\left(C,\frac{1}{2}\right).$$ So, let consider the following equivalences.

$\begin{array}{l} [\begin{array}{l} I_{k} & O_{k \times n - k} \\ O_{n - k \times k}^{T} & O_{k \times k} \end{array}] ⋄ [\begin{array}{l} A_{k \times k} & A_{k \times n - k} \\ A_{k \times n - k}^{T} & A_{n - k \times n - k} \end{array}] = \frac{1}{2} [\begin{array}{l} A_{k \times k} & A_{k \times n - k} \\ A_{k \times n - k}^{T} & A_{n - k \times n - k} \end{array}] \\ ⇕ \\ \frac{1}{2} ([\begin{array}{l} I_{k} & O_{k \times n - k} \\ O_{k \times n - k}^{T} & O_{n - k \times n - k} \end{array}] [\begin{array}{l} A_{k \times k} & A_{k \times n - k} \\ A_{k \times n - k}^{T} & A_{n - k \times n - k} \end{array}] \\ + [\begin{array}{l} A_{k \times k} & A_{k \times n - k} \\ A_{k \times n - k}^{T} & A_{n - k \times n - k} \end{array}] [\begin{array}{l} I_{k} & O_{k \times n - k} \\ O_{k \times n - k}^{T} & O_{n - k \times n - k} \end{array}]) = \frac{1}{2} [\begin{array}{l} A_{k \times k} & A_{k \times n - k} \\ A_{k \times n - k}^{T} & A_{n - k \times n - k} \end{array}] \\ ⇕ \\ \frac{1}{2} [\begin{array}{l} A_{k \times k} & A_{k \times n - k} \\ O_{k \times n - k}^{T} & O_{n - k \times n - k} \end{array}] + \frac{1}{2} [\begin{array}{l} A_{k \times k} & O_{k \times n - k} \\ A_{k \times n - k}^{T} & O_{n - k \times n - k} \end{array}] = [\begin{array}{l} \frac{1}{2} A_{k \times k} & \frac{1}{2} A_{k \times n - k} \\ \frac{1}{2} A_{k \times n - k}^{T} & \frac{1}{2} A_{n - k \times n - k} \end{array}] \\ ⇕ \\ [\begin{array}{l} A_{k \times k} & \frac{1}{2} A_{k \times n - k} \\ \frac{1}{2} A_{k \times n - k}^{T} & O_{n - k \times n - k} \end{array}] = [\begin{array}{l} \frac{1}{2} A_{k \times k} & \frac{1}{2} A_{k \times n - k} \\ \frac{1}{2} A_{k \times n - k}^{T} & \frac{1}{2} A_{n - k \times n - k} \end{array}] \\ ⇕ \\ (A_{k \times k} = O_{k \times k}) \land (A_{n - k \times n - k} = O_{n - k \times n - k}) . \end{array}$ $Mathematical equation: $$ \begin{array}{l}\left[\begin{array}{ll}{I}_k& {O}_{k\times n-k}\\ {O}_{n-k\times k}^T& {O}_{k\times k}\end{array}\right]\diamond \left[\begin{array}{ll}{A}_{k\times k}& {A}_{k\times n-k}\\ {A}_{k\times n-k}^T& {A}_{n-k\times n-k}\\ & \end{array}\right]=\frac{1}{2}\left[\begin{array}{ll}{A}_{k\times k}& {A}_{k\times n-k}\\ {A}_{k\times n-k}^T& {A}_{n-k\times n-k}\\ & \end{array}\right]\\ \Updownarrow \\ \frac{1}{2}\left(\left[\begin{array}{ll}{I}_k& {O}_{k\times n-k}\\ {O}_{k\times n-k}^T& {O}_{n-k\times n-k}\end{array}\right]\left[\begin{array}{ll}{A}_{k\times k}& {A}_{k\times n-k}\\ {A}_{k\times n-k}^T& {A}_{n-k\times n-k}\end{array}\right]\right.\\ +\left.\left[\begin{array}{ll}{A}_{k\times k}& {A}_{k\times n-k}\\ {A}_{k\times n-k}^T& {A}_{n-k\times n-k}\end{array}\right]\left[\begin{array}{ll}{I}_k& {O}_{k\times n-k}\\ {O}_{k\times n-k}^T& {O}_{n-k\times n-k}\end{array}\right]\right)=\frac{1}{2}\left[\begin{array}{ll}{A}_{k\times k}& {A}_{k\times n-k}\\ {A}_{k\times n-k}^T& {A}_{n-k\times n-k}\end{array}\right]\\ \Updownarrow \\ \frac{1}{2}\left[\begin{array}{ll}{A}_{k\times k}& {A}_{k\times n-k}\\ {O}_{k\times n-k}^T& {O}_{n-k\times n-k}\end{array}\right]+\frac{1}{2}\left[\begin{array}{ll}{A}_{k\times k}& {O}_{k\times n-k}\\ {A}_{k\times n-k}^T& {O}_{n-k\times n-k}\end{array}\right]=\left[\begin{array}{ll}\frac{1}{2}{A}_{k\times k}& \frac{1}{2}{A}_{k\times n-k}\\ \frac{1}{2}{A}_{k\times n-k}^T& \frac{1}{2}{A}_{n-k\times n-k}\end{array}\right]\\ \Updownarrow \\ \left[\begin{array}{ll}{A}_{k\times k}& \frac{1}{2}{A}_{k\times n-k}\\ \frac{1}{2}{A}_{k\times n-k}^T& {O}_{n-k\times n-k}\end{array}\right]=\left[\begin{array}{ll}\frac{1}{2}{A}_{k\times k}& \frac{1}{2}{A}_{k\times n-k}\\ \frac{1}{2}{A}_{k\times n-k}^T& \frac{1}{2}{A}_{n-k\times n-k}\end{array}\right]\\ \Updownarrow \\ ({A}_{k\times k}={O}_{k\times k})\wedge ({A}_{n-k\times n-k}={O}_{n-k\times n-k}).\end{array} $$$

Hence, we have $A (C, \frac{1}{2}) = {[\begin{array}{l} O_{k \times k} & A_{k \times n - k} \\ A_{k \times n - k}^{T} & O_{n - k \times n - k} \end{array}] : A_{k \times n - k} \in M_{k \times n - k} (R)} .$ $Mathematical equation: $ \mathcal{A}(C,\frac{1}{2})=\left\{\left[\begin{array}{ll}{O}_{k\times k}& {A}_{k\times n-k}\\ {A}_{k\times n-k}^T& {O}_{n-k\times n-k}\end{array}\right]:{A}_{k\times n-k}\in {M}_{k\times n-k}(\mathbb{R})\right\}.$$ Now, let’s calculate $A (C, 0)$ $Mathematical equation: $ \mathcal{A}(C,0)$$ . So let consider the following calculations:

$\begin{array}{l} [\begin{array}{l} I_{k} & O_{k \times n - k} \\ O_{n - k \times k}^{T} & O_{k \times k} \end{array}] ⋄ [\begin{array}{l} A_{k \times k} & A_{k \times n - k} \\ A_{k \times n - k}^{T} & A_{n - k \times n - k} \end{array}] = 0 [\begin{array}{l} A_{k \times k} & A_{k \times n - k} \\ A_{k \times n - k}^{T} & A_{n - k \times n - k} \end{array}] \\ ⇕ \\ \frac{1}{2} ([\begin{array}{l} I_{k} & O_{k \times n - k} \\ O_{k \times n - k}^{T} & O_{n - k \times n - k} \end{array}] [\begin{array}{l} A_{k \times k} & A_{k \times n - k} \\ A_{k \times n - k}^{T} & A_{n - k \times n - k} \end{array}] \\ + [\begin{array}{l} A_{k \times k} & A_{k \times n - k} \\ A_{k \times n - k}^{T} & A_{n - k \times n - k} \end{array}] [\begin{array}{l} I_{k} & O_{k \times n - k} \\ O_{k \times n - k}^{T} & O_{n - k \times n - k} \end{array}]) = [\begin{array}{l} O_{k \times k} & O_{k \times n - k} \\ O_{k \times n - k}^{T} & O_{n - k \times n - k} \end{array}] \\ ⇕ \\ \frac{1}{2} [\begin{array}{l} A_{k \times k} & A_{k \times n - k} \\ O_{k \times n - k}^{T} & O_{n - k \times n - k} \end{array}] + \frac{1}{2} [\begin{array}{l} A_{k \times k} & O_{k \times n - k} \\ A_{k \times n - k}^{T} & O_{n - k \times n - k} \end{array}] = [\begin{array}{l} O_{k \times k} & O_{k \times n - k} \\ O_{k \times n - k}^{T} & O_{n - k \times n - k} \end{array}] \\ ⇕ \\ [\begin{array}{l} A_{k \times k} & \frac{1}{2} A_{k \times n - k} \\ \frac{1}{2} A_{k \times n - k}^{T} & O_{n - k \times n - k} \end{array}] = [\begin{array}{l} O_{k \times k} & O_{k \times n - k} \\ O_{k \times n - k}^{T} & O_{n - k \times n - k} \end{array}] \\ ⇕ \\ (A_{k \times k} = O_{k \times k}) \land (A_{k \times n - k} = O_{k \times n - k}) . \end{array}$ $Mathematical equation: $$ \begin{array}{l}\left[\begin{array}{ll}{I}_k& {O}_{k\times n-k}\\ {O}_{n-k\times k}^T& {O}_{k\times k}\end{array}\right]\diamond \left[\begin{array}{ll}{A}_{k\times k}& {A}_{k\times n-k}\\ {A}_{k\times n-k}^T& {A}_{n-k\times n-k}\\ & \end{array}\right]=0\left[\begin{array}{ll}{A}_{k\times k}& {A}_{k\times n-k}\\ {A}_{k\times n-k}^T& {A}_{n-k\times n-k}\\ & \end{array}\right]\\ \Updownarrow \\ \frac{1}{2}\left(\left[\begin{array}{ll}{I}_k& {O}_{k\times n-k}\\ {O}_{k\times n-k}^T& {O}_{n-k\times n-k}\end{array}\right]\left[\begin{array}{ll}{A}_{k\times k}& {A}_{k\times n-k}\\ {A}_{k\times n-k}^T& {A}_{n-k\times n-k}\end{array}\right]\right.\\ +\left.\left[\begin{array}{ll}{A}_{k\times k}& {A}_{k\times n-k}\\ {A}_{k\times n-k}^T& {A}_{n-k\times n-k}\end{array}\right]\left[\begin{array}{ll}{I}_k& {O}_{k\times n-k}\\ {O}_{k\times n-k}^T& {O}_{n-k\times n-k}\end{array}\right]\right)=\left[\begin{array}{ll}{O}_{k\times k}& {O}_{k\times n-k}\\ {O}_{k\times n-k}^T& {O}_{n-k\times n-k}\end{array}\right]\\ \Updownarrow \\ \frac{1}{2}\left[\begin{array}{ll}{A}_{k\times k}& {A}_{k\times n-k}\\ {O}_{k\times n-k}^T& {O}_{n-k\times n-k}\end{array}\right]+\frac{1}{2}\left[\begin{array}{ll}{A}_{k\times k}& {O}_{k\times n-k}\\ {A}_{k\times n-k}^T& {O}_{n-k\times n-k}\end{array}\right]=\left[\begin{array}{ll}{O}_{k\times k}& {O}_{k\times n-k}\\ {O}_{k\times n-k}^T& {O}_{n-k\times n-k}\end{array}\right]\\ \Updownarrow \\ \left[\begin{array}{ll}{A}_{k\times k}& \frac{1}{2}{A}_{k\times n-k}\\ \frac{1}{2}{A}_{k\times n-k}^T& {O}_{n-k\times n-k}\end{array}\right]=\left[\begin{array}{ll}{O}_{k\times k}& {O}_{k\times n-k}\\ {O}_{k\times n-k}^T& {O}_{n-k\times n-k}\end{array}\right]\\ \Updownarrow \\ ({A}_{k\times k}={O}_{k\times k})\wedge ({A}_{k\times n-k}={O}_{k\times n-k}).\end{array} $$$

Hence we have $A (C, 0) = {[\begin{array}{l} O_{k \times k} & O_{k \times n - k} \\ O_{k \times n - k}^{T} & A_{n - k \times n - k} \end{array}] : A_{n - k \times n - k} \in Sym (n - k, R)}$ $Mathematical equation: $ \mathcal{A}(C,0)=\left\{\left[\begin{array}{ll}{O}_{k\times k}& {O}_{k\times n-k}\\ {O}_{k\times n-k}^T& {A}_{n-k\times n-k}\end{array}\right]:{A}_{n-k\times n-k}\in \enspace \mathrm{Sym}\enspace (n-k,\mathbb{R})\right\}$$ .

For, other hand if we consider a Jordan frame $S = {g_{1}, g_{2}, \dots, g_{r}}$ $Mathematical equation: $ S=\{{g}_1,{g}_2,\cdots,{g}_r\}$$ of an Euclidean Jordan algebra $A$ $Mathematical equation: $ \mathcal{A}$$ and considering the spaces $A_{ii} = {x \in A : L_{⋄} (g_{i}) x = x}$ $Mathematical equation: $ {\mathcal{A}}_{{ii}}=\{x\in \mathcal{A}:{L}_{\diamond }({g}_i)x=x\}$$ and the spaces $A_{ij} = {x \in A : L_{⋄} (g_{i}) x = \frac{1}{2} x \land L_{⋄} (g_{j}) x = \frac{1}{2} x}$ $Mathematical equation: $ {\mathcal{A}}_{{ij}}=\{x\in \mathcal{A}:{L}_{\diamond }({g}_i)x=\frac{1}{2}x\wedge {L}_{\diamond }({g}_j)x=\frac{1}{2}x\}$$ then we obtain the decomposition of $A$ $Mathematical equation: $ \mathcal{A}$$ as an orthogonal direct sum of the vector spaces $A_{ii}$ $Mathematical equation: $ {\mathcal{A}}_{{ii}}$$ s and $A_{ij}$ $Mathematical equation: $ {\mathcal{A}}_{{ij}}$$ s in the following way: $A = \sum_{i = 1}^{r} A_{ii} + \sum_{1 \leq i < j \leq r} A_{ij} .$ $Mathematical equation: $ \mathcal{A}={\sum }_{i=1}^r {\mathcal{A}}_{{ii}}+{\sum }_1 {\mathcal{A}}_{{ij}}.$$

In the case when the Euclidean symmetric Jordan algebra is $A = Sym (n, R)$ $Mathematical equation: $ \mathcal{A}=\enspace \mathrm{Sym}\enspace (n,\mathbb{R})$$ and we consider the Jordan frame of $A$ $Mathematical equation: $ \mathcal{A}$$ , $S = {E_{11}, E_{22}, \dots, E_{nn}}$ $Mathematical equation: $ S=\{{E}_{11},{E}_{22},\cdots,{E}_{{nn}}\}$$ we obtain the following spaces $A_{ii} = {A \in M_{n} (R) : \exists α \in R : A = α E_{ii}}$ $Mathematical equation: $ {\mathcal{A}}_{{ii}}=\{A\in {M}_n(\mathbb{R}):\exists \alpha \in \mathbb{R}:A=\alpha {E}_{{ii}}\}$$ and the spaces $A_{ij} = {A \in M_{n} (R) : \exists x_{ij} \in R : A = x_{ij} (E_{ij} + E_{ji})},$ $Mathematical equation: $ {\mathcal{A}}_{{ij}}=\{A\in {M}_n(\mathbb{R}):\exists {x}_{{ij}}\in \mathbb{R}:A={x}_{{ij}}({E}_{{ij}}+{E}_{{ji}})\},$$ where the matrices E _iis are the matrices with 1 in the entry ii and with the others entries zero and the matrix E _ij is the matrix with 1 in the entry ij and zero on the others entries. Therefore the Pierce decomposition of $A$ $Mathematical equation: $ \mathcal{A}$$ relatively to the Jordan frame of $A$ $Mathematical equation: $ \mathcal{A}$$ is $A = \sum_{i = 1}^{n} A_{ii} + \sum_{i \leq 1 < i < j \leq n}^{n} A_{ij} .$ $Mathematical equation: $ \mathcal{A}={\sum }_{i=1}^n {A}_{{ii}}+{\sum }_{i\le 1 < i < j\le n}^n {A}_{{ij}}.$$ Therefore, we can write, any matrix of the $Sym (n, R)$ $Mathematical equation: $ \mathrm{Sym}\enspace (n,\mathbb{R})$$ in the form $A = \sum_{i = 1}^{n} a_{ii} E_{ii} + \sum_{{1 \leq i < j \leq n}}^{n} a_{ij} (E_{ij} + E_{ji}) .$ $Mathematical equation: $ A={\sum }_{i=1}^n {a}_{{ii}}{E}_{{ii}}+{\sum }_{\{1\le i < j\le n\}}^n {a}_{{ij}}({E}_{{ij}}+{E}_{{ji}}).$$

Let consider the spin Euclidean Jordan algebra $A_{n + 1}$ $Mathematical equation: $ {\mathcal{A}}_{n+1}$$ and let consider the idempotent $c = \frac{1}{2} [\begin{array}{l} 1 \\ \overset{̅}{x} \end{array}]$ $Mathematical equation: $ c=\frac{1}{2}\left[\begin{array}{l}1\\ \bar{x}\end{array}\right]$$ with $| | \overset{̅}{x} | | = 1 .$ $Mathematical equation: $ ||\bar{x}||=1.$$ We, will obtain the spaces $A_{n + 1} (c, 1)$ $Mathematical equation: $ {\mathcal{A}}_{n+1}(c,1)$$ , $A_{n + 1} (c, \frac{1}{2})$ $Mathematical equation: $ {\mathcal{A}}_{n+1}\left(c,\frac{1}{2}\right)$$ and $A_{n + 1} (c, 0) .$ $Mathematical equation: $ {\mathcal{A}}_{n+1}(c,0).$$ We have, the following equivalences

$\begin{array}{l} c ⋄ [\begin{array}{l} a_{1} \\ \overset{̅}{a} \end{array}] = [\begin{array}{l} a_{1} \\ \overset{̅}{a} \end{array}] \Leftrightarrow \frac{1}{2} [\begin{array}{l} 1 \\ \overset{̅}{x} \end{array}] ⋄ [\begin{array}{l} a_{1} \\ \overset{̅}{a} \end{array}] = [\begin{array}{l} a_{1} \\ \overset{̅}{a} \end{array}] \\ \Leftrightarrow [\begin{array}{l} \frac{1}{2} a_{1} + \frac{1}{2} \overset{̅}{x} | \overset{̅}{a} \\ \frac{1}{2} \overset{̅}{a} + \frac{1}{2} a_{1} \overset{̅}{x} \end{array}] = [\begin{array}{l} a_{1} \\ \overset{̅}{a} \end{array}] \\ \Leftrightarrow a_{1} = \overset{̅}{x} | \overset{̅}{a} \land \overset{̅}{a} = (\overset{̅}{x} | \overset{̅}{a}) \overset{̅}{x} \\ \Leftrightarrow [\begin{array}{l} a_{1} \\ \overset{̅}{a} \end{array}] = (2 \overset{̅}{x} | \overset{̅}{a}) \frac{1}{2} [\begin{array}{l} 1 \\ \overset{̅}{x} \end{array}] \Leftrightarrow [\begin{array}{l} a_{1} \\ \overset{̅}{a} \end{array}] = (2 \overset{̅}{x} | \overset{̅}{a}) c . \end{array}$ $Mathematical equation: $$ \begin{array}{l}c\diamond \left[\begin{array}{l}{a}_1\\ \bar{a}\end{array}\right]=\left[\begin{array}{l}{a}_1\\ \bar{a}\end{array}\right]\iff \frac{1}{2}\left[\begin{array}{l}1\\ \bar{x}\\ \end{array}\right]\diamond \left[\begin{array}{l}{a}_1\\ \bar{a}\end{array}\right]=\left[\begin{array}{l}{a}_1\\ \bar{a}\end{array}\right]\\ \iff \left[\begin{array}{l}\frac{1}{2}{a}_1+\frac{1}{2}\bar{x}|\bar{a}\\ \frac{1}{2}\bar{a}+\frac{1}{2}{a}_1\bar{x}\end{array}\right]=\left[\begin{array}{l}{a}_1\\ \bar{a}\end{array}\right]\\ \iff {a}_1=\bar{x}|\bar{a}\wedge \bar{a}=(\bar{x}|\bar{a})\bar{x}\\ \iff \left[\begin{array}{l}{a}_1\\ \bar{a}\end{array}\right]=(2\bar{x}|\bar{a})\frac{1}{2}\left[\begin{array}{l}1\\ \bar{x}\end{array}\right]\iff \left[\begin{array}{l}{a}_1\\ \bar{a}\end{array}\right]=(2\bar{x}|\bar{a})c.\end{array} $$$

So, we conclude that $A_{n + 1} (c, 1) = {α c, α \in R} .$ $Mathematical equation: $ {\mathcal{A}}_{n+1}(c,1)=\{\alpha c,\alpha \in \mathbb{R}\}.$$ Now, we will calculate $A_{n + 1} (c, 0) .$ $Mathematical equation: $ {\mathcal{A}}_{n+1}(c,0).$$

$\begin{array}{l} c ⋄ [\begin{array}{l} a_{1} \\ \overset{̅}{a} \end{array}] = 0 [\begin{array}{l} a_{1} \\ \overset{̅}{a} \end{array}] \Leftrightarrow \frac{1}{2} [\begin{matrix} 1 \\ \overset{̅}{x} \end{matrix}] ⋄ [\begin{array}{l} a_{1} \\ \overset{̅}{a} \end{array}] = [\begin{array}{l} 0 \\ \overset{̅}{0} \end{array}] \\ \Leftrightarrow [\begin{array}{l} \frac{1}{2} a_{1} + \frac{1}{2} \overset{̅}{x} | \overset{̅}{a} \\ \frac{1}{2} \overset{̅}{a} + \frac{1}{2} a_{1} \overset{̅}{x} \end{array}] = [\begin{array}{l} 0 \\ \overset{̅}{0} \end{array}] \Leftrightarrow a_{1} = - \overset{̅}{x} | \overset{̅}{a} \land \overset{̅}{a} = (\overset{̅}{x} | \overset{̅}{a}) \overset{̅}{x} \\ \Leftrightarrow [\begin{array}{l} a_{1} \\ \overset{̅}{a} \end{array}] = (- 2 \overset{̅}{x} | \overset{̅}{a}) \frac{1}{2} [\begin{matrix} 1 \\ - \overset{̅}{x} \end{matrix}] . \end{array}$ $Mathematical equation: $$ \begin{array}{l}c\diamond \left[\begin{array}{l}{a}_1\\ \bar{a}\end{array}\right]=0\left[\begin{array}{l}{a}_1\\ \bar{a}\end{array}\right]\iff \frac{1}{2}\left[\begin{array}{c}1\\ \bar{x}\end{array}\right]\diamond \left[\begin{array}{l}{a}_1\\ \bar{a}\end{array}\right]=\left[\begin{array}{l}0\\ \bar{0}\end{array}\right]\\ \iff \left[\begin{array}{l}\frac{1}{2}{a}_1+\frac{1}{2}\bar{x}|\bar{a}\\ \frac{1}{2}\bar{a}+\frac{1}{2}{a}_1\bar{x}\end{array}\right]=\left[\begin{array}{l}0\\ \bar{0}\end{array}\right]\iff {a}_1=-\bar{x}|\bar{a}\wedge \bar{a}=(\bar{x}|\bar{a})\bar{x}\\ \iff \left[\begin{array}{l}{a}_1\\ \bar{a}\end{array}\right]=(-2\bar{x}|\bar{a})\frac{1}{2}\left[\begin{array}{c}1\\ -\bar{x}\end{array}\right].\end{array} $$$ Therefore, we obtain, $A_{n + 1} (c, 0) = {α \frac{1}{2} [\begin{matrix} 1 \\ - \overset{̅}{x} \end{matrix}], α \in R} .$ $Mathematical equation: $ {\mathcal{A}}_{n+1}(c,0)=\left\{\alpha \frac{1}{2}\left[\begin{array}{c}1\\ -\bar{x}\end{array}\right],\alpha \in \mathbb{R}\right\}.$$ We, can rewrite $A_{n + 1} (c, 0) = {α [\begin{array}{l} 1 & O_{1 \times n} \\ O_{1 \times n}^{T} & - I_{n} \end{array}] c, α \in R} .$ $Mathematical equation: $ {\mathcal{A}}_{n+1}(c,0)=\left\{\alpha \left[\begin{array}{ll}1& {O}_{1\times n}\\ {O}_{1\times n}^T& -{I}_n\end{array}\right]c,\alpha \in \mathbb{R}\right\}.$$ Now, we will obtain $A_{n + 1} (c, \frac{1}{2}) .$ $Mathematical equation: $ {\mathcal{A}}_{n+1}\left(c,\frac{1}{2}\right).$$ So, we have

$\begin{matrix} c ⋄ [\begin{array}{l} a_{1} \\ \overset{̅}{a} \end{array}] = \frac{1}{2} [\begin{array}{l} a_{1} \\ \overset{̅}{a} \end{array}] \Leftrightarrow \frac{1}{2} [\begin{array}{l} 1 \\ \overset{̅}{x} \end{array}] ⋄ [\begin{array}{l} a_{1} \\ \overset{̅}{a} \end{array}] = [\begin{array}{l} \frac{1}{2} a_{1} \\ \frac{1}{2} \overset{̅}{a} \end{array}] \\ \Leftrightarrow [\begin{array}{l} \frac{1}{2} a_{1} + \frac{1}{2} \overset{̅}{x} | \overset{̅}{a} \\ \frac{1}{2} \overset{̅}{a} + \frac{1}{2} a_{1} \overset{̅}{x} \end{array}] = [\begin{array}{l} \frac{1}{2} a_{1} \\ \frac{1}{2} \overset{̅}{a} \end{array}] \Leftrightarrow \overset{̅}{x} | \overset{̅}{a} = 0 \land a_{1} \overset{̅}{x} = \overset{̅}{0} \\ \Leftrightarrow \\ (a_{1} = 0) \land (\overset{̅}{x} | \overset{̅}{a} = 0) . \end{matrix}$ $Mathematical equation: $$ \begin{array}{c}c\diamond \left[\begin{array}{l}{a}_1\\ \bar{a}\end{array}\right]=\frac{1}{2}\left[\begin{array}{l}{a}_1\\ \bar{a}\end{array}\right]\iff \frac{1}{2}\left[\begin{array}{l}1\\ \bar{x}\\ \end{array}\right]\diamond \left[\begin{array}{l}{a}_1\\ \bar{a}\end{array}\right]=\left[\begin{array}{l}\frac{1}{2}{a}_1\\ \frac{1}{2}\bar{a}\end{array}\right]\\ \iff \left[\begin{array}{l}\frac{1}{2}{a}_1+\frac{1}{2}\bar{x}|\bar{a}\\ \frac{1}{2}\bar{a}+\frac{1}{2}{a}_1\bar{x}\end{array}\right]=\left[\begin{array}{l}\frac{1}{2}{a}_1\\ \frac{1}{2}\bar{a}\end{array}\right]\iff \bar{x}|\bar{a}=0\wedge {a}_1\bar{x}=\bar{0}\\ \iff \\ ({a}_1=0)\wedge (\bar{x}|\bar{a}=0).\end{array} $$$

Therefore, we obtain,

$A_{n + 1} (c, \frac{1}{2}) = {α [\begin{array}{l} 0 \\ \overset{̅}{a} \end{array}] : \overset{̅}{x} | \overset{̅}{a} = 0, α \in R} .$ $Mathematical equation: $$ {\mathcal{A}}_{n+1}\left(c,\frac{1}{2}\right)=\left\{\alpha \left[\begin{array}{l}0\\ \bar{a}\end{array}\right]:\bar{x}|\bar{a}=0,\alpha \in \mathbb{R}\right\}. $$$

Now, we will analyse the Pierce decomposition of $A_{n + 1}$ $Mathematical equation: $ {\mathcal{A}}_{n+1}$$ relatively to the Jordan frame

${c_{1}, c_{2}} = {\frac{1}{2} [\begin{matrix} 1 \\ 1 \\ 0_{n - 1} \end{matrix}], \frac{1}{2} [\begin{matrix} 1 \\ - 1 \\ 0_{n - 1} \end{matrix}]} .$ $Mathematical equation: $$ \{{c}_1,{c}_2\}=\left\{\frac{1}{2}\left[\begin{array}{c}1\\ 1\\ {0}_{n-1}\end{array}\right],\frac{1}{2}\left[\begin{array}{c}1\\ -1\\ {0}_{n-1}\end{array}\right]\right\}. $$$

As in the previous example, we have $(A_{n + 1})_{11} = A_{n + 1} (c_{1}, 1) = {α c_{1} : α \in R},$ $Mathematical equation: $ ({\mathcal{A}}_{n+1}{)}_{11}={\mathcal{A}}_{n+1}({c}_1,1)=\{\alpha {c}_1:\alpha \in \mathbb{R}\},$$ and $(A_{n + 1})_{22} = A_{n + 1} (c_{2}, 1) = {α c_{2} : α \in R} .$ $Mathematical equation: $ ({\mathcal{A}}_{n+1}{)}_{22}={\mathcal{A}}_{n+1}({c}_2,1)=\{\alpha {c}_2:\alpha \in \mathbb{R}\}.$$ Now, we will calculate the vector space

$(A_{n + 1})_{12} = {x \in A_{n + 1} : (L_{⋄} (c_{1}) x = \frac{1}{2} x) \land (L_{⋄} (c_{2}) x = \frac{1}{2} x)}$ $Mathematical equation: $$ ({\mathcal{A}}_{n+1}{)}_{12}=\left\{x\in {\mathcal{A}}_{n+1}:({L}_{\diamond }({c}_1)x=\frac{1}{2}x)\wedge ({L}_{\diamond }({c}_2)x=\frac{1}{2}x)\right\} $$$

So, we have the following calculations

$\begin{array}{l} \frac{1}{2} [\begin{array}{l} 1 \\ 1 \\ \overset{̅}{0} \end{array}] ⋄ [\begin{array}{l} x_{1} \\ x_{2} \\ \overset{̅}{x} \end{array}] = \frac{1}{2} [\begin{array}{l} x_{1} \\ x_{2} \\ \overset{̅}{x} \end{array}] \Leftrightarrow [\begin{array}{l} \frac{1}{2} x_{1} + \frac{1}{2} x_{2} + \overset{̅}{0} | \overset{̅}{x} \\ \frac{1}{2} [\begin{array}{l} x_{2} \\ \overset{̅}{x} \end{array}] + \frac{x_{1}}{2} [\begin{matrix} 1 \\ \overset{̅}{0} \end{matrix}] \end{array}] = \frac{1}{2} [\begin{array}{l} x_{1} \\ x_{2} \\ \overset{̅}{x} \end{array}] \\ \Leftrightarrow [\begin{array}{l} \frac{1}{2} x_{1} + \frac{1}{2} x_{2} \\ \frac{1}{2} x_{2} + \frac{1}{2} x_{1} \\ \frac{1}{2} \overset{̅}{x} \end{array}] = [\begin{array}{l} \frac{1}{2} x_{1} \\ \frac{1}{2} x_{2} \\ \frac{1}{2} \overset{̅}{x} \end{array}] \\ (x_{1} = 0) \land (x_{2} = 0) . \end{array}$ $Mathematical equation: $$ \begin{array}{l}\frac{1}{2}\left[\begin{array}{l}1\\ 1\\ \bar{0}\end{array}\right]\diamond \left[\begin{array}{l}{x}_1\\ {x}_2\\ \bar{x}\end{array}\right]=\frac{1}{2}\left[\begin{array}{l}{x}_1\\ {x}_2\\ \bar{x}\end{array}\right]\iff \left[\begin{array}{l}\frac{1}{2}{x}_1+\frac{1}{2}{x}_2+\bar{0}|\bar{x}\\ \frac{1}{2}\left[\begin{array}{l}{x}_2\\ \bar{x}\end{array}\right]+\frac{{x}_1}{2}\left[\begin{array}{c}1\\ \bar{0}\end{array}\right]\end{array}\right]=\frac{1}{2}\left[\begin{array}{l}{x}_1\\ {x}_2\\ \bar{x}\end{array}\right]\\ \iff \left[\begin{array}{l}\frac{1}{2}{x}_1+\frac{1}{2}{x}_2\\ \frac{1}{2}{x}_2+\frac{1}{2}{x}_1\\ \frac{1}{2}\bar{x}\\ \end{array}\right]=\left[\begin{array}{l}\frac{1}{2}{x}_1\\ \frac{1}{2}{x}_2\\ \frac{1}{2}\bar{x}\end{array}\right]\\ \left({x}_1=0\right)\wedge \left({x}_2=0\right).\end{array} $$$

Therefore $A_{n + 1} (c_{1}, \frac{1}{2}) = {[\begin{array}{l} 0 \\ 0 \\ \overset{̅}{x} \end{array}] : \overset{̅}{x} \in R^{n - 1}} .$ $Mathematical equation: $ {\mathcal{A}}_{n+1}({c}_1,\frac{1}{2})=\{\left[\begin{array}{l}0\\ 0\\ \bar{x}\end{array}\right]:\bar{x}\in {\mathbb{R}}^{n-1}\}.$$ Now, we will calculate $A_{n + 1} (c_{2}, \frac{1}{2}) .$ $Mathematical equation: $ {\mathcal{A}}_{n+1}\left({c}_2,\frac{1}{2}\right).$$

$\begin{array}{l} \frac{1}{2} [\begin{matrix} 1 \\ - 1 \\ \overset{̅}{0} \end{matrix}] ⋄ [\begin{matrix} x_{1} \\ x_{2} \\ \overset{̅}{x} \end{matrix}] = \frac{1}{2} [\begin{matrix} x_{1} \\ x_{2} \\ \overset{̅}{x} \end{matrix}] \Leftrightarrow [\begin{matrix} \frac{1}{2} x_{1} - \frac{1}{2} x_{2} + \overset{̅}{0} | \overset{̅}{x} \\ \frac{1}{2} [\begin{array}{l} x_{2} \\ \overset{̅}{x} \end{array}] + x_{1} [\begin{matrix} - \frac{1}{2} \\ \overset{̅}{0} \end{matrix}] \end{matrix}] = \frac{1}{2} [\begin{matrix} x_{1} \\ x_{2} \\ \overset{̅}{x} \end{matrix}] \\ \Leftrightarrow [\begin{matrix} \frac{1}{2} x_{1} - \frac{1}{2} x_{2} \\ \frac{1}{2} x_{2} - \frac{1}{2} x_{1} \\ \overset{̅}{x} \end{matrix}] = [\begin{matrix} \frac{1}{2} x_{1} \\ \frac{1}{2} x_{2} \\ \overset{̅}{x} \end{matrix}] \Leftrightarrow (x_{1} = 0) \land (x_{2} = 0) . \end{array}$ $Mathematical equation: $$ \begin{array}{l}\frac{1}{2}\left[\begin{array}{c}1\\ -1\\ \bar{0}\end{array}\right]\diamond \left[\begin{array}{c}{x}_1\\ {x}_2\\ \bar{x}\end{array}\right]=\frac{1}{2}\left[\begin{array}{c}{x}_1\\ {x}_2\\ \bar{x}\end{array}\right]\iff \left[\begin{array}{c}\frac{1}{2}{x}_1-\frac{1}{2}{x}_2+\bar{0}|\bar{x}\\ \frac{1}{2}\left[\begin{array}{l}{x}_2\\ \bar{x}\end{array}\right]+{x}_1\left[\begin{array}{c}-\frac{1}{2}\\ \bar{0}\end{array}\right]\end{array}\right]=\frac{1}{2}\left[\begin{array}{c}{x}_1\\ {x}_2\\ \bar{x}\end{array}\right]\\ \iff \left[\begin{array}{c}\frac{1}{2}{x}_1-\frac{1}{2}{x}_2\\ \frac{1}{2}{x}_2-\frac{1}{2}{x}_1\\ \bar{x}\\ \end{array}\right]=\left[\begin{array}{c}\frac{1}{2}{x}_1\\ \frac{1}{2}{x}_2\\ \bar{x}\end{array}\right]\iff ({x}_1=0)\wedge ({x}_2=0).\end{array} $$$

Hence, we have also that $A_{n + 1} (c_{2}, \frac{1}{2}) = A_{n + 1} (c_{1}, \frac{1}{2}) .$ $Mathematical equation: $ {\mathcal{A}}_{n+1}\left({c}_2,\frac{1}{2}\right)={\mathcal{A}}_{n+1}\left({c}_1,\frac{1}{2}\right).$$ Therefore we have

$(A_{n + 1})_{12} = {[\begin{matrix} 0 \\ 0 \\ \overset{̅}{x} \end{matrix}] : \overset{̅}{x} \in R^{n - 1}} .$ $Mathematical equation: $$ ({\mathcal{A}}_{n+1}{)}_{12}=\left\{\left[\begin{array}{c}0\\ 0\\ \bar{x}\end{array}\right]:\bar{x}\in {\mathbb{R}}^{n-1}\right\}. $$$

Therefore for any element $x = [\begin{array}{l} x_{1} \\ x_{2} \\ \overset{̅}{x} \end{array}]$ $Mathematical equation: $ x=\left[\begin{array}{l}{x}_1\\ {x}_2\\ \bar{x}\end{array}\right]$$ of $A_{n + 1}$ $Mathematical equation: $ {\mathcal{A}}_{n+1}$$ we conclude that

$x = (x_{1} + x_{2}) c_{1} + (x_{1} - x_{2}) c_{2} + [\begin{array}{l} 0 \\ 0 \\ \overset{̅}{x} \end{array}] .$ $Mathematical equation: $$ x=({x}_1+{x}_2){c}_1+({x}_1-{x}_2){c}_2+\left[\begin{array}{l}0\\ 0\\ \bar{x}\end{array}\right]. $$$

A brief introduction to strongly regular graphs

An undirect graph X is a pair of sets (V(X), E(X)) with $V (X) = {v_{1}, \dots, v_{n - 1}, v_{n}}$ $Mathematical equation: $ V(X)=\left\{{v}_1,\cdots,{v}_{n-1}\right.\left.,{v}_n\right\}$$ and E(X), the set of edges of X, a subset of V(X) × V(X). For simplicity, we will denote an edge between the vertices a and b by ab. The order of the graph X is the number of vertices of $X, | V (X) |$ $Mathematical equation: $ X,\enspace |V(X)|$$ and we call the dimension of $X, | E (X) |$ Mathematical equation: $ X,|E(X)|$ to the number of edges of X. One calls a graph a simple graph if it has no multiple edges (more than one edge between two vertices) and if it has no loops.

Sometimes we make a sketch to represent a graph X like the one presented on the Figure 1.

Figure 1

A simple graph X.

An edge is incident on a vertice v of a graph X if v is one of its extreme points. Two vertices of a graph X are adjacent if they are connected by an edge. The adjacency matrix of a simple graph X of order n is a square matrix of order n, A such that $A = [a_{ij}]$ $Mathematical equation: $ A=[{a}_{{ij}}]$$ where $a_{ij} = 1$ $Mathematical equation: $ {a}_{{ij}}=1$$ if $v_{i} v_{j} \in E (X)$ $Mathematical equation: $ {v}_i{v}_j\in E(X)$$ and 0 otherwise. The adjacency matrix of a simple graph is a symmetric matrix and we must observe that the diagonal entries of this matrix are null. The number of edges incident to a vertice v of a simple graph is called the degree of v. And, we call a simple graph a regular graph if each of its vertices have the same degree and we say that a graph G is a k-regular graph if each of its vertices have degree k.

The complement graph of a graph X denoted by $\overset{̅}{X}$ $Mathematical equation: $ \bar{X}$$ is a graph with the same set of vertices of X and such that two distinct vertices are adjacent vertices of $\overset{̅}{X}$ $Mathematical equation: $ \bar{X}$$ if and only if they are non adjacent vertices of X.

Along this paper we consider only non-empty, simple and non complete graphs.

Strongly regular graphs were firstly introduced by R. C. Bose in the paper [33].

A graph X is called a (n, k; λ, μ)-strongly regular graph if is k-regular and any pair of adjacent vertices have λ common neighbors and any pair of non-adjacent vertices have μ common adjacent vertices.

The adjacency matrix A of a (n, k; λ, μ)-strongly regular X satisfies the equation $A^{2} = k I_{n} + λ A + μ (J_{n} - A - I_{n})$ $Mathematical equation: $ {A}^2=k{I}_n+\lambda A+\mu ({J}_n-A-{I}_n)$$ , where J _n is the all ones real matrix of order n.

The eigenvalues θ, τ, k and the multiplicities m _θ and m _τ of θ and τ respectively of a (n, k; λ, μ)-strongly regular graph X, see, for instance [34, 35], are defined by the equalities (5):

$\begin{matrix} θ = (λ - μ + \sqrt{(λ - μ)^{2} + 4 (k - μ)}) / 2, \\ τ = (λ - μ - \sqrt{(λ - μ)^{2} + 4 (k - μ)}) / 2, \\ \begin{matrix} m_{θ} = \frac{| τ | n + τ - k}{n (θ - τ)}, \\ m_{τ} = \frac{θ n + k - θ}{n (θ - τ)} . \end{matrix} \end{matrix}$ $Mathematical equation: $$ \begin{array}{c}\theta =(\lambda -\mu +\sqrt{(\lambda -\mu {)}^2+4(k-\mu )})/2,\\ \tau =(\lambda -\mu -\sqrt{(\lambda -\mu {)}^2+4(k-\mu )})/2,\\ \begin{array}{c}{m}_{\theta }=\frac{|\tau |n+\tau -k}{n(\theta -\tau )},\\ {m}_{\tau }=\frac{\theta n+k-\theta }{n(\theta -\tau )}.\end{array}\end{array} $$$ (5)

Therefore necessary conditions for the parameters of a (n, k; λ, μ)-strongly regular graph are that $\frac{| τ | n + τ - k}{n (θ - τ)}$ $Mathematical equation: $ \frac{|\tau |n+\tau -k}{n(\theta -\tau )}$$ and $\frac{θ n + k - θ}{n (θ - τ)}$ $Mathematical equation: $ \frac{\theta n+k-\theta }{n(\theta -\tau )}$$ must be natural numbers, they are known as the integrability conditions of a strongly regular graph, see [34].

We note that a graph X is a (n, k; λ, μ)-strongly regular graph if and only if its complement graph $\overset{̅}{X}$ $Mathematical equation: $ \bar{X}$$ is a (n, n − k − 1; n − 2 − 2k + μ, n − 2k + λ). Now we will present some admissibility conditions on the parameters of a (n, k; λ, μ)-strongly regular graph. The parameters of a (n, k; λ, μ)-strongly regular graph X verifies the admissibility condition (6).

$k (k - 1 - λ) = μ (n - k - 1) .$ $Mathematical equation: $$ k(k-1-\lambda )=\mu (n-k-1). $$$ (6)

The inequalities (7) are known as the Krein conditions of X, see [36].

$\begin{matrix} (k + θ) (τ + 1)^{2} \geq (θ + 1) (k + θ + 2 θ τ), \\ (k + τ) (θ + 1)^{2} \geq (τ + 1) (k + τ + 2 θ τ) . \end{matrix}$ $Mathematical equation: $$ \begin{array}{c}(k+\theta )(\tau +1{)}^2\ge (\theta +1)(k+\theta +2{\theta \tau }),\\ (k+\tau )(\theta +1{)}^2\ge (\tau +1)(k+\tau +2{\theta \tau }).\end{array} $$$ (7)

Given a graph X, we call a path in X between two vertices v ₁ and v _k+1 to a non null sequence of distinct vertices, exceptionally the first vertice and the last vertice can be equal, and distinct edges, W = v ₁ e ₁ v ₂ e ₂ v ₃⋯v _k e _k v _k+1 whose terms are vertices and edges alternated and such that for 1 ≤ i ≤ k the vertices v _i and $v_{i + 1}$ $Mathematical equation: $ {v}_{i+1}$$ define the edge e _i . The path is a closed path or a cycle if and only if the only repeated vertices are the initial vertice and the final vertice.

A graph Y′ is a subgraph of a graph Y and we write $Y' ⊑ Y$ $Mathematical equation: $ Y\mathrm{\prime}\sqsubseteq Y$$ if $V (Y') ⊑ V (Y)$ $Mathematical equation: $ V(Y\mathrm{\prime})\sqsubseteq V(Y)$$ and $E (Y') ⊑ E (Y) .$ $Mathematical equation: $ E(Y\mathrm{\prime})\sqsubseteq E(Y).$$ If $Y' ⊑ Y$ $Mathematical equation: $ Y\mathrm{\prime}\sqsubseteq Y$$ and $Y' \neq Y,$ $Mathematical equation: $ Y\mathrm{\prime}\ne Y,$$ we say that Y′ is a proper subgraph of Y. We must observe that for any non empty subset V′ of V(Y) we can construct a subgraph of Y whose set of vertices is $V'$ $Mathematical equation: $ V\mathrm{\prime}$$ and whose set of edges is formed by the edges of E(Y) whose extreme points are vertices in V′ which we call the induced subgraph of Y and which we denote by $Y (V') .$ $Mathematical equation: $ Y(V\mathrm{\prime}).$$ Two vertices v ₁ and v ₂ of a graph X are connected if there is a path between v ₁ and v ₂ in X. This relation between vertices is a relation of equivalence in the set of vertices of the graph X, V(X), whereby there exists a partition of V(X) in non empty subsets V ₁, V ₂,⋯, V _I of V(X) such that two vertices are connected if and only if they belong to the same set V _i for a given $i \in {1,2, \dots, l} .$ $Mathematical equation: $ i\in \{\mathrm{1,2},\cdots,l\}.$$ The subgraphs X(V ₁), X(V ₂), ⋯, X(V _l) are called the connected components of X. If X has only one component then we say that the graph X is connected otherwise the graph X is a disconnected graph. A (n, k; λ, μ)-strongly regular graph X is primitive if and only X and $\overset{̅}{X}$ $Mathematical equation: $ \bar{X}$$ are connected graphs. Otherwise is called an imprimitive strongly regular graph. To characterize the connected graphs we will present the definition of reducible and of irreducible matrix. Let n be a natural number greater or equal 2 and $A$ Mathematical equation: $ A$ a matrix in $M_{n} (R) .$ $Mathematical equation: $ {M}_n(\mathbb{R}).$$ We say that the matrix $A$ is a reducible matrix if there exists a permutation matrix $P$ Mathematical equation: $ P$ such that

$P^{T} AP = [\begin{array}{l} C_{k \times k} & O_{k \times n - k} \\ D_{n - k \times k} & E_{n - k \times n - k} \end{array}]$ $Mathematical equation: $$ {P}^T{AP}=\left[\begin{array}{ll}{C}_{k\times k}& {O}_{k\enspace \times n-k}\\ {D}_{n-k\enspace \times k}& {E}_{n-k\enspace \times n-k}\\ & \end{array}\right] $$$ (8)where k is such that 1 ≤ k ≤ n−1, if doesn’t exist such matrix P we say that the matrix A is irreducible. If A is a reducible symmetric of order n then $D_{n - k \times k} = O_{n - k \times k} .$ $Mathematical equation: $ {D}_{n-k\enspace \times k}={\mathrm{O}}_{n-k\enspace \times k}.$$ From the Theorem of Frobenius we know that if A is a real square irreducible matrix with non negative entries then A has an eigenvector u such that Au = ru with all entries positive and such that $| λ | \leq r$ $Mathematical equation: $ |\lambda |\le r$$ for any eigenvalue of A and r is a simple positive eigenvalue of A. Now since a graph is connected if and only if its matrix of adjacency is irreducible then, if a graph X is a connected strongly regular graph then the greater eigenvalue of its adjacency matrix A is a simple eigenvalue of A with an eigenvector with all components positive. Hence the regularity of a connected strongly regular graph X is a simple eigenvalue of the adjacency matrix of X $.$ Mathematical equation: $.$

Finally, since from now on, we only consider primitive strongly regular graphs, we note that a (n, k; λ, μ)-strongly regular graph is imprimitive if and only if μ = 0 or μ = k. From now on, we only consider (n, k; λ, μ)- strongly regular graphs with k − 1 ≥ μ > 0. The multiplicities of the eigenvalues θ and τ of a primitive strongly regular graph X of order n satisfy the conditions (10) known has the absolute bounds

$\frac{m_{θ} (m_{θ} + 3)}{2} \geq n$ $Mathematical equation: $$ \frac{{m}_{\theta }({m}_{\theta }+3)}{2}\ge n $$$ (9)

$\frac{m_{τ} (m_{τ} + 3)}{2} \geq n .$ $Mathematical equation: $$ \frac{{m}_{\tau }({m}_{\tau }+3)}{2}\ge n. $$$ (10)and they also satisfy the equalities $1 + m_{θ} + m_{τ} = n$ $Mathematical equation: $ 1+{m}_{\theta }+{m}_{\tau }=n$$ and $k + m_{θ} θ + m_{τ} τ = 0 .$ $Mathematical equation: $ k+{m}_{\theta }\theta +{m}_{\tau }\tau =0.$$

Some relations on the parameters of a strongly regular graph

Let m be a natural number. We denote the set of real matrices of order m by $M_{m} (R)$ $Mathematical equation: $ {M}_m(\mathbb{R})$$ and the set of symmetric of $M_{m} (R)$ $Mathematical equation: $ {M}_m(\mathbb{R})$$ by $Sym (m, R)$ $Mathematical equation: $ \mathrm{Sym}\enspace (m,\mathbb{R})$$ . For any two matrices $H = [h_{ij}]$ $Mathematical equation: $ H=[{h}_{{ij}}]$$ and $L = [l_{ij}]$ $Mathematical equation: $ L=[{l}_{{ij}}]$$ of $M_{m} (R) .$ $Mathematical equation: $ {M}_m(\mathbb{R}).$$ we define the Hadamard product of H and L as being the matrix $H \circ L = [h_{ij} l_{ij}]$ $Mathematical equation: $ H\circ L=[{h}_{{ij}}{l}_{{ij}}]$$ and the Kronecker product of matrices H and L as being the matrix $H \otimes L = [h_{ij} L] .$ $Mathematical equation: $ H\otimes L=[{h}_{{ij}}L].$$ For any matrix P of $M_{m} (R)$ $Mathematical equation: $ {M}_m(\mathbb{R})$$ and for any nonnegative integer number j we define the Schur (Hadamard) power of order j of P, as being the matrix $P^{j \circ}$ $Mathematical equation: $ {P}^{j\circ }$$ in the following way: $P^{0 \circ} = J_{n}, P^{1 \circ} = P$ $Mathematical equation: $ {P}^{0\circ }={J}_n,{P}^{1\circ }=P$$ for any natural number j ≥ 2 we define $P^{j + 1 \circ} = P \circ P^{j \circ} .$ $Mathematical equation: $ {P}^{j+1\circ }=P\circ {P}^{j\circ }.$$

In this section we will establish some inequalities over the parameters and over the spectra of a primitive strongly regular graph.

Let’s consider a primitive (n, k; λ, μ)-strongly regular graph G such that $\frac{n}{2} > k > μ > 0$ $Mathematical equation: $ \frac{n}{2}>k>\mu >0$$ and with μ < λ, and let A be its adjacency matrix with the distinct eigenvalues τ, θ and k $.$ Mathematical equation: $.$ Now, we consider the Euclidean Jordan algebra $A = Sym (n, R)$ $Mathematical equation: $ \mathcal{A}=\enspace \mathrm{Sym}\enspace (n,\mathbb{R})$$ with the Jordan product $x ⋄ y = \frac{xy + yx}{2}$ $Mathematical equation: $ x\diamond y=\frac{{xy}+{yx}}{2}$$ and with the inner product $x | y = trace (x ⋄ y),$ $Mathematical equation: $ x|y=\mathrm{trace}\enspace (x\diamond y),$$ where xy and yx are the usual products of x by y and the usual product of y by x. Now we consider the Euclidean Jordan subalgebra $A$ $Mathematical equation: $ \mathcal{A}$$ of $Sym (n, R)$ $Mathematical equation: $ \mathrm{Sym}\enspace (n,\mathbb{R})$$ spanned by I _n and the natural powers of A. We have that $rank (A) = 3$ $Mathematical equation: $ \mathrm{rank}\enspace (A)=3$$ since has three distinct eigenvalues and is a three dimensional real Euclidean Jordan algebra. Let $B = {E_{1}, E_{2}, E_{3}}$ $Mathematical equation: $ \mathcal{B}=\{{E}_1,{E}_2,{E}_3\}$$ be the unique Jordan frame of $A$ $Mathematical equation: $ \mathcal{A}$$ associated to A, where

$E_{1} = \frac{1}{n} I_{n} + \frac{1}{n} A + \frac{1}{n} (J_{n} - A - I_{n}),$ $Mathematical equation: $$ {E}_1=\frac{1}{n}{I}_n+\frac{1}{n}A+\frac{1}{n}\left({J}_n-A-{I}_n\right), $$$

$E_{2} = \frac{| τ | n + τ - k}{n (θ - τ)} I_{n} + \frac{n + τ - k}{n (θ - τ)} A + \frac{τ - k}{n (θ - τ)} (J_{n} - A - I_{n}),$ $Mathematical equation: $$ {E}_2=\frac{|\tau |n+\tau -k}{n(\theta -\tau )}{I}_n+\frac{n+\tau -k}{n(\theta -\tau )}A+\frac{\tau -k}{n(\theta -\tau )}({J}_n-A-{I}_n), $$$

$E_{3} = \frac{θ n + k - θ}{n (θ - τ)} I_{n} + \frac{- n + k - θ}{n (θ - τ)} A + \frac{k - θ}{n (θ - τ)} (J_{n} - A - I_{n}) .$ $Mathematical equation: $$ {E}_3=\frac{\theta n+k-\theta }{n\left(\theta -\tau \right)}{I}_n+\frac{-n+k-\theta }{n\left(\theta -\tau \right)}A+\frac{k-\theta }{n\left(\theta -\tau \right)}\left({J}_n-A-{I}_n\right). $$$

Let’s consider the real positive number x such that x ≤ 1, and let’s consider the binomial Hadamard series $S_{z} = \sum_{l = 0}^{+ \infty} (- 1)^{l} (\begin{matrix} - z \\ l \end{matrix}) {(\frac{(A^{2} - θ^{2} I_{n})^{2 \circ}}{k^{4}})}^{l \circ} .$ $Mathematical equation: $ {S}_z={\sum }_{l=0}^{+\mathrm{\infty }} (-1{)}^l\left(\begin{array}{c}-z\\ l\end{array}\right){\left(\frac{({A}^2-{\theta }^2{I}_n{)}^{2\circ }}{{k}^4}\right)}^{l\circ }.$$ The second spectral decomposition of S _z relatively to the Jordan frame $B$ $Mathematical equation: $ \mathcal{B}$$ is $S_{z} = \sum_{i = 1}^{3} q_{iz} E_{i} .$ $Mathematical equation: $ {S}_z={\sum }_{i=1}^3 {q}_{{iz}}{E}_i.$$ Now, we show that the eigenvalues q _iz of S _z are positive.

Since $(- 1)^{l} (\begin{matrix} - z \\ l \end{matrix}) = (- 1)^{l} \frac{(- z) (- z - 1) (- z - 2) \dots (- z - l + 1)}{l!}$ $Mathematical equation: $ (-1{)}^l\left(\begin{array}{c}-z\\ l\end{array}\right)=(-1{)}^l\frac{(-z)(-z-1)(-z-2)\cdots (-z-l+1)}{l!}$$ then

$(- 1)^{l} (\begin{matrix} - z \\ l \end{matrix}) = (- 1)^{2 l} \frac{(z) (z + 1) (z + 2) \dots (z + l - 1)}{l!} \geq 0 .$ $Mathematical equation: $$ (-1{)}^l\left(\begin{array}{c}-z\\ l\end{array}\right)=(-1{)}^{2l}\frac{(z)(z+1)(z+2)\cdots (z+l-1)}{l!}\ge 0. $$$

Now, we have $S_{nz} = \sum_{l = 0}^{n} (- 1)^{l} (\begin{matrix} - z \\ l \end{matrix}) {(\frac{(A^{2} - θ^{2} I_{n})^{2 \circ}}{k^{4}})}^{l \circ} .$ $Mathematical equation: $ {S}_{{nz}}={\sum }_{l=0}^n (-1{)}^l\left(\begin{array}{c}-z\\ l\end{array}\right){\left(\frac{({A}^2-{\theta }^2{I}_n{)}^{2\circ }}{{k}^4}\right)}^{l\circ }.$$ Since $A^{2} = k I_{n} + λ A + μ (J_{n} - A - I_{n})$ $Mathematical equation: $ {A}^2=k{I}_n+\lambda A+\mu ({J}_n-A-{I}_n)$$ then we conclude that

$\begin{matrix} \frac{(A^{2} - θ^{2} I_{n})^{2 \circ}}{k^{4}} = \frac{(k I_{n} + λ A + μ (J_{n} - A - I_{n}) - θ^{2} I_{n})^{2 \circ}}{k^{4}} \\ = \frac{((k - θ^{2}) I_{n} + λ A + μ (J_{n} - A - I_{n}))^{2 \circ}}{k^{4}} \\ \begin{matrix} = \frac{(k - θ^{2})^{2} I_{n} + λ^{2} A + μ^{2} (J_{n} - A - I_{n})}{k^{4}} \\ = \frac{(k - θ^{2})^{2}}{k^{4}} I_{n} + \frac{λ^{2}}{k^{4}} A + \frac{μ^{2}}{k^{4}} (J_{n} - A - I_{n}) . \end{matrix} \end{matrix}$ $Mathematical equation: $$ \begin{array}{c}\frac{({A}^2-{\theta }^2{I}_n{)}^{2\circ }}{{k}^4}=\frac{(k{I}_n+\lambda A+\mu ({J}_n-A-{I}_n)-{\theta }^2{I}_n{)}^{2\circ }}{{k}^4}\\ =\frac{((k-{\theta }^2){I}_n+\lambda A+\mu ({J}_n-A-{I}_n){)}^{2\circ }}{{k}^4}\\ \begin{array}{c}=\frac{(k-{\theta }^2{)}^2{I}_n+{\lambda }^2A+{\mu }^2({J}_n-A-{I}_n)}{{k}^4}\\ =\frac{(k-{\theta }^2{)}^2}{{k}^4}{I}_n+\frac{{\lambda }^2}{{k}^4}A+\frac{{\mu }^2}{{k}^4}({J}_{\mathrm{n}}-A-{I}_n).\end{array}\end{array} $$$

Let’s consider the second spectral decomposition $S_{nz} = q_{n 1 z} E_{1} + q_{n 2 z} E_{2} + q_{n 3 z} E_{3} .$ $Mathematical equation: $ {S}_{{nz}}={q}_{n1z}{E}_1+{q}_{n2z}{E}_2+{q}_{n3z}{E}_3.$$ Since λ > μ then we have that $| τ | < θ$ $Mathematical equation: $ |\tau | < \theta $$ and therefore $| τ |^{2} < θ^{2}$ $Mathematical equation: $ |\tau {|}^2 < {\theta }^2$$ and since $θ^{2} \leq k^{2}$ $Mathematical equation: $ {\theta }^2\le {k}^2$$ then the eigenvalues of $\frac{A^{2} - θ^{2} I_{n}}{k^{4}}$ $Mathematical equation: $ \frac{{A}^2-{\theta }^2{I}_n}{{k}^4}$$ are positive. Since for any two matrices C and D of $M_{n} (R)$ $Mathematical equation: $ {M}_n(\mathbb{R})$$ we have $λ_{\min} (C) λ_{\min} (D) \leq λ_{\min} (C \circ D)$ $Mathematical equation: $ {\lambda }_{\mathrm{min}}(C){\lambda }_{\mathrm{min}}(D)\le {\lambda }_{\mathrm{min}}(C\circ D)$$ and since B is a Jordan frame of $A$ $Mathematical equation: $ \mathcal{A}$$ that is a basis of $A$ $Mathematical equation: $ \mathcal{A}$$ and $A$ $Mathematical equation: $ \mathcal{A}$$ is closed for the Schur product of matrices we deduce that the eigenvalues of ${(\frac{(A^{2} - θ^{2} I_{n})^{2 \circ}}{k^{4}})}^{l \circ}$ $Mathematical equation: $ {\left(\frac{({A}^2-{\theta }^2{I}_n{)}^{2\circ }}{{k}^4}\right)}^{l\circ }$$ are positive. So we conclude that the eigenvalues q _niz for i = 1, ⋯, 3 of S _nz are all positive.

Since $q_{1 z} = \lim_{n \to + \infty} q_{n 1 z}, q_{2 z} = \lim_{n \to + \infty} q_{n 2 z}, q_{3 z} = \lim_{n \to + \infty} q_{n 3 z}$ $Mathematical equation: $ {q}_{1z}={\mathrm{lim}}_{n\to +\mathrm{\infty }}{q}_{n1z},\hspace{0.5em}{q}_{2z}={\mathrm{lim}}_{n\to +\mathrm{\infty }}{q}_{n2z},\hspace{0.5em}{q}_{3z}={\mathrm{lim}}_{n\to +\mathrm{\infty }}{q}_{n3z}$$ then we have $q_{1 z} \geq 0, q_{2 z} \geq 0$ $Mathematical equation: $ {q}_{1z}\ge 0,{q}_{2z}\ge 0$$ and $q_{3 z} \geq 0 .$ $Mathematical equation: $ {q}_{3z}\ge 0.$$ We must say that $S_{z} E_{1} = q_{1 z} E_{1}, S_{z} E_{2} = q_{2 z} E_{2}$ $Mathematical equation: $ {S}_z{E}_1={q}_{1z}{E}_1,{S}_z{E}_2={q}_{2z}{E}_2$$ and $S_{z} E_{3} = q_{3 z} E_{3},$ $Mathematical equation: $ {S}_z{E}_3={q}_{3z}{E}_3,$$ hence we have: $q_{1 z} = \frac{1}{(\frac{k^{4} - (k - θ^{2})^{2}}{k^{4}})^{z}} + \frac{1}{(\frac{k^{4} - λ^{2}}{k^{4}})^{z}} k + \frac{1}{(\frac{k^{4} - μ^{2}}{k^{4}})^{z}} (n - k - 1),$ $Mathematical equation: $ {q}_{1z}=\frac{1}{(\frac{{k}^4-(k-{\theta }^2{)}^2}{{k}^4}{)}^z}+\frac{1}{(\frac{{k}^4-{\lambda }^2}{{k}^4}{)}^z}k+\frac{1}{(\frac{{k}^4-{\mu }^2}{{k}^4}{)}^z}(n-k-1),$$ $q_{2 z} = \frac{1}{(\frac{k^{4} - (k - θ^{2})^{2}}{k^{4}})^{z}} + \frac{1}{(\frac{k^{4} - λ^{2}}{k^{4}})^{z}} θ + \frac{1}{(\frac{k^{4} - μ^{2}}{k^{4}})^{z}} (- θ - 1)$ $Mathematical equation: $ {q}_{2z}=\frac{1}{(\frac{{k}^4-(k-{\theta }^2{)}^2}{{k}^4}{)}^z}+\frac{1}{(\frac{{k}^4-{\lambda }^2}{{k}^4}{)}^z}\theta +\frac{1}{(\frac{{k}^4-{\mu }^2}{{k}^4}{)}^z}(-\theta -1)$$ and $q_{3 z} = \frac{1}{(\frac{k^{4} - (k - θ^{2})^{2}}{k^{4}})^{z}} + \frac{1}{(\frac{k^{4} - λ^{2}}{k^{4}})^{z}} τ + \frac{1}{(\frac{k^{4} - μ^{2}}{k^{4}})^{z}} (- τ - 1) .$ $Mathematical equation: $ {q}_{3z}=\frac{1}{(\frac{{k}^4-(k-{\theta }^2{)}^2}{{k}^4}{)}^z}+\frac{1}{(\frac{{k}^4-{\lambda }^2}{{k}^4}{)}^z}\tau +\frac{1}{(\frac{{k}^4-{\mu }^2}{{k}^4}{)}^z}(-\tau -1).$$

Let’s consider the element $S_{3 z} = E_{3} \circ S_{z}$ $Mathematical equation: $ {S}_{3z}={E}_3\circ {S}_z$$ of $A .$ $Mathematical equation: $ \mathcal{A}.$$ Since the eigenvalues of E ₃ and of S _z are positive and since $λ_{\min} (E_{3}) λ_{\min} (S_{z}) \leq λ_{\min} (E_{3} \circ S_{x})$ $Mathematical equation: $ {\lambda }_{\mathrm{min}}({E}_3){\lambda }_{\mathrm{min}}({S}_z)\le {\lambda }_{\mathrm{min}}({E}_3\circ {S}_x)$$ then the eigenvalues of $E_{3} \circ S_{z}$ $Mathematical equation: $ {E}_3\circ {S}_z$$ are positive. Now since $k < \frac{n}{2}$ $Mathematical equation: $ k < \frac{n}{2}$$ and λ > μ and by an asymptotical analysis of the spectrum of $E_{3} \circ S_{z}$ $Mathematical equation: $ {E}_3\circ {S}_z$$ we will deduce the inequalities (16) and (21) of the Theorems 3 and 4 respectively are verified.

Now, we consider the second spectral decomposition $E_{3} \circ S_{z} = q_{3 z}^{1} E_{1} + q_{3 z}^{2} E_{2} + q_{3 z}^{3} E_{3} .$ $Mathematical equation: $ {E}_3\circ {S}_z={q}_{3z}^1{E}_1+{q}_{3z}^2{E}_2+{q}_{3z}^3{E}_3.$$ Then, we have

$q_{3 z}^{1} = \frac{θ n + k - θ}{n (θ - τ)} \frac{1}{{(\frac{k^{4} - (k - θ^{2})^{2}}{k^{4}})}^{z}} + \frac{- n + k - θ}{n (θ - τ)} \frac{1}{{(\frac{k^{4} - λ^{2}}{k^{4}})}^{z}} k + \frac{k - θ}{n (θ - τ)} \frac{1}{{(\frac{k^{4} - μ^{2}}{k^{4}})}^{z}} (n - k - 1),$ $Mathematical equation: $$ {q}_{3z}^1=\frac{\theta n+k-\theta }{n(\theta -\tau )}\frac{1}{{\left(\frac{{k}^4-(k-{\theta }^2{)}^2}{{k}^4}\right)}^z}+\frac{-n+k-\theta }{n(\theta -\tau )}\frac{1}{{\left(\frac{{k}^4-{\lambda }^2}{{k}^4}\right)}^z}k+\frac{k-\theta }{n\left(\theta -\tau \right)}\frac{1}{{\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)}^z}\left(n-k-1\right), $$$ (11)

$q_{3 z}^{2} = \frac{θ n + k - θ}{n (θ - τ)} \frac{1}{{(\frac{k^{4} - (k - θ^{2})^{2}}{k^{4}})}^{z}} + \frac{- n + k - θ}{n (θ - τ)} \frac{1}{{(\frac{k^{4} - λ^{2}}{k^{4}})}^{z}} τ + \frac{k - θ}{n (θ - τ)} \frac{1}{{(\frac{k^{4} - μ^{2}}{k^{4}})}^{z}} (- τ - 1),$ $Mathematical equation: $$ {q}_{3z}^2=\frac{\theta n+k-\theta }{n\left(\theta -\tau \right)}\frac{1}{{\left(\frac{{k}^4-(k-{\theta }^2{)}^2}{{k}^4}\right)}^z}+\frac{-n+k-\theta }{n\left(\theta -\tau \right)}\frac{1}{{\left(\frac{{k}^4-{\lambda }^2}{{k}^4}\right)}^z}\tau +\frac{k-\theta }{n(\theta -\tau )}\frac{1}{{\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)}^z}(-\tau -1), $$$ (12)

$q_{3 z}^{3} = \frac{θ n + k - θ}{n (θ - τ)} \frac{1}{{(\frac{k^{4} - (k - θ^{2})^{2}}{k^{4}})}^{z}} + \frac{- n + k - θ}{n (θ - τ)} \frac{1}{{(\frac{k^{4} - λ^{2}}{k^{4}})}^{z}} θ + \frac{k - θ}{n (θ - τ)} \frac{1}{{(\frac{k^{4} - μ^{2}}{k^{4}})}^{z}} (- θ - 1) .$ $Mathematical equation: $$ {q}_{3z}^3=\frac{\theta n+k-\theta }{n(\theta -\tau )}\frac{1}{{\left(\frac{{k}^4-(k-{\theta }^2{)}^2}{{k}^4}\right)}^z}+\frac{-n+k-\theta }{n(\theta -\tau )}\frac{1}{{\left(\frac{{k}^4-{\lambda }^2}{{k}^4}\right)}^z}\theta +\frac{k-\theta }{n(\theta -\tau )}\frac{1}{{\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)}^z}(-\theta -1). $$$ (13)

Now, since $\frac{θ n + k - θ}{n (θ - τ)} + \frac{- n + k - θ}{n (θ - τ)} k + \frac{k - θ}{n (θ - τ)} (n - k - 1) = 0$ $Mathematical equation: $ \frac{\theta n+k-\theta }{n(\theta -\tau )}+\frac{-n+k-\theta }{n(\theta -\tau )}k+\frac{k-\theta }{n(\theta -\tau )}(n-k-1)=0$$ then $\frac{k - θ}{n (θ - τ)} (n - k - 1) = - \frac{θ n + k - θ}{n (θ - τ)} - \frac{- n + k - θ}{n (θ - τ)} k$ $Mathematical equation: $ \frac{k-\theta }{n(\theta -\tau )}(n-k-1)=-\frac{\theta n+k-\theta }{n(\theta -\tau )}-\frac{-n+k-\theta }{n(\theta -\tau )}k$$ and consequently the parameter $q_{3 z}^{1}$ $Mathematical equation: $ {q}_{3z}^1$$ takes the form:

$q_{3 z}^{1} = \frac{θ n + k - θ}{n (θ - τ)} (\frac{{(\frac{k^{4} - μ^{2}}{k^{4}})}^{z} - {(\frac{k^{4} - (k - θ^{2})^{2}}{k^{4}})}^{z}}{{(1 - \frac{(k - θ^{2})^{2}}{k^{4}})}^{z} {(1 - \frac{μ^{2}}{k^{4}})}^{z}}) + \frac{- n + k - θ}{n (θ - τ)} (\frac{{(\frac{k^{4} - μ^{2}}{k^{4}})}^{z} - {(\frac{k^{4} - λ^{2}}{k^{4}})}^{z}}{{(1 - \frac{λ^{2}}{k^{4}})}^{z} {(1 - \frac{μ^{2}}{k^{4}})}^{z}} -) k,$ $Mathematical equation: $$ {q}_{3z}^1=\frac{\theta n+k-\theta }{n(\theta -\tau )}\left(\frac{{\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)}^z-{\left(\frac{{k}^4-(k-{\theta }^2{)}^2}{{k}^4}\right)}^z}{{\left(1-\frac{(k-{\theta }^2{)}^2}{{k}^4}\right)}^z{\left(1-\frac{{\mu }^2}{{k}^4}\right)}^z}\right)+\frac{-n+k-\theta }{n(\theta -\tau )}\left(\frac{{\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)}^z-{\left(\frac{{k}^4-{\lambda }^2}{{k}^4}\right)}^z}{{\left(1-\frac{{\lambda }^2}{{k}^4}\right)}^z{\left(1-\frac{{\mu }^2}{{k}^4}\right)}^z}-\right)k, $$$ (14)and $q_{3 z}^{3}$ $Mathematical equation: $ {q}_{3z}^3$$ takes the form:

$q_{3 z}^{3} = \frac{θ n + k - θ}{n (θ - τ)} (\frac{{(\frac{k^{4} - μ^{2}}{k^{4}})}^{z} - {(\frac{k^{4} - (k - θ^{2})^{2}}{k^{4}})}^{x}}{{(\frac{k^{4} - (k - θ^{2})^{2}}{k^{4}})}^{z} {(\frac{k^{4} - μ^{2}}{k^{4}})}^{z}}) + \frac{- n + k - θ}{n (θ - τ)} (\frac{{(\frac{k^{4} - μ^{2}}{k^{4}})}^{z} - {(\frac{k^{4} - λ^{2}}{k^{4}})}^{z}}{{(\frac{k^{4} - λ^{2}}{k^{4}})}^{z} {(\frac{k^{4} - μ^{2}}{k^{4}})}^{z}}) θ .$ $Mathematical equation: $$ {q}_{3z}^3=\frac{\theta n+k-\theta }{n(\theta -\tau )}\left(\frac{{\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)}^z-{\left(\frac{{k}^4-(k-{\theta }^2{)}^2}{{k}^4}\right)}^x}{{\left(\frac{{k}^4-(k-{\theta }^2{)}^2}{{k}^4}\right)}^z{\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)}^z}\right)+\frac{-n+k-\theta }{n(\theta -\tau )}\left(\frac{{\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)}^z-{\left(\frac{{k}^4-{\lambda }^2}{{k}^4}\right)}^z}{{\left(\frac{{k}^4-{\lambda }^2}{{k}^4}\right)}^z{\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)}^z}\right)\theta. $$$ (15)

Using the fact that $k < \frac{n}{2}$ $Mathematical equation: $ k < \frac{n}{2}$$ then we conclude that $\frac{n - k + θ}{θ n + k - θ} > \frac{1}{2 θ + 1} .$ $Mathematical equation: $ \frac{n-k+\theta }{\theta n+k-\theta }>\frac{1}{2\theta +1}.$$

Theorem 3. Let n, k, μ and λ be natural numbers with n − 1 > k > μ and X be a (n, k; λ, μ)-primitive strongly regular graph with distinct eigenvalues k, θ and τ $.$ Mathematical equation: $.$ If $k < \frac{n}{2}$ $Mathematical equation: $ k < \frac{n}{2}$$ and $λ > μ$ $Mathematical equation: $ \lambda >\mu $$ then

${(\frac{k^{4} - μ^{2}}{k^{4} - (k - θ^{2})^{2}})}^{2 θ + 1} > {(\frac{k^{4} - μ^{2}}{k^{4} - λ^{2}})}^{k}$ $Mathematical equation: $$ {\left(\frac{{k}^4-{\mu }^2}{{k}^4-(k-{\theta }^2{)}^2}\right)}^{2\theta +1}>{\left(\frac{{k}^4-{\mu }^2}{{k}^4-{\lambda }^2}\right)}^k $$$ (16) Proof. Since $q_{3 z}^{1} \geq 0$ $Mathematical equation: $ {q}_{3z}^1\ge 0$$ and recurring to the equality (14) we deduce the equality (17).

$\frac{θ n + k - θ}{n (θ - τ)} \cdot (\frac{{(\frac{k^{4} - μ^{2}}{k^{4}})}^{z} - {(\frac{k^{4} - (k - θ^{2})^{2}}{k^{4}})}^{z}}{{(\frac{k^{4} - (k - θ^{2})^{2}}{k^{4}})}^{z} {(\frac{k^{4} - μ^{2}}{k^{4}})}^{z}}) + \frac{- n + k - θ}{n (θ - τ)} k (\frac{{(\frac{k^{4} - μ^{2}}{k^{4}})}^{z} - {(\frac{k^{4} - λ^{2}}{k^{4}})}^{z}}{{(\frac{k^{4} - λ^{2}}{k^{4}})}^{z} {(\frac{k^{4} - μ^{2}}{k^{4}})}^{z}}) \geq 0 .$ $Mathematical equation: $$ \frac{\theta n+k-\theta }{n\left(\theta -\tau \right)}\cdot \left(\frac{{\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)}^z-{\left(\frac{{k}^4-(k-{\theta }^2{)}^2}{{k}^4}\right)}^z}{{\left(\frac{{k}^4-(k-{\theta }^2{)}^2}{{k}^4}\right)}^z{\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)}^z}\right)+\frac{-n+k-\theta }{n(\theta -\tau )}k\left(\frac{{\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)}^z-{\left(\frac{{k}^4-{\lambda }^2}{{k}^4}\right)}^z}{{\left(\frac{{k}^4-{\lambda }^2}{{k}^4}\right)}^z{\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)}^z}\right)\ge 0. $$$ (17)

Making, some algebraic manipulation of equality (17) we obtain the inequality (18).

$\frac{{(\frac{k^{4} - λ^{2}}{k^{4}})}^{z}}{{(\frac{k^{4} - (k - θ^{2})^{2}}{k^{4}})}^{z}} \geq \frac{n - k + θ}{θ n + k - θ} k (\frac{{(\frac{k^{4} - μ^{2}}{k^{4}})}^{z} - {(\frac{k^{4} - λ^{2}}{k^{4}})}^{z}}{{(\frac{k^{4} - μ^{2}}{k^{4}})}^{z} - {(\frac{k^{4} - (k - θ^{2})^{2}}{k^{4}})}^{z}}) .$ $Mathematical equation: $$ \frac{{\left(\frac{{k}^4-{\lambda }^2}{{k}^4}\right)}^z}{{\left(\frac{{k}^4-(k-{\theta }^2{)}^2}{{k}^4}\right)}^z}\ge \frac{n-k+\theta }{\theta n+k-\theta }k\left(\frac{{\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)}^z-{\left(\frac{{k}^4-{\lambda }^2}{{k}^4}\right)}^z}{{\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)}^z-{\left(\frac{{k}^4-(k-{\theta }^2{)}^2}{{k}^4}\right)}^z}\right). $$$ (18)

Calculating the limits of the expressions of both hand sides of (18) when x approaches zero we obtain the equality (19).

$1 \geq \frac{n - k + θ}{θ n + k - θ} k (\frac{\ln (\frac{k^{4} - μ^{2}}{k^{4}}) - \ln (\frac{k^{4} - λ^{2}}{k^{4}})}{\ln (\frac{k^{4} - μ^{2}}{k^{4}}) - \ln (\frac{k^{4} - (k - θ^{2})^{2}}{k^{4}})}) .$ $Mathematical equation: $$ 1\ge \frac{n-k+\theta }{\theta n+k-\theta }k\left(\frac{\mathrm{ln}\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)-\mathrm{ln}\left(\frac{{k}^4-{\lambda }^2}{{k}^4}\right)}{\mathrm{ln}\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)-\mathrm{ln}\left(\frac{{k}^4-(k-{\theta }^2{)}^2}{{k}^4}\right)}\right). $$$ (19)

Now since when $k < \frac{n}{2}$ $Mathematical equation: $ k < \frac{n}{2}$$ we have $\frac{n - k + θ}{θ n + k - θ} > \frac{1}{2 θ + 1}$ $Mathematical equation: $ \frac{n-k+\theta }{\theta n+k-\theta }>\frac{1}{2\theta +1}$$ and after some algebraic manipulation of the inequality (19) we obtain the inequality (20).

$1 > \frac{k}{2 θ + 1} (\frac{\ln (\frac{k^{4} - μ^{2}}{k^{4} - λ^{2}})}{\ln (\frac{k^{4} - μ^{2}}{k^{4} - (k - θ^{2})^{2}})}) .$ $Mathematical equation: $$ 1>\frac{k}{2\theta +1}\left(\frac{\mathrm{ln}\left(\frac{{k}^4-{\mu }^2}{{k}^4-{\lambda }^2}\right)}{\mathrm{ln}\left(\frac{{k}^4-{\mu }^2}{{k}^4-(k-{\theta }^2{)}^2}\right)}\right). $$$ (20)

And, finally we conclude that ${(\frac{k^{4} - μ^{2}}{k^{4} - (k - θ^{2})^{2}})}^{2 θ + 1} > {(\frac{k^{4} - μ^{2}}{k^{4} - λ^{2}})}^{k} .$ $Mathematical equation: $ {\left(\frac{{k}^4-{\mu }^2}{{k}^4-(k-{\theta }^2{)}^2}\right)}^{2\theta +1}>{\left(\frac{{k}^4-{\mu }^2}{{k}^4-{\lambda }^2}\right)}^k.$$

Theorem 4. Let n, k, μ and λ be natural numbers and let X be a (n, k; λ, μ)-primitive strongly regular graph with $n - 1 > k > μ, \frac{n}{2} > k, μ < λ$ $Mathematical equation: $ n-1>k>\mu,\frac{n}{2}>k,\mu < \lambda $$ and with the distinct eigenvalues θ, τ and k $.$ Mathematical equation: $.$ Then

${(\frac{k^{4} - μ^{2}}{k^{4} - (k - θ^{2})^{2}})}^{2 θ + 1} > {(\frac{k^{4} - μ^{2}}{k^{4} - λ^{2}})}^{θ} .$ $Mathematical equation: $$ {\left(\frac{{k}^4-{\mu }^2}{{k}^4-(k-{\theta }^2{)}^2}\right)}^{2\theta +1}>{\left(\frac{{k}^4-{\mu }^2}{{k}^4-{\lambda }^2}\right)}^{\theta }. $$$ (21) Proof. Now since $q_{3 z}^{3} \geq 0$ $Mathematical equation: $ {q}_{3z}^3\ge 0$$ and recurring to the equality (15) we obtain that

$\frac{θ n + k - θ}{n (θ - τ)} \cdot (\frac{{(\frac{k^{4} - μ^{2}}{k^{4}})}^{z} - {(\frac{k^{4} - (k - θ^{2})^{2}}{k^{4}})}^{z}}{{(\frac{k^{4} - (k - θ^{2})^{2}}{k^{4}})}^{z} {(\frac{k^{4} - μ^{2}}{k^{4}})}^{z}}) + \frac{- n + k - θ}{n (θ - τ)} (\frac{{(\frac{k^{4} - μ^{2}}{k^{4}})}^{z} - {(\frac{k^{4} - λ^{2}}{k^{4}})}^{z}}{{(\frac{k^{4} - λ^{2}}{k^{4}})}^{z} {(\frac{k^{4} - μ^{2}}{k^{4}})}^{z}}) θ \geq 0 .$ $Mathematical equation: $$ \frac{\theta n+k-\theta }{n(\theta -\tau )}\cdot \left(\frac{{\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)}^z-{\left(\frac{{k}^4-(k-{\theta }^2{)}^2}{{k}^4}\right)}^z}{{\left(\frac{{k}^4-(k-{\theta }^2{)}^2}{{k}^4}\right)}^z{\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)}^z}\right)+\frac{-n+k-\theta }{n(\theta -\tau )}\left(\frac{{\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)}^z-{\left(\frac{{k}^4-{\lambda }^2}{{k}^4}\right)}^z}{{\left(\frac{{k}^4-{\lambda }^2}{{k}^4}\right)}^z{\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)}^z}\right)\theta \ge 0. $$$ (22)

From (22) we deduce the inequality (23).

$\frac{{(\frac{k^{4} - λ^{2}}{k^{4}})}^{z}}{{(\frac{k^{4} - (k - θ^{2})^{2}}{k^{4}})}^{z}} \geq \frac{n - k + θ}{θ n + k - θ} (\frac{{(\frac{k^{4} - μ^{2}}{k^{4}})}^{z} - {(\frac{k^{4} - λ^{2}}{k^{4}})}^{z}}{{(\frac{k^{4} - μ^{2}}{k^{4}})}^{z} - {(\frac{k^{4} - (k - θ^{2})^{2}}{k^{4}})}^{z}}) θ .$ $Mathematical equation: $$ \frac{{\left(\frac{{k}^4-{\lambda }^2}{{k}^4}\right)}^z}{{\left(\frac{{k}^4-(k-{\theta }^2{)}^2}{{k}^4}\right)}^z}\ge \frac{n-k+\theta }{\theta n+k-\theta }\left(\frac{{\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)}^z-{\left(\frac{{k}^4-{\lambda }^2}{{k}^4}\right)}^z}{{\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)}^z-{\left(\frac{{k}^4-(k-{\theta }^2{)}^2}{{k}^4}\right)}^z}\right)\theta. $$$ (23)

Calculating the limits of the expressions of both hand sides of (23) when x approaches zero we obtain the equality (24).

$1 \geq \frac{n - k + θ}{θ n + k - θ} θ (\frac{\ln (\frac{k^{4} - μ^{2}}{k^{4}}) - \ln (1 - \frac{k^{4} - λ^{2}}{k^{4}})}{\ln (\frac{k^{4} - μ^{2}}{k^{4}}) - \ln (\frac{k^{4} - (k - θ^{2})^{2}}{k^{4}})}) .$ $Mathematical equation: $$ 1\ge \frac{n-k+\theta }{\theta n+k-\theta }\theta \left(\frac{\mathrm{ln}\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)-\mathrm{ln}\left(1-\frac{{k}^4-{\lambda }^2}{{k}^4}\right)}{\mathrm{ln}\left(\frac{{k}^4-{\mu }^2}{{k}^4}\right)-\mathrm{ln}\left(\frac{{k}^4-(k-{\theta }^2{)}^2}{{k}^4}\right)}\right). $$$ (24)

Now, since when $k < \frac{n}{2}$ $Mathematical equation: $ k < \frac{n}{2}$$ we have $\frac{n - k + θ}{θ n + k - θ} > \frac{1}{2 θ + 1}$ $Mathematical equation: $ \frac{n-k+\theta }{\theta n+k-\theta }>\frac{1}{2\theta +1}$$ and after some algebraic manipulation of the inequality (24) we obtain the inequality (25).

$1 > \frac{θ}{2 θ + 1} (\frac{\ln (\frac{k^{4} - μ^{2}}{k^{4} - λ^{2}})}{\ln (\frac{k^{4} - μ^{2}}{k^{4} - (k - θ^{2})^{2}})}) .$ $Mathematical equation: $$ 1>\frac{\theta }{2\mathrm{\theta }+1}\left(\frac{\mathrm{ln}\left(\frac{{k}^4-{\mu }^2}{{k}^4-{\lambda }^2}\right)}{\mathrm{ln}\left(\frac{{k}^4-{\mu }^2}{{k}^4-(k-{\theta }^2{)}^2}\right)}\right). $$$ (25)And, finally we conclude that ${(\frac{k^{4} - μ^{2}}{k^{4} - (k - θ^{2})^{2}})}^{2 θ + 1} > {(\frac{k^{4} - μ^{2}}{k^{4} - λ^{2}})}^{θ} .$ $Mathematical equation: $ {\left(\frac{{k}^4-{\mu }^2}{{k}^4-(k-{\theta }^2{)}^2}\right)}^{2\theta +1}>{\left(\frac{{k}^4-{\mu }^2}{{k}^4-{\lambda }^2}\right)}^{\theta }.$$

Preliminares on quaternions and octonions

This section is a brief introduction on quaternions and octonions. Good readable texts on this algebraic structures can be found on the works [37, 38]. Now, we consider the real linear space $A$ Mathematical equation: $ A$ of quaternions spanned by the basis $B = {1, i, j, k}$ $Mathematical equation: $ B=\{1,i,j,k\}$$ , where the elements of B verify the following rules of multiplication $i^{2} = j^{2} = k^{2} = - 1$ $Mathematical equation: $ {i}^2={j}^2={k}^2=-1$$ and

$ij = - ji = k;$
$jk = - kj = i;$
$ki = - ik = j .$

So, we can write

A = {α_{0} 1 + α_{1} i + α_{2} j + α_{3} k, α_{0}, α_{1}, α_{2}, α_{3} \in R} .

$Mathematical equation: $ \mathcal{A}=\{{\alpha }_01+{\alpha }_1i+{\alpha }_2j+{\alpha }_3k,{\alpha }_0,{\alpha }_1,{\alpha }_2,{\alpha }_3\in \mathbb{R}\}.$$ Given a quaternion

x = x_{0} 1 + x_{1} i + x_{2} j + x_{3} k

$Mathematical equation: $ x={x}_01+{x}_1i+{x}_2j+{x}_3k$$ we call to x ₀ the real part of x we denote it by Re (x) and we call to

x_{1} i + x_{2} j + x_{3} k

$Mathematical equation: $ {x}_1i+{x}_2j+{x}_3k$$ the imaginary part of x and we write

Im (x) = x_{1} i + x_{2} j + x_{3} k .

$Mathematical equation: $ \mathrm{Im}\enspace (x)={x}_1i+{x}_2j+{x}_3k.$$

One says that a quaternion x is a pure quaternion if Re (x)=0 and if x = Re (x) one says that the quaternion x is a real number.

For discovering the equalities for multiplication describe above we must use the diagram of Fano, see Figure 2.

Figure 2

Diagram of Fano for quaternions.

When we multiply to elements of the set ${i, j, k}$ $Mathematical equation: $ \{i,j,k\}$$ we use the rule: when we multiply two elements in clockwise sense we get the next element, so for instance we have jk = i, but if we multiply them in the counterclockwise sense we obtain the next but with minus sign kj = −i.

And, therefore we obtain Table 1 of multiplication of two elements of A.

Table 1

Table of multiplication of quaternions.

If we consider $x = x_{0} 1 + x_{1} i + x_{2} j + x_{3} k$ $Mathematical equation: $ x={x}_01+{x}_1i+{x}_2j+{x}_3k$$ and $y = y_{0} 1 + y_{1} i + y_{2} j + y_{3} k$ $Mathematical equation: $ y={y}_01+{y}_1i+{y}_2j+{y}_3k$$ we obtain the following expression for the product x $*$ $Mathematical equation: $ \mathrm{*}$$ y,

$\begin{array}{l} x * y = x_{0} y_{0} - (x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3}) + x_{0} (y_{1} i + y_{2} j + y_{3} k) + y_{0} (x_{1} i + x_{2} j + x_{3} k) + (x_{2} y_{3} - x_{3} y_{2}) i + (x_{3} y_{1} - x_{1} y_{3}) j + (x_{1} y_{2} - x_{2} y_{1}) k . \end{array}$ $Mathematical equation: $$ \begin{array}{l}x\mathrm{*}y={x}_0{y}_0-({x}_1{y}_1+{x}_2{y}_2+{x}_3{y}_3)+{x}_0({y}_1i+{y}_2j+{y}_3k)+{y}_0({x}_1i+{x}_2j+{x}_3k)+({x}_2{y}_3-{x}_3{y}_2)i+({x}_3{y}_1-{x}_1{y}_3)j+({x}_1{y}_2-{x}_2{y}_1)\mathrm{k}.\\ \\ \end{array} $$$

Using the notation $Re (x) = x_{0}$ $Mathematical equation: $ \mathrm{Re}\enspace (x)={x}_0$$ and $Im (x) = x_{1} i + x_{2} j + x_{3} k$ $Mathematical equation: $ \mathrm{Im}\enspace (x)={x}_1i+{x}_2j+{x}_3k$$ we conclude that

$\begin{array}{l} x * y = Re (x) Re (y) - Im (x) | Im (y) + Re (x) Im (y) + Re (y) Im (x) + Im (x) \times Im (y) . \end{array}$ $Mathematical equation: $$ \begin{array}{l}x\mathrm{*}y=\enspace \mathrm{Re}\enspace (x)\enspace \mathrm{Re}\enspace (y)-\enspace \mathrm{Im}\enspace (x)|\enspace \mathrm{Im}\enspace (y)+\enspace \mathrm{Re}\enspace (x)\enspace \mathrm{Im}\enspace (y)+\enspace \mathrm{Re}\enspace (y)\enspace \mathrm{Im}\enspace (x)+\enspace \mathrm{Im}\enspace (x)\times \enspace \mathrm{Im}\enspace (y).\end{array} $$$

For a quaternion $x = x_{0} 1 + x_{1} i + x_{2} j + x_{3} k$ $Mathematical equation: $ x={x}_01+{x}_1i+{x}_2j+{x}_3k$$ we define we define $| | x | |$ Mathematical equation: $ ||x||$ by the the equality $| | x | | = \sqrt{x_{0}^{2} + x_{1}^{2} + x_{2}^{2} + x_{3}^{2}} .$ $Mathematical equation: $ ||x||=\sqrt{{x}_0^2+{x}_1^2+{x}_2^2+{x}_3^2}.$$ Now, we will show that $\overset{̅}{x * y} = \overset{̅}{y} * \overset{̅}{x} .$ $Mathematical equation: $ \overline{x\mathrm{*}y}=\bar{y}\mathrm{*}\bar{x}.$$ Hence, we have the following calculations:

$\begin{array}{l} \overset{̅}{x * y} = \overset{̅}{(Re (x) + Im (x)) * (Re (y) + Im (y))} \\ = \overset{̅}{Re (x) Re (y) - Im (x) | Im (y) + Re (x) Im (y) + Re (y) Im (x) + Im (x) \times Im (y)} \\ = Re (x) Re (y) - Re (x) Im (y) - Re (y) Im (x) - Im (x) | Im (y) - Im (x) \times Im (y) \\ = (Re (y) - Im (y)) * (Re (x) - Im (x)) \\ = (\overset{̅}{Re (y) + Im (y)}) * (\overset{̅}{Re (x) + Im (x)}) \\ = \overset{̅}{y} * \overset{̅}{x} . \end{array}$ $Mathematical equation: $$ \begin{array}{l}\overline{x\mathrm{*}y}=\overline{\left(\enspace \mathrm{Re}\enspace (x)+\enspace \mathrm{Im}\enspace (x)\right)\mathrm{*}\left(\mathrm{Re}\enspace (y)+\enspace \mathrm{Im}\enspace (y)\right)}\\ =\overline{\enspace \mathrm{Re}\enspace (x)\enspace \mathrm{Re}\enspace (y)-\enspace \mathrm{Im}\enspace (x)|\enspace \mathrm{Im}\enspace (y)+\enspace \mathrm{Re}\enspace (x)\enspace \mathrm{Im}\enspace (y)+\enspace \mathrm{Re}\enspace (y)\enspace \mathrm{Im}\enspace (x)+\enspace \mathrm{Im}\enspace (x)\times \enspace \mathrm{Im}\enspace (y)}\\ =\enspace \mathrm{Re}\enspace (x)\enspace \mathrm{Re}\enspace (y)-\enspace \mathrm{Re}\enspace (x)\enspace \mathrm{Im}\enspace (y)-\enspace \mathrm{Re}\enspace (y)\enspace \mathrm{Im}\enspace (x)-\enspace \mathrm{Im}\enspace (x)|\enspace \mathrm{Im}\enspace (y)-\enspace \mathrm{Im}\enspace (x)\times \enspace \mathrm{Im}\enspace (y)\\ =\left(\mathrm{Re}\enspace (y)-\enspace \mathrm{Im}\enspace (y)\right)\mathrm{*}\left(\mathrm{Re}\enspace (x)-\enspace \mathrm{Im}\enspace (x)\right)\\ =\left(\overline{\enspace \mathrm{Re}\enspace (y)+\enspace \mathrm{Im}\enspace (y)}\right)\mathrm{*}\left(\overline{\enspace \mathrm{Re}\enspace (x)+\enspace \mathrm{Im}\enspace (x)}\right)\\ =\bar{y}\mathrm{*}\bar{x}.\end{array} $$$

In a similar way, we deduce that $| | x | | = \sqrt{x * \overset{̅}{x}} .$ $Mathematical equation: $ ||x||=\sqrt{x\mathrm{*}\bar{x}}.$$ The inverse of a nonzero quaternion x is define as an element x ⁻¹ such that x $*$ $Mathematical equation: $ \mathrm{*}$$ x ⁻¹ = x ⁻¹ $*$ $Mathematical equation: $ \mathrm{*}$$ x = 1. But since $x * \overset{̅}{x} = \overset{̅}{x} * x = x | \overset{̅}{x} = | | x | |^{2} = x^{2 *},$ $Mathematical equation: $ x\mathrm{*}\bar{x}=\bar{x}\mathrm{*}x=x|\bar{x}=||x|{|}^2={x}^{2\mathrm{*}},$$ where x ^2* = x $*$ $Mathematical equation: $ \mathrm{*}$$ x. then, $x^{- 1} = \frac{\overset{̅}{x}}{| | x | |^{2}} .$ $Mathematical equation: $ {x}^{-1}=\frac{\bar{x}}{||x|{|}^2}.$$

Now, we will introduce the real linear space of octonions. $A = {x_{0} + x_{1} f_{1} + x_{2} f_{2} + x_{3} f_{3} + x_{4} f_{4} + x_{5} f_{5} + x_{6} f_{6} + x_{7} f_{7}, x_{i} \in R, x_{0} \in R, x_{i} \in R, i = 1, \dots, 7}$ $Mathematical equation: $ \mathcal{A}=\{{x}_0+{x}_1{f}_1+{x}_2{f}_2+{x}_3{f}_3+{x}_4{f}_4+{x}_5{f}_5+{x}_6{f}_6+{x}_7{f}_7,{x}_i\in \mathbb{R},{x}_0\in \mathbb{R},{x}_i\in \mathbb{R},i=1,\cdots,7\}$$ where the elements of the basis $B = {1, f_{1}, f_{2}, f_{3}, f_{4}, f_{5}, f_{6}, f_{7}}$ $Mathematical equation: $ B=\{1,{f}_1,{f}_2,{f}_3,{f}_4,{f}_5,{f}_6,{f}_7\}$$ of $A$ Mathematical equation: $ A$ satisfy the rules of multiplication presented in Table 2 below and deduced using the diagram presented in Figure 3.

Figure 3

Diagram of Fano for octonions.

Table 2

Table of multiplication of octonions.

Now, we fulfill the table recurring to the diagram of Fano, for instance we have f ₂ f ₄ = f ₆ but we have f ₄ f ₂ = −f ₆ since the sense from f ₄ to f ₂ is contrary to the sense of line that contains f ₄, f ₂ and f ₆.

The conjugate of the octonion $x = x_{0} + x_{1} f_{1} + x_{2} f_{2} + x_{3} f_{3} + x_{4} f_{4} + x_{5} f_{5} + x_{6} f_{6} + x_{7} f_{7}$ $Mathematical equation: $ x={x}_0+{x}_1{f}_1+{x}_2{f}_2+{x}_3{f}_3+{x}_4{f}_4+{x}_5{f}_5+{x}_6{f}_6+{x}_7{f}_7$$ is $\overset{̅}{x} = x_{0} - x_{1} f_{1} - x_{2} f_{2} - x_{3} f_{3} - x_{4} f_{4} - x_{5} f_{5} - x_{6} f_{6} - x_{7} f_{7}$ $Mathematical equation: $ \bar{x}={x}_0-{x}_1{f}_1-{x}_2{f}_2-{x}_3{f}_3-{x}_4{f}_4-{x}_5{f}_5-{x}_6{f}_6-{x}_7{f}_7$$ and the real part of the octonion x is $Re (x) = x_{0}$ $Mathematical equation: $ \mathrm{Re}\enspace (x)={x}_0$$ and the imaginary part of the octonion is $Im (x) = x_{1} f_{1} + x_{2} f_{2} + \dots + x_{7} f_{7} .$ $Mathematical equation: $ \mathrm{Im}\enspace (x)={x}_1{f}_1+{x}_2{f}_2+\cdots +{x}_7{f}_7.$$ We define

$| | x | | | = \sqrt{x * \overset{̅}{x}} = \sqrt{x_{0}^{2} + x_{1}^{2} + x_{2}^{2} + \dots + x_{7}^{2}} .$ $Mathematical equation: $$ ||x|||=\sqrt{x\mathrm{*}\bar{x}}=\sqrt{{x}_0^2+{x}_1^2+{x}_2^2+\cdots +{x}_7^2}. $$$

Using the table of multiplication 2 we obtain for $x = x_{0} + x_{1} f_{1} + x_{2} f_{2} + x_{3} f_{3} + x_{4} f_{4} + x_{5} f_{5} + x_{6} f_{6} + x_{7} f_{7}$ $Mathematical equation: $ x={x}_0+{x}_1{f}_1+{x}_2{f}_2+{x}_3{f}_3+{x}_4{f}_4+{x}_5{f}_5+{x}_6{f}_6+{x}_7{f}_7$$ and for $y = y_{0} + y_{1} f_{1} + y_{2} f_{2} + y_{3} f_{3} + y_{4} f_{4} + y_{5} f_{5} + y_{6} f_{6} + y_{7} f_{7}$ $Mathematical equation: $ y={y}_0+{y}_1{f}_1+{y}_2{f}_2+{y}_3{f}_3+{y}_4{f}_4+{y}_5{f}_5+{y}_6{f}_6+{y}_7{f}_7$$ we obtain

$\begin{array}{l} x * y = x_{0} y_{0} + x_{0} (y_{1} f_{1} + y_{2} f_{2} + y_{3} f_{3} + y_{4} f_{4} + y_{5} f_{5} + y_{6} f_{6} + y_{7} f_{7}) + \\ + y_{0} (x_{1} f_{1} + x_{2} f_{2} + x_{3} f_{3} + x_{4} f_{4} + x_{5} f_{5} + x_{6} f_{6} + x_{7} f_{7}) + \\ - x_{1} y_{1} - x_{2} y_{2} - x_{3} y_{3} - x_{4} y_{4} - x_{5} f_{5} - x_{6} y_{6} - x_{7} y_{7} + \\ + (x_{2} y_{3} - x_{3} y_{2}) f_{1} + (x_{3} y_{1} - x_{1} y_{3}) f_{2} + \dots + (x_{1} y_{2} - x_{2} y_{1}) f_{7} . \end{array}$ $Mathematical equation: $$ \begin{array}{l}x\mathrm{*}y={x}_0{y}_0+{x}_0\left({y}_1{f}_1+{y}_2{f}_2+{y}_3{f}_3+{y}_4{f}_4+{y}_5{f}_5+{y}_6{f}_6+{y}_7{f}_7\right)+\\ +{y}_0\left({x}_1{f}_1+{x}_2{f}_2+{x}_3{f}_3+{x}_4{f}_4+{x}_5{f}_5+{x}_6{f}_6+{x}_7{f}_7\right)+\\ -{x}_1{y}_1-{x}_2{y}_2-{x}_3{y}_3-{x}_4{y}_4-{x}_5{f}_5-{x}_6{y}_6-{x}_7{y}_7+\\ +\left({x}_2{y}_3-{x}_3{y}_2\right){f}_1+\left({x}_3{y}_1-{x}_1{y}_3\right){f}_2+\cdots +\left({x}_1{y}_2-{x}_2{y}_1\right){f}_7.\end{array} $$$

We must note that: $x * y = Re (x) Re (y) + Re (x) Im (y) + Re (y) Im (x) - Im (x) | Im (y) + Im (x) \times Im (y) .$ $Mathematical equation: $ x\mathrm{*}y=\enspace \mathrm{Re}\enspace (x)\enspace \mathrm{Re}\enspace (y)+\enspace \mathrm{Re}\enspace (x)\enspace \mathrm{Im}\enspace (y)+\enspace \mathrm{Re}\enspace (y)\enspace \mathrm{Im}\enspace (x)-\enspace \mathrm{Im}\enspace (x)|\enspace \mathrm{Im}\enspace (y)+\enspace \mathrm{Im}\enspace (x)\times \enspace \mathrm{Im}\enspace (y).$$ Proceeding like we have done for the quaternions we deduce that $\overset{̅}{x * y} = \overset{̅}{y} * \overset{̅}{x} .$ $Mathematical equation: $ \overline{x\mathrm{*}y}=\bar{y}\mathrm{*}\bar{x}.$$ We define the norm of an octonion as being $| | x | | = \sqrt{x_{0}^{2} + x_{1}^{2} + x_{2}^{2} + \dots + x_{7}^{2}} .$ $Mathematical equation: $ ||x||=\sqrt{{x}_0^2+{x}_1^2+{x}_2^2+\cdots +{x}_7^2}.$$ We, must say, again that $| | x | | = \sqrt{x * \overset{̅}{x}} .$ $Mathematical equation: $ ||x||=\sqrt{x\mathrm{*}\bar{x}}.$$ Let x be an octonion such that $| | x | | \neq 0$ $Mathematical equation: $ ||x||\ne 0$$ then the inverse of x is $x^{- 1} = \frac{\overset{̅}{x}}{| | x | |^{2}},$ $Mathematical equation: $ {x}^{-1}=\frac{\bar{x}}{||x|{|}^2},$$ since x $*$ $Mathematical equation: $ \mathrm{*}$$ x ⁻¹ = x ⁻¹ $*$ $Mathematical equation: $ \mathrm{*}$$ x = 1.

Acknowledgments

Luís Vieira was partially supported by CMUP (UID/MAT/00144/2019), which is funded by FCT with national (MCTES) and European structural funds through the programs FEDER, under the partnership agreement PT 2020.

The author are grateful to the anonymous referee for a careful checking of the details and for helpful comments that improved this paper.

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Cite this article as: Vieira L 2019. Euclidean Jordan algebras and some conditions over the spectraof a strongly regular graph. 4open, 2, 21.

All Tables

Table 1

Table of multiplication of quaternions.

In the text

Table 2

Table of multiplication of octonions.

In the text

All Figures

	Figure 1 A simple graph X.
In the text

	Figure 2 Diagram of Fano for quaternions.
In the text

	Figure 3 Diagram of Fano for octonions.
In the text

[R1] Faybusovich L (1997), Linear systems in Jordan algebras and primal-dual interior-point algorithms. J Comput Appl Math 86, 149–175. [Google Scholar]

[R2] Faybusovich L (1997), Euclidean Jordan algebras and Interior-point Algorithms. Positivity 1, 331–357. [Google Scholar]

[R3] Faybusovich L (2002), A jordan algebraic approach to potential reduction algorithms. Math Zeitschr 239, 117–129. [CrossRef] [Google Scholar]

[R4] Faybusovich L, Tsuchiya T (2003), Primal-dual algorithms and infinite-dimensional Jordan algebras of finite rank. Math Program 97, 3, 471–493. [Google Scholar]

[R5] Schmieta SH, Alizadeh F (2001), Associative and Jordan algebras and polynomial time Interior point alghoritms for symmetric cones. Math Oper Res 26, 3, 543–564. [CrossRef] [Google Scholar]

[R6] Schmieta SH, Alizadeh F (2003), Extension of commutative class of primal dual interior point algorithms to symmetric cones. Math Program 96, 409–438. [Google Scholar]

[R7] Alizadeh F, Goldfarb D (2003), Second order cone programming. Math Program 95, 1, 3–51. [Google Scholar]

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