The Eigenvalue Decomposition in a Jordan Algebra

Introduction

A Jordan algebra is commutative and need not be associative, but every element of it generates an associative subalgebra, and that subalgebra carries the idempotent calculus of the associative theory. This article reads the eigenvalue decomposition of an element $x$ of a Jordan algebra,

$$ x=\sum_{i=1}^{r}\lambda_i e_i , $$

with the $\lambda_i$ scalars and the $e_i$ pairwise orthogonal idempotents resolving the identity. The decomposition exists exactly when the minimal polynomial of $x$ splits with distinct roots; in a formally real algebra this always happens, and the eigenvalues are real. The two facts that distinguish the Jordan theory from The Eigenvalue Decomposition of an Element are that the eigenvalues are roots of a minimal polynomial internal to the subalgebra generated by $x$, and that the multiplication operator $L_x$ has, besides the eigenvalues $\lambda_i$, the midpoints $(\lambda_i+\lambda_j)/2$. The middle value $\tfrac12$ is the Peirce eigenvalue of an idempotent, and it is the fingerprint of the Jordan identity.

The base is a field $k$ of characteristic not two, $J$ is a finite-dimensional unital Jordan algebra over $k$, and the product is written $\bullet$ with square $x^2=x\bullet x$. The axioms, the Peirce decomposition and the cubic $2L_e^3-3L_e^2+L_e=0$ are from Jordan Algebras; the formally real algebras, their positive cone and their spectral theorem are from Jordan Algebras and the Positive Cone; the simple algebras and the Albert algebra are Special and Exceptional Jordan Algebras; the Hermitian Jordan algebra of a sesqualgebra is The Hermitian Jordan Algebra; the spin factors and the Clifford envelope are Spin Factors and the Clifford Envelope with Inner Conjugation; and the associative original is The Eigenvalue Decomposition of an Element.

The Minimal Polynomial and the Spectrum

The Subalgebra Generated by an Element

Proposition. For every $x\in J$ the subalgebra $k[x]$ generated by $x$ is commutative and associative, and it is the linear span of the powers $x^n$, $n\geq0$.

Proof. A Jordan algebra is power-associative (Jordan Algebras, §Power Associativity): the subalgebra generated by one element is associative, so the powers are well defined and $x^m\bullet x^n=x^{m+n}$. The span of the powers is therefore closed under the product, contains $1=x^0$ and $x$, and is the smallest such subalgebra.

Remark. The subalgebra $k[x]$ is the whole of the Jordan content of a single element. Every idempotent, every polynomial and every eigenvalue attached to $x$ lives in $k[x]$; the ambient algebra enters only through the multiplication operator $L_x$, which acts on the Peirce spaces of the resolution.

Eigenvalues

Definition. A scalar $\lambda\in k$ is an eigenvalue of $x$ when $x-\lambda1$ is not invertible in the unital algebra $k[x]$. The spectrum of $x$ is the set $\sigma(x)$ of its eigenvalues.

Proposition. Let $m_x$ be the monic generator of the ideal $\{p\in k[t]:p(x)=0\}$, the minimal polynomial of $x$. Then

$$ \boxed{\;\lambda\in\sigma(x) \quad\Longleftrightarrow\quad m_x(\lambda)=0\;} . $$

Proof. The evaluation map $k[t]\to k[x]$, $p\mapsto p(x)$, has kernel $(m_x)$ and induces an isomorphism $k[x]\cong k[t]/(m_x)$. Under it the element $x-\lambda1$ corresponds to the class of $t-\lambda$. That class is invertible in $k[t]/(m_x)$ exactly when $t-\lambda$ is coprime to $m_x$, that is, exactly when $m_x(\lambda)\neq0$. Hence non-invertibility is the vanishing of $m_x$ at $\lambda$.

Corollary. The spectrum is finite, of cardinality at most $\deg m_x$, and it is nonempty when $k$ is algebraically closed. It may be empty over a field that is not algebraically closed.

Remark. The definition is the polynomial one, not the invertibility in the quadratic sense of the ambient Jordan algebra. The two notions agree on the spectral resolution, and the polynomial one is the one that the idempotent calculus uses. The caution matters over a non-formally-real algebra, where an element may be a zero divisor of the ambient algebra without having a polynomial eigenvalue.

Formal Reality Forces Real Squarefree Spectrum

Theorem. Let $J$ be a finite-dimensional formally real Jordan algebra over $\mathbb{R}$ and let $x\in J$. Then $m_x$ has only real roots and is squarefree.

Proof. Suppose first that $m_x$ has a repeated factor. Write $m_x=q^2r$ with $q$ irreducible, and put $y=q(x)r(x)$. Then

$$ y^2=q(x)^2r(x)^2=m_x(x)r(x)=0 . $$

The element $y$ is nonzero: if $q(x)r(x)=0$, then $m_x$ divides $qr$, which is impossible because $\deg(qr)<\deg(q^2r)=\deg m_x$. Thus $y$ is a nonzero nilpotent, and $y^2=0$ contradicts formal reality. Hence $m_x$ is squarefree.

Suppose now that $m_x$ has a nonreal root. Since $m_x$ has real coefficients, it has an irreducible quadratic factor $q(t)=t^2+at+b$ with discriminant $a^2-4b<0$, and because $m_x$ is squarefree $q$ is coprime to $r=m_x/q$. The Chinese remainder theorem gives

$$ \mathbb{R}[x]\cong\mathbb{R}[t]/(m_x)\cong\mathbb{R}[t]/(q)\times\mathbb{R}[t]/(r), $$

and $\mathbb{R}[t]/(q)\cong\mathbb{C}$. Let $z$ be the image of $x$ in the first factor, a nonreal complex number. Put

$$ y=\frac{z-\operatorname{Re}z}{\lvert\operatorname{Im}z\rvert}, $$

a polynomial in $z$ with $y^2=-1$. The element of $\mathbb{R}[x]$ corresponding to the pair $(y,0)$ is a polynomial in $x$, and its square is $(-1,0)=-1$. Lifting it to $J$ gives an element, again written $y$, with $y^2+1^2=0$. Formal reality gives $y=0$ and $1=0$, a contradiction. Hence every root of $m_x$ is real.

Remark. The two halves of the proof use the two consequences of formal reality: there is no element with square $-1$, and there is no nonzero nilpotent. Both follow from the single condition that a finite sum of squares vanishes only when every term does, since $y^2+1=0$ is such a sum and $y^2=0$ is such a sum. The order of the two halves matters: the squarefree part is what makes the quadratic factor coprime to the rest in the second part.

The Spectral Resolution

The Definition

Definition. A spectral resolution of $x$ is a writing

$$ x=\sum_{i=1}^{r}\lambda_i e_i $$

with $\lambda_1,\dots,\lambda_r\in k$ pairwise distinct, $e_1,\dots,e_r$ nonzero idempotents of $J$ satisfying

$$ e_i\bullet e_j=\delta_{ij}e_i, \qquad \sum_{i=1}^{r}e_i=1 . $$

The $e_i$ are the spectral idempotents or Peirce idempotents of $x$, and the condition on the sum is the resolution of the identity. The number $r$ is the number of distinct eigenvalues.

The Existence Theorem

Theorem. Let $x\in J$. The following are equivalent:

$x$ has a spectral resolution;
$m_x$ splits over $k$ with distinct roots;
$m_x(t)=\prod_{i=1}^{r}(t-\lambda_i)$ with $\lambda_i\in k$ pairwise distinct.

When they hold, the eigenvalues are the $\lambda_i$, the spectral idempotents are the Lagrange polynomials

$$ e_i=\prod_{j\neq i}\frac{x-\lambda_j}{\lambda_i-\lambda_j}, $$

and the resolution is unique.

Proof. Assume 2. The algebra $k[x]$ is finite dimensional, commutative and associative, so The Eigenvalue Decomposition of an Element, §Existence and Uniqueness, applied to the element $x$ of $k[x]$, gives a decomposition $x=\sum_i\lambda_i e_i$ with $e_i\in k[x]$ idempotent, pairwise orthogonal and summing to $1$. The product of $k[x]$ is the Jordan product, so this is a spectral resolution in $J$. The formula for $e_i$ is the Lagrange formula in $k[x]$.

Assume 1. The polynomial $p(t)=\prod_i(t-\lambda_i)$ vanishes at $x$: on the component $e_j$ it has the factor $t-\lambda_j$, and $x$ acts on that component as $\lambda_j$. Hence $m_x$ divides $p$. A proper divisor of $p$ omits some factor $t-\lambda_i$; evaluated at $x$ on $e_i$ it gives $\prod_{j\neq i}(\lambda_i-\lambda_j)\neq0$, so it does not vanish at $x$. Hence $m_x=p$. This proves 3 and the equality of the spectra.

The uniqueness is the uniqueness of the Lagrange interpolation: if $x=\sum_i\lambda_i e_i=\sum_i\lambda_i f_i$ are two resolutions with the same eigenvalues, then applying the polynomial $h_i$ of the Lagrange formula to both writings gives $e_i=h_i(x)=f_i$. The eigenvalues are the roots of $m_x$, hence are determined by $x$.

Remark. The theorem is the associative theorem applied to the commutative algebra $k[x]$, and the only Jordan input is the power associativity that makes $k[x]$ associative. The spectral idempotents are polynomials in $x$ with coefficients in $k$; in particular they commute with $x$ and with one another, and the resolution is internal to the commutative algebra $k[x]$.

The Spectral Theorem in a Formally Real Algebra

Theorem. Let $J$ be a finite-dimensional formally real Jordan algebra over $\mathbb{R}$. Then every $x\in J$ has a spectral resolution

$$ \boxed{\;x=\sum_{i=1}^{r}\lambda_i e_i\;} $$

with $\lambda_i\in\mathbb{R}$ and $e_i$ pairwise orthogonal idempotents. The eigenvalues are the roots of $m_x$, they are real and simple, and the resolution is unique up to the order of the terms.

Proof. By §Formal Reality Forces Real Squarefree Spectrum, $m_x$ has only real roots and is squarefree. The existence theorem applies. The uniqueness is its uniqueness clause.

Corollary (the positive cone). For $x=\sum_i\lambda_i e_i$ in a formally real algebra, $x$ lies in the positive cone $J_+$ if and only if $\lambda_i\geq0$ for every $i$.

Proof. If $x=y^2$ and $y=\sum_j\mu_j f_j$ is the spectral resolution of $y$, then $x=\sum_j\mu_j^2 f_j$, and the uniqueness of the resolution gives the eigenvalues $\mu_j^2\geq0$ (after collecting the equal values). Conversely, if every $\lambda_i\geq0$, then $y=\sum_i\sqrt{\lambda_i}e_i$ satisfies $y^2=\sum_i\lambda_i e_i=x$.

Corollary (spectral mapping). For every polynomial $p\in\mathbb{R}[t]$,

$$ \boxed{\;p(x)=\sum_{i=1}^{r}p(\lambda_i)e_i\;} . $$

In particular $x^n=\sum_i\lambda_i^n e_i$ for $n\geq0$, with the convention $\lambda_i^0=1$ and $\sum_i e_i=1$. The spectral idempotent $e_i$ is the value at $x$ of the indicator polynomial that is $1$ at $\lambda_i$ and $0$ at the other eigenvalues.

Proof. The polynomial acts on each component $e_i$ by its value at $\lambda_i$; summing gives the formula. The indicator polynomial is the Lagrange polynomial of the existence theorem.

The Trace Form

Theorem. Let $T(x,y)=\operatorname{tr}(L_{x\bullet y})$ be the trace form of Jordan Algebras, §The Trace Form. For a spectral resolution $x=\sum_i\lambda_i e_i$ the idempotents are pairwise orthogonal for $T$,

$$ T(e_i,e_j)=0 \quad (i\neq j), $$

and in a formally real algebra

$$ T(e_i,e_i)=\dim J_{ii}+\frac12\sum_{j\neq i}\dim J_{ij}>0 . $$

Proof. For $i\neq j$, $e_i\bullet e_j=0$, so $T(e_i,e_j)=\operatorname{tr}(L_{e_i\bullet e_j})=0$. For the diagonal, $T(e_i,e_i)=\operatorname{tr}(L_{e_i})$; the operator $L_{e_i}$ has eigenvalues $1$ on $J_{ii}$, $\tfrac12$ on $\bigoplus_{j\neq i}J_{ij}$ and $0$ on the remaining Peirce spaces, with the dimensions as multiplicities. The trace is the displayed sum, and it is positive in a formally real algebra because $T$ is positive definite and $e_i\neq0$.

Remark. The trace form is the Jordan analogue of the Frobenius form of an associative algebra. In $H_n(\mathbb{R})$ it is $T(x,y)=\frac{n+1}{2}\operatorname{tr}(xy)$, and the positivity on the idempotents is the positivity of the rank. The trace form is not used in the existence theorem; it records the orthogonality of the spectral resolution and the dimensions of the spectral pieces.

The Multiplication Operator and the Peirce Midpoints

The Peirce Decomposition of a Complete Family

Let $x=\sum_i\lambda_i e_i$ be a spectral resolution. The idempotents $e_i$ are pairwise orthogonal and sum to $1$, so they form a complete family in the sense of Jordan Algebras, §The Peirce Spaces. The Peirce spaces are

$$ J_{ii}=\{y:e_i\bullet y=y\}, \qquad J_{ij}=\left\{y:e_i\bullet y=\tfrac12 y,\ e_j\bullet y=\tfrac12 y\right\} \quad (i\neq j), $$

and the algebra splits as

$$ J=\bigoplus_{i=1}^{r}J_{ii}\oplus\bigoplus_{i

The multiplication table is the one of Jordan Algebras: the diagonal spaces are subalgebras with units $e_i$, the off-diagonal spaces are modules over the pair, $J_{ii}\bullet J_{jj}=0$ for $i\neq j$, and $J_{ij}\bullet J_{kl}\subseteq J_{il}$ when the indices match and vanishes otherwise.

The Action of the Element

Theorem. On the Peirce spaces of the spectral resolution,

$$ L_x y=\lambda_i y \quad (y\in J_{ii}), \qquad L_x y=\frac{\lambda_i+\lambda_j}{2}\,y \quad (y\in J_{ij},\ i\neq j). $$

Consequently the eigenvalues of the operator $L_x$ are the numbers $\lambda_i$ and the midpoints $(\lambda_i+\lambda_j)/2$ for $i

Proof. For $y\in J_{ii}$ one has $e_i\bullet y=y$ and $e_k\bullet y=0$ for $k\neq i$; hence $x\bullet y=\sum_k\lambda_k e_k\bullet y=\lambda_i y$. For $y\in J_{ij}$ with $i\neq j$ one has $e_i\bullet y=e_j\bullet y=\tfrac12 y$ and $e_k\bullet y=0$ for $k\notin\{i,j\}$; hence

$$ x\bullet y=\lambda_i\tfrac12 y+\lambda_j\tfrac12 y=\frac{\lambda_i+\lambda_j}{2}y . $$

The spectrum of $L_x$ is therefore contained in the set of the $\lambda_i$ and the midpoints $(\lambda_i+\lambda_j)/2$, and a displayed value is an eigenvalue as soon as the corresponding Peirce space is nonzero. The different midpoints may coincide, and a midpoint may coincide with an eigenvalue; the list is a list with multiplicities, not a set of distinct values.

Remark. The operator $L_x$ is not the left multiplication of an associative algebra: it is symmetric for the trace form, and its eigenvalues are the pairwise midpoints. The midpoint $\tfrac12$ is the value that the Peirce calculus assigns to the mixed space, and it is the reason a single idempotent has the three eigenvalues $1,\tfrac12,0$ rather than the two eigenvalues $1,0$ of an associative projection. In the associative case the mixed space splits into $eA(1-e)$ and $(1-e)Ae$, on which $L_e$ acts by $1$ and $0$; the symmetrisation combines the two into one Jordan Peirce space on which $L_e$ acts by $\tfrac12$.

The Case of One Idempotent

Corollary. Let $e\in J$ be an idempotent. Then $L_e$ satisfies

$$ 2L_e^3-3L_e^2+L_e=0, $$

and its eigenvalues lie in $\{0,\tfrac12,1\}$. The Peirce spaces $J_1(e)$, $J_{1/2}(e)$ and $J_0(e)$ are the eigenspaces.

Proof. This is the cubic of Jordan Algebras, §The Cubic Identity for an Idempotent, whose eigenvalues are the three roots of $2t^3-3t^2+t=t(2t-1)(t-1)$. The spectral resolution of $e$ is $e=1\cdot e+0\cdot(1-e)$, and the theorem above with $\lambda_1=1$, $\lambda_0=0$ gives the three values.

Remark. The corollary is the smallest case of the midpoint phenomenon: the two eigenvalues $1$ and $0$ of the resolution produce the third eigenvalue $\tfrac12$ on the mixed space. The number of eigenvalues of $L_x$ is at most $r(r+1)/2$, against the $r$ eigenvalues of $x$ itself; the extra values are the midpoints.

Examples

The Hermitian Matrices

Let $J=H_n(\mathbb{R})$ with $x\bullet y=\tfrac12(xy+yx)$. A symmetric matrix $x$ has the spectral resolution $x=\sum_i\lambda_i P_i$ of the usual spectral theorem, where the $P_i$ are the orthogonal projections onto the eigenspaces of the distinct eigenvalues. The idempotents are Hermitian, pairwise orthogonal and sum to the identity. The Peirce spaces $J_{ii}$ are the symmetric matrices supported on the $i$-th eigenspace, the off-diagonal $J_{ij}$ are the symmetric matrices carrying the two eigenspaces, and $L_x$ acts by $\lambda_i$ on $J_{ii}$ and by $(\lambda_i+\lambda_j)/2$ on $J_{ij}$. The trace form is $T(x,y)=\frac{n+1}{2}\operatorname{tr}(xy)$, and the spectral resolution is orthogonal for it.

The Spin Factor

Let $J=JSpin(V)$ be the spin factor of a positive definite quadratic form $Q$ on a finite-dimensional real space $V$, with elements $(\alpha,v)$ and square

$$ (\alpha,v)^2=(\alpha^2+Q(v),\,2\alpha v), $$

as in Spin Factors and the Clifford Envelope with Inner Conjugation. For $v\neq0$ put $s=\sqrt{Q(v)}$. The minimal polynomial of $x=(\alpha,v)$ is

$$ m_x(t)=t^2-2\alpha t+\alpha^2-Q(v)=(t-\alpha)^2-s^2, $$

with the two real roots $\lambda_\pm=\alpha\pm s$. The spectral idempotents are

$$ e_\pm=\frac12\left(1\pm\frac{v}{s}\right), $$

and one checks $e_\pm^2=e_\pm$, $e_+\bullet e_-=0$ and $e_++e_-=1$. The resolution is $x=\lambda_+e_++\lambda_-e_-$. The Peirce middle space is the orthogonal complement of the line $\mathbb{R}v$ in $V$, of dimension $\dim V-1$, and $L_x$ acts on it by the midpoint $\alpha$. The eigenvalues of $L_x$ are therefore $\alpha+s$, $\alpha-s$ and $\alpha$, the last with multiplicity $\dim V-1$. For $v=0$ the element is the scalar $\alpha1$ and the resolution has the single term $\alpha1$.

The Albert Algebra

Let $J=H_3(\mathbb{O})$ be the exceptional Albert algebra. It is formally real of degree three, so every element has a spectral resolution with at most three real eigenvalues and three pairwise orthogonal primitive idempotents. The resolution is the one of Jordan Algebras and the Positive Cone, §Spectral Resolution; the Peirce spaces are the three diagonal corners and the three off-diagonal spaces of the Freudenthal triple system, and the multiplication operator has the three eigenvalues $\lambda_1,\lambda_2,\lambda_3$ and the three midpoints $(\lambda_1+\lambda_2)/2$, $(\lambda_1+\lambda_3)/2$, $(\lambda_2+\lambda_3)/2$. The explicit matrices, the determinant and the trace are those of Special and Exceptional Jordan Algebras and Determinants in Jordan Matrix Algebras; no new computation is made here.

A Non-Formally-Real Algebra

Let $J=M_2(\mathbb{R})^+$ with the symmetrised product. The algebra is not formally real: the matrix unit $E_{12}$ is nonzero with square zero. Its minimal polynomial is $t^2$, which is not squarefree, and it has no spectral resolution. The rotation matrix

$$ R=\begin{pmatrix}0&-1\\1&0\end{pmatrix} $$

has minimal polynomial $t^2+1$, which has no real root; it has no spectral resolution over $\mathbb{R}$ either. Over $\mathbb{C}$ the polynomial splits as $(t-i)(t+i)$, and the spectral resolution exists in $M_2(\mathbb{C})^+$, with the two idempotents

$$ e_\pm=\frac12\left(1\mp iR\right), $$

which are not Hermitian for the transpose involution. The example is the boundary between the Jordan spectral theorem and the associative one: formal reality is exactly what rules out the two failures, the nilpotent and the nonreal pair.

The Special Case of an Associative Algebra

Let $J=A^+$ for a finite-dimensional associative algebra $A$ and let $x\in A$ be semisimple. The spectral resolution of $x$ in $J$ is the associative eigenvalue decomposition of $x$ in $A$, because the subalgebra generated by $x$ is the same in both readings and the idempotents are the same. The difference between the two theories is not in the resolution of the element but in the multiplication operator: $L_x$ on $J$ has the midpoints, while the left multiplication of $A$ has only the eigenvalues. If $x$ is not semisimple, neither theory gives a resolution.

Summary

The eigenvalue decomposition of an element $x$ of a Jordan algebra is the writing $x=\sum_i\lambda_i e_i$ with distinct scalars $\lambda_i$ and pairwise orthogonal idempotents $e_i$ summing to the identity. It exists exactly when the minimal polynomial $m_x$, taken in the commutative associative algebra $k[x]$ generated by $x$, splits over $k$ with distinct roots. The spectral idempotents are the Lagrange polynomials in $x$, the eigenvalues are the roots of $m_x$, and the resolution is unique.

In a finite-dimensional formally real Jordan algebra over $\mathbb{R}$ the minimal polynomial of every element has only real roots and is squarefree, so every element has a spectral resolution with real eigenvalues. The positive cone is described by the signs of the eigenvalues, the polynomial calculus acts by the spectral mapping, and the trace form is orthogonal on the spectral idempotents.

The multiplication operator $L_x$ acts by $\lambda_i$ on the diagonal Peirce space $J_{ii}$ and by the midpoint $(\lambda_i+\lambda_j)/2$ on the off-diagonal space $J_{ij}$. For a single idempotent the three values are $1,\tfrac12,0$, the roots of the cubic $2L_e^3-3L_e^2+L_e=0$. The midpoint is the structural difference from the associative theory, where the two mixed Peirce spaces carry the eigenvalues $1$ and $0$ separately.

The associative original is The Eigenvalue Decomposition of an Element; the Jordan axioms and the Peirce decomposition are Jordan Algebras; the formal reality, the positive cone and the spectral theorem are Jordan Algebras and the Positive Cone; the classification and the Albert algebra are Special and Exceptional Jordan Algebras; the spin factors are Spin Factors and the Clifford Envelope with Inner Conjugation; and the Hermitian case is The Hermitian Jordan Algebra and The Eigenvalue Decomposition of a Hermitian Element.

Summary of Notation

Symbol	Meaning
$k$	a field of characteristic not two
$J$	a finite-dimensional unital Jordan algebra over $k$
$\bullet$	the Jordan product, commutative with the Jordan identity
$x^2$	the square $x\bullet x$; the powers are the Jordan powers
$L_x$	the multiplication operator $y\mapsto x\bullet y$
$k[x]$	the commutative associative subalgebra generated by $x$
$m_x$	the minimal polynomial of $x$ in $k[x]$
$\sigma(x)$	the spectrum, the set of roots of $m_x$ in $k$
$x=\sum_i\lambda_i e_i$	the spectral resolution
$e_i$	the spectral idempotents, $e_i\bullet e_j=\delta_{ij}e_i$, $\sum_i e_i=1$
$J_{ii}$, $J_{ij}$	the Peirce spaces of the complete family $\{e_i\}$
$J_1(e),J_{1/2}(e),J_0(e)$	the Peirce spaces of a single idempotent
$T(x,y)=\operatorname{tr}(L_{x\bullet y})$	the trace form
$J_+$	the positive cone of a formally real algebra
$r$	the number of distinct eigenvalues