The $3\times3$ Eigenvalue Problem with Real Eigenvalues

We now turn to the $3\times3$ case. It is not immediately obvious that $3\times3$ octonionic Hermitian matrices have a well-defined characteristic equation. We therefore first review some of the properties of these matrices before turning to the eigenvalue problem. As in the $2\times2$ case, over the octonions there will be solutions of the eigenvalue problem with eigenvalues which are not real; we consider here only the real eigenvalue problem.

Recall from § 11.2 that the $3\times3$ octonionic Hermitian matrices, which we call Jordan matrices, form the exceptional Jordan algebra, also known as the Albert algebra, under the Jordan product.

Remarkably, Jordan matrices satisfy the usual characteristic equation [Harvey] \begin{equation} \AAA^3 - (\tr \AAA) \, \AAA^2 + \sigma(\AAA) \, \AAA - (\det \AAA) \, \III = 0 \label{Char} \end{equation} where $\sigma(\AAA)$ is defined by \begin{equation} \sigma(\AAA) = {1\over2} \left( (\tr \AAA)^2 - \tr (\AAA^2) \right) = \tr(\AAA*\AAA) \label{Sigma} \end{equation} and where the determinant of $\AAA$ was defined abstractly in terms of the Jordan and Freudenthal products in § 11.2. Concretely, if \begin{equation} \AAA = \begin{pmatrix}p& a& \bar{b}\\ \bar{a}& m& c\\ b& \bar{c}& n\\\end{pmatrix} \label{Three} \end{equation} with $p,m,n\in\RR$ and $a,b,c\in\OO$ then \begin{align} \tr \AAA &= p + m + n \label{trace}\\ \sigma(\AAA) &= pm + pn + mn - |a|^2 - |b|^2 - |c|^2 \label{sigma}\\ \det \AAA &= pmn + b(ac) + \bar{b(ac)} - n|a|^2 - m|b|^2 - p|c|^2 \label{det}\\ \end{align}

An $n\times n$ Hermitian matrix over any of the normed division algebras can be rewritten as a symmetric $kn\times kn$ real matrix, where $k$ denotes the dimension of the underlying division algebra. It is therefore clear that a $3\times3$ octonionic Hermitian matrix must have $8\times3=24$ real eigenvalues. However, as we now show, instead of having (a maximum of) 3 distinct real eigenvalues, each with multiplicity 8, we show that there are (a maximum of) 6 distinct real eigenvalues, each with multiplicity 4.

The reason for this peculiar property is that, somewhat surprisingly, a (real) eigenvalue $\lambda$ of a Jordan matrix $A$ does not in general satisfy the characteristic equation (\ref{Char}). 1) To see this, consider the ordinary eigenvalue equation \begin{equation} \AAA v = \lambda v \label{Orig} \end{equation} with $\AAA$ as in (\ref{Three}), $\lambda\in\RR$, and where \begin{equation} v=\begin{pmatrix}x\\ y\\ z\\\end{pmatrix} \end{equation} Explicit computation yields \begin{align} (\lambda-p) x &= ay + \bar{b}z \label{EI}\\ (\lambda-m) y &= cz + \bar{a}x \label{EII}\\ (\lambda-n) z &= bx + \bar{c}y \label{EIII} \end{align} so that \begin{equation} (\lambda-p) (\lambda-m) y = (\lambda-p)(cz + \bar{a}x) = (\lambda-p) cz + \bar{a} (ay + \bar{b}z) \end{equation} which implies \begin{equation} \left[ (\lambda-p) (\lambda-m) - |a|^2 \right] y = \bar{a} (\bar{b}z) + (\lambda-p) cz \label{EIV} \end{equation} Assume first that $\lambda\ne p$. Using (\ref{EI}) and (\ref{EIV}) in (\ref{EIII}) leads to \begin{align} &\left[ (\lambda-p)(\lambda-m) - |a|^2 \right] (\lambda-p) (\lambda-n) z \nonumber\\ &= \left[ (\lambda-p) (\lambda-m) - |a|^2 \right] (\lambda-p) (bx+\bar{c}y) \nonumber\\ &= \left[ (\lambda-p) (\lambda-m) - |a|^2 \right] b (ay+\bar{b}z) + (\lambda-p)\bar{c}\! \left[ \bar{a}(\bar{b}z) + (\lambda-p) cz \right] \nonumber\\ &= b \left[a \left(\bar{a} (\bar{b}z) + (\lambda-p) cz \right) \right] + \left[ (\lambda-p) (\lambda-m) - |a|^2 \right] b (\bar{b}z) \nonumber\\ &\qquad+ (\lambda-p)\bar{c}\! \left[ \bar{a}(\bar{b}z) + (\lambda-p) cz \right] \nonumber\\ &= (\lambda-p) \bigg[ (\lambda-m) |b|^2 z + (\lambda-p) |c|^2 z + b \Big( a (cz) \Big) + \bar{c} \left( \bar{a} (\bar{b}z) \right) \bigg] \end{align} Expanding this out and comparing with (\ref{trace})–(\ref{det}) results finally in \begin{align} \Big[ \det(\lambda I - A) \Big] z &\equiv \left[ \lambda^3 - (\tr A) \, \lambda^2 + \sigma(A) \, \lambda - \det A \right] z \nonumber\\ &= b \Big( a (cz) \Big) + \bar{c} \left( \bar{a} (\bar{b}z) \right) - \left[ b(ac) + (\bar{c}\,\bar{a})\bar{b} \right] z \label{CharLam} \end{align} Now consider the case $\lambda=p$. We still have (\ref{EIV}), which here takes the form \begin{equation} -|a|^2 y = \bar{a} (\bar{b}z) \label{FI} \end{equation} Inserting this into (\ref{EII}), we can solve for $x$, obtaining \begin{equation} -|a|^2 x = a (cz) + (p-m) \bar{b}z \label{FII} \end{equation} Finally, inserting (\ref{FI}) and (\ref{FII}) in (\ref{EIII}) yields \begin{equation} -\left( |a|^2 (p-n) + |b|^2 (p-m) \right) z = b \Big( a (cz) \Big) + \bar{c} \left( \bar{a} (\bar{b}z) \right) \end{equation} Comparing with (\ref{trace})–(\ref{det}) and using $\lambda=p$, we see that (\ref{CharLam}) still holds, and thus holds in general.

If $a$, $b$, $c$, and $z$ associate, the right-hand side of (\ref{CharLam}) vanishes, and $\lambda$ does indeed satisfy the characteristic equation (\ref{Char}); this will not happen in general. However, since the left-hand side of (\ref{CharLam}) is a real multiple of $z$, this must also be true of the right-hand side, so that \begin{equation} b \Big( a (cz) \Big) + \bar{c} \left( \bar{a} (\bar{b}z) \right) - \left[ b(ac) + (\bar{c}\,\bar{a})\bar{b} \right] z = rz \label{RReal} \end{equation} for some $r\in\RR$, which can be solved to yield a quadratic equation for $r$ as well as constraints on $z$.

Lemma: The real eigenvalues of the $3\times3$ octonionic Hermitian matrix $\AAA$ satisfy the modified characteristic equation \begin{equation} \det(\lambda I - \AAA) = \lambda^3 - (\tr \AAA) \, \lambda^2 + \sigma(\AAA) \, \lambda - \det \AAA = r \label{Lameq} \end{equation} where $r$ is either of the two roots of \begin{equation} r^2 + 4\Phi(a,b,c) \, r - \Big| [a,b,c] \Big|^2 = 0 \label{Req} \end{equation} with $a,b,c$ as defined by (\ref{Three}) and where \begin{equation} \Phi(a,b,c) = {1\over2} \, \Re( [a,\bar{b}] c) \label{PhiEq} \end{equation}

Proof: These results were obtained using Mathematica to solve (\ref{RReal}) by brute force for real $r$ and octonionic $z$ given generic octonions $a$, $b$, $c$  %* [arXiv:math.RA/9807133] .

Furthermore, provided that $[a,b,c]\ne0$, each of $x$, $y$, and $z$ can be shown to admit an expansion in terms of 4 real parameters.

Corollary: With $\AAA$ and $r$ as above, and assuming $[a,b,c]\ne0$, \begin{equation} z = (\alpha a + \beta b + \gamma c + \delta) \left( 1 + {[a,b,c] \, r \over \Big| [a,b,c] \Big|^2} \right) \label{Zeq} \end{equation} with $\alpha,\beta,\gamma,\delta\in\RR$. Similar expansions hold for $x$ and $y$.

The real parameters $\alpha,\beta,\gamma,\delta$ may be freely specified for one (nonzero) component, say $z$; the remaining components $x,y$ have a similar form which is then fully determined by (\ref{EI})–(\ref{EIII}).

Corollary: The real eigenvalues of $\bar{\AAA}$ are the same as those of $\AAA$.

Proof: Direct computation (or (\ref{trace})–(\ref{det})) shows that \begin{equation} \det \bar{\AAA} = \det \AAA - 4 \Phi(a,b,c) \end{equation} But $-4\Phi(a,b,c)$ is precisely the sum of the roots of (\ref{Req}), and replacing $\AAA$ by $\bar{\AAA}$ merely flips the sign of $r$, that is $r[\bar\AAA]=-r[\AAA]$. Thus, the 2 possible values of $\det \AAA + r[\AAA]$ are precisely the same as those for $\det\bar{\AAA} + r[\bar\AAA]$. Since $\tr{\bar{\AAA}}=\tr{\AAA}$ and $\sigma(\bar{\AAA})=\sigma(\AAA)$, (\ref{Lameq}) is unchanged.

The solutions of (\ref{Lameq}) are real, since the corresponding $24\times24$ real symmetric matrix has 24 real eigenvalues. We will refer to the 3 real solutions of (\ref{Lameq}) corresponding to a single value of $r$ as a family of eigenvalues of $\AAA$. There are thus 2 families of real eigenvalues, each corresponding to 4 independent (over $\RR$) eigenvectors.

We note several intriguing properties of these results. If $\AAA$ is in fact complex, then the only solution of (\ref{Req}) is $r=0$, and we recover the usual characteristic equation with a unique set of 3 (real) eigenvalues. If $\AAA$ is quaternionic, then one solution of (\ref{Req}) is $r=0$, leading to the standard set of 3 real eigenvalues and their corresponding quaternionic eigenvectors. However, unless $a$, $b$, $c$ involve only two independent imaginary quaternionic directions (in which case $\Phi(a,b,c)=0=[a,b,c]$), there will also be a nonzero solution for $r$, leading to a second set of 3 real eigenvalues. From the preceding corollary, we see that this second set of eigenvalues consists precisely of the usual ($r=0$) eigenvalues of $\bar{\AAA}$! Furthermore, since \begin{equation} \AAA(\ell v)=\ell(\bar{\AAA}v) \end{equation} if the components of $\AAA$ and $v$ are in $\HH$ and for imaginary $\ell\in\OO$ orthogonal to $\HH$, the eigenvectors of $\AAA$ corresponding to $r\ne0$ are precisely $\ell$ times the quaternionic ($r=0$) eigenvectors of $\bar{\AAA}$. In this sense, the octonionic eigenvalue problem for quaternionic $\AAA$ is equivalent to the quaternionic eigenvalue problem for both $\AAA$ and $\bar{\AAA}$ together. Finally, if $\AAA$ is octonionic (so that in particular $[a,b,c]\ne0$), then there are two distinct solutions for $r$, and hence two different sets of real eigenvalues, with corresponding eigenvectors. Note that if $\det{\AAA}=0\ne[a,b,c]$ then all of the eigenvalues of $\AAA$ will be nonzero!

The final surprise lies with the orthogonality condition for eigenvectors $v,w$ corresponding to different eigenvalues. As in the $2\times2$ case considered in § 13.3, it is not true that $v^\dagger w=0$, although the real part of this expression does vanish, that is, \begin{equation} v^\dagger w + w^\dagger v = 0 \end{equation} However, just as in the $2\times2$ case, what is needed to ensure a decomposition of the form \begin{equation} \AAA = \sum_{m=1}^3 \lambda_m v_m v_m^\dagger \label{DecompGen} \end{equation} is (\ref{Ortho}), and a lengthy, direct computation verifies that (\ref{Ortho}) holds provided that both eigenvectors correspond to the same value of $r$.

Lemma: If $v$ and $w$ are eigenvectors of the $3\times3$ octonionic Hermitian matrix $\AAA$ corresponding to different real eigenvalues in the same family (same $r$ value), then $v$ and $w$ are mutually orthogonal in the sense of (\ref{Ortho}), that is, $(vv^\dagger)w=0$.

Proof: The modified characteristic equation (\ref{Lameq}) can be used to eliminate cubic and higher powers of $\lambda$ from any expression. Furthermore, given two distinct eigenvalues $\lambda_1\ne\lambda_2$, subtracting the two versions of (\ref{Lameq}) and factoring the result leads to the equation \begin{equation} (\lambda_1^2+\lambda_1\lambda_2+\lambda_2^2) - \tr{\AAA} (\lambda_1+\lambda_2) + \sigma(\AAA) = 0 \end{equation} which can be used to eliminate quadratic terms in one of the eigenvalues. 2)

For Jordan matrices, we thus obtain two decompositions of the form (\ref{DecompGen}), corresponding to the two sets of real eigenvalues. For each, the eigenvectors are fixed up to orthogonal transformations which preserve the form (\ref{Zeq}) of $z$.

Theorem: Let $\AAA$ be a $3\times3$ octonionic Hermitian matrix. Then $\AAA$ can be expanded as in (\ref{DecompGen}), where $\{v_1, v_2, v_3\}$ are orthonormal (as per (\ref{Ortho})) eigenvectors of $\AAA$ corresponding to the real eigenvalues $\lambda_m$, which belong to the same family (same $r$ value).

Note in particular that for some quaternionic matrices with determinant equal to zero, one and only one of these two decompositions will contain the eigenvalue zero.

In the $2\times2$ case, \begin{equation} (vv^\dagger)(vv^\dagger) = (v^\dagger v)(vv^\dagger) \label{Idem} \end{equation} which tells us that, for normalized $v$, $vv^\dagger$ squares to itself, and hence is idempotent. The analogous decomposition in the $2\times2$ case is thus an idempotent decomposition. But (\ref{Idem}) fails in the $3\times3$ case, so that the decomposition in Theorem 1 is therefore not an idempotent decomposition.

It is nevertheless straightforward to show that if $u$, $v$, and $w$ are orthonormal in the sense of (\ref{Ortho}) then \begin{equation} uu^\dagger + vv^\dagger + ww^\dagger = \III \label{ThreeIdent} \end{equation} since the left-hand side has eigenvalue $1$ with multiplicity $3$. This permits us to view $\{u,v,w\}$ as a basis of $\OO^3$ in the following sense

Lemma: Let $u,v,w\in\OO^3$ be orthonormal in the sense of (\ref{Ortho}) and let $g$ be any vector in $\OO^3$. Then \begin{equation} g = (uu^\dagger) \, g + (vv^\dagger) \, g + (ww^\dagger) \, g \end{equation}

Proof: This follows immediately from (\ref{ThreeIdent}).

However, another consequence of the failure of (\ref{Idem}) in the $3\times3$ case is that the Gram-Schmidt orthogonalization procedure no longer works. It appears to be fortuitous that we are nevertheless able to find orthonormal eigenvectors in the $3\times3$ case with repeated eigenvalues; we suspect that this might fail in general, perhaps already in the $4\times4$ case with an eigenvalue of multiplicity 3.

We can relate our notion of orthonormality to the usual one by noting that $n$ vectors in $\OO^n$ which are orthonormal in the sense (\ref{Ortho}) satisfy \begin{equation} vv^\dagger + … + ww^\dagger=\III \end{equation} If we define a matrix $\UUU$ whose columns are just $v,…,w$, then this statement is equivalent to \begin{equation} \UUU \UUU^\dagger = \III \end{equation} Over the quaternions, left matrix inverses are the same as right matrix inverses, and we would also have \begin{equation} \UUU^\dagger \UUU = \III \end{equation} or equivalently \begin{equation} v^\dagger v = 1 = … = w^\dagger w; \qquad v^\dagger w = 0 = … \end{equation} which is just the standard notion of orthogonality. These two notions of orthogonality fail to be equivalent over the octonions; we have been led to view the former as more fundamental.

We can now rewrite the eigenvalue equation in the form \begin{equation} \AAA \UUU = \UUU \DDD \label{Matrix} \end{equation} where $\DDD$ is a diagonal matrix whose entries are the real eigenvalues. 3) Multiplying (\ref{Matrix}) on the left by $\UUU^\dagger$ yields \begin{equation} \UUU^\dagger (\AAA\UUU) = \UUU^\dagger (\UUU\DDD) = (\UUU^\dagger \UUU) \DDD \end{equation} (since $\DDD$ is real), but this does not lead to a diagonalization of $\AAA$ since, as noted above, $\UUU^\dagger \UUU$ is not in general equal to the identity matrix. However, Theorem 1 can be rewritten as \begin{equation} \AAA = \UUU \DDD \UUU^\dagger \end{equation} so that in this sense $\AAA$ is diagonalizable. Furthermore, multiplication of (\ref{Matrix}) on the right by $\UUU^\dagger$ shows that \begin{equation} (\AAA \UUU) \UUU^\dagger = (\UUU\DDD) \UUU^\dagger = \AAA = \AAA (\UUU\UUU^\dagger) \end{equation} and this assertion of associativity can be taken as a restatement of Theorem 1. In the $2\times2$ case, this associativity holds for a single eigenvector $v$, that is, \begin{equation} (\AA v) v^\dagger = \AA (vv^\dagger) \label{Assoc} \end{equation} which leads to the elegant one-line derivation \begin{equation} \AA = \AA \left( \sum_{m=1}^2 v_m v_m^\dagger \right) = \sum_{m=1}^2 \AA (v_m v_m^\dagger) = \sum_{m=1}^2 (\AA v_m) v_m^\dagger = \sum_{m=1}^2 \lambda_m v_m v_m^\dagger \end{equation} However, (\ref{Assoc}) fails in the $3\times3$ case, and we are unaware of a correspondingly elegant proof of Theorem 1.

Our original proof of the $3\times3$ orthogonality result (\ref{MainOrtho}) used Mathematica to explicitly perform a horrendous, but exact, algebraic computation. Although Okubo did later give an analytic proof of this result, the Mathematica computation nevertheless establishes a result which would otherwise have remained merely a conjecture. This is a good example of being able to use the computer to verify one's intuition when it may not be possible to do so otherwise. This issue is further discussed in [arXiv:math.RA/9807133] .

1) Ogievetskii constructed a $6$th order polynomial satisfied by the real eigenvalues, which he called the characteristic equation. This polynomial is presumably equivalent to the modified characteristic equations (for both values of $r$) given below.
2) We used Mathematica to implement these simplifications in a brute force verification of (\ref{Ortho}) in this context, which ran for 6 hours on a SUN Sparc20 with 224 Mb of RAM [arXiv:math.RA/9807133] . An amusing footnote to this story is that the subsequent version of Mathematica was unable to finish this computation on the same hardware, although later versions (on better hardware) can reproduce the computation much more quickly.
3) Since the eigenvalues are real, it doesn't matter if we put them on the right.

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