In § 13.2–13.5, we considered both the left and right eigenvalue problems for $2\times2$ and $3\times3$ octonionic Hermitian matrices. In the $3\times3$ octonionic case, even if we assume that the eigenvalues are real, the eigenvalue problem does not quite behave as expected. For this case, there are 6, rather than 3, real eigenvalues, which come in 2 independent families, each consisting of 3 real eigenvalues which satisfy a modified characteristic equation rather than the usual one. Furthermore, the corresponding eigenvectors are not orthogonal in the usual sense, but do satisfy a generalized notion of orthogonality (see also [arXiv:math.RA/9807133] ). Finally, all such matrices admit a decomposition in terms of (the “squares” of) orthonormal eigenvectors. However, due to associativity problems, these matrices are not idempotents (matrices which square to themselves).

We describe here a related eigenvalue problem for $3\times3$ Hermitian octonionic matrices which does have the standard properties: There are 3 real eigenvalues, which solve the usual characteristic equation, and which lead to a decomposition in terms of orthogonal “eigenvectors” which are indeed (primitive) idempotents.

We consider the eigenmatrix problem \begin{equation} \label{JEigen} \AAA \circ \VVV = \lambda \VVV \end{equation} where $\VVV$ is itself an octonionic Hermitian matrix and $\circ$ denotes the Jordan product \begin{equation} \label{JProd} \AAA \circ \BBB = {1\over2} \left( \AAA\BBB + \BBB\AAA \right) \end{equation} which is commutative but not associative. We further restrict $\VVV$ to be a (primitive) idempotent; as discussed below, this ensures that the Jordan eigenvalue problem (\ref{JEigen}) reduces to the traditional eigenvalue problem (\ref{right}) in the non-octonionic cases. Since $\AAA$ and $\VVV$ are Jordan matrices, the left-hand side of (\ref{JEigen}) is Hermitian, which forces $\lambda$ to be real.

Suppose first that $\AAA$ is diagonal. Then the diagonal elements $p$, $m$, $n$ are clearly eigenvalues, with obvious diagonal eigenmatrices. But there are also other “eigenvalues”, namely the averages $(p+m)/2$, $(m+n)/2$, $(n+p)/2$. However, the corresponding eigenmatrices — which are related to Peirce decompositions [Jacobson, Schafer] — have only zeros on the diagonal. Thus, by (\ref{NormZero}), they can not satisfy (\ref{Cayley}), and hence can not be written in the form (\ref{ISq}). To exclude this case, we therefore restrict $\VVV$ in (\ref{JEigen}) to the Cayley plane (\ref{Cayley}), which ensures that the eigenmatrices $\VVV$ are primitive idempotents; they really do correspond to “eigenvectors” $v$. Recall that this forces the components of $\VVV$ to lie in a quaternionic subalgebra of $\OO$ (which depends on $\VVV$) even though the components of $\AAA$ may not.

Next consider the characteristic equation (\ref{Char}) in the form \begin{equation} \label{CharDet} -\det(\AAA-\lambda\,\III) = \lambda^3 - (\tr\,\AAA)\,\lambda^2 + \sigma(\AAA)\,\lambda - (\det \AAA)\,\III = 0 \end{equation} It is not at first obvious that all solutions $\lambda$ of (\ref{CharDet}) are real. To see that this is indeed the case, we note that $\AAA$ can be rewritten as a $24\times24$ real symmetric matrix, whose eigenvalues are of course real. However, as discussed in [arXiv:math.RA/9807126] . these latter eigenvalues do not satisfy the characteristic equation (\ref{CharDet})! Rather, they satisfy a modified characteristic equation of the form \begin{equation} \label{Cubic} \det(\AAA-\lambda\,\III) + r = 0 \end{equation} where $r$ is either of the roots of a quadratic equation which depends on $\AAA$. As shown explicitly using Mathematica in Figure 5 of [arXiv:math.RA/9807133] ), not only are these roots real, but they have opposite signs (or at least one is zero). But, as can be seen immediately using elementary graphing techniques, if the cubic equation (\ref{Cubic}) has 3 real roots for both a positive and a negative value of $r$, it also has 3 real roots for all values of $r$ in between, including $r=0$. This shows that (\ref{CharDet}) does indeed have 3 real roots.

Alternatively, since $F_4$ preserves both the determinant and the trace (and therefore also $\sigma$) [Harvey] , it leaves the characteristic equation invariant. Since $F_4$ can be used to diagonalize $\AAA$ (see § 13.7), and since the resulting diagonal elements clearly satisfy the characteristic equation, we have another, indirect, proof that the characteristic equation has 3 real roots. Furthermore, this shows that these roots correspond precisely to the 3 real eigenvalues whose eigenmatrices lie in the Cayley plane. We therefore reserve the word “eigenvalue” for the three solutions of the characteristic equation (\ref{CharDet}), explicitly excluding their averages. The above argument shows that these correspond to solutions $\VVV$ of (\ref{JEigen}) which lie in the Cayley plane; we will verify this explicitly below.

Restricting the eigenvalues in this way corresponds to the traditional eigenvalue problem in the following sense. If the components of $\AAA$ and $v\ne0$ lie in a quaternionic subalgebra of the octonions, then the Jordan eigenvalue problem (\ref{JEigen}) together with the restriction (\ref{Cayley}) becomes \begin{equation} \label{CEigen} \AAA \, vv^\dagger + vv^\dagger \AAA = 2 \lambda \,vv^\dagger \end{equation} Multiplying (\ref{CEigen}) on the right by $v$ and simplifying the result using the trace of (\ref{CEigen}) leads immediately to $Av=\lambda v$ (with $\lambda\in\RR$), that is, the Jordan eigenvalue equation implies the ordinary eigenvalue equation in this context. Since the converse is immediate, the Jordan eigenvalue problem (\ref{JEigen}) (with $\VVV$ restricted to the Cayley plane but $\AAA$ octonionic) is seen to be a reasonable generalization of the ordinary eigenvalue problem.

We now show how to construct eigenmatrices $\VVV$ of (\ref{JEigen}), restricted to lie in the Cayley plane, and with real eigenvalues $\lambda$ satisfying the characteristic equation (\ref{CharDet}). From the definition of the determinant, we have for real $\lambda$ satisfying (\ref{CharDet}) \begin{equation} 0 = \det(\AAA-\lambda\,\III) = (\AAA-\lambda\,\III) \circ \Big( (\AAA-\lambda\,\III)*(\AAA-\lambda\,\III) \Big) \end{equation} Thus, setting \begin{equation} \Qcal_\lambda = (\AAA-\lambda\,\III)*(\AAA-\lambda\,\III) \end{equation} we have \begin{equation} \label{QEigen} (\AAA-\lambda\,\III) \circ \Qcal_\lambda = 0 \end{equation} so that $\Qcal_\lambda$ is a solution of (\ref{JEigen}).

Due to the identity (\ref{Springer}), we have \begin{equation} \label{QSq} \Qcal_\lambda * \Qcal_\lambda=0 \end{equation} If $\Qcal_\lambda\ne0$, we can renormalize $\Qcal_\lambda$ by defining \begin{equation} \Pcal_\lambda = {\Qcal_\lambda\over\tr(\Qcal_\lambda)} \end{equation} Each resulting $\Pcal_\lambda$ is in the Cayley plane, and is hence a primitive idempotent. Due to (\ref{QSq}), we can write \begin{equation} \Pcal_\lambda=v_\lambda^{\phantom\dagger} v_\lambda^\dagger \end{equation} and we call $v_\lambda$ the (generalized) eigenvector of $\AAA$ with eigenvalue $\lambda$. Note that $v_\lambda$ does not in general satisfy either (\ref{wrong}) or (\ref{right}). Rather, we have \begin{equation} \AAA \circ v_\lambda^{\phantom\dagger} v_\lambda^\dagger = \lambda \, v_\lambda^{\phantom\dagger} v_\lambda^\dagger \end{equation} as well as \begin{equation} v_\lambda^\dagger v_\lambda^{\phantom\dagger} = 1 \end{equation}

Writing out all the terms and using the identities \begin{align} \tilde\AAA = \AAA -\tr(\AAA)\,\III &= -2 \, \III*\AAA \label{TID}\\ (\tilde \AAA\circ \AAA)\circ(\AAA*\AAA) &= (\det \AAA) \, \tilde \AAA \label{ID} \end{align} one computes directly that \begin{equation} \Qcal_\lambda\circ(\AAA\circ \Qcal_\mu) = (\Qcal_\lambda\circ \AAA)\circ \Qcal_\mu \end{equation} If $\lambda$, $\mu$ are solutions of the characteristic equation (\ref{CharDet}), then using (\ref{QEigen}) leads to \begin{equation} \mu\,(\Qcal_\lambda\circ \Qcal\mu) = \lambda\,(\Qcal_\lambda\circ \Qcal\mu) \end{equation} If we now assume $\lambda\ne\mu$ and $\Qcal_\lambda\ne0\ne \Qcal_\mu$, this shows that eigenmatrices corresponding to different eigenvalues are orthogonal in the sense \begin{equation} \Pcal_\lambda\circ \Pcal_\mu = 0 \end{equation} where we have normalized the eigenmatrices.

We now turn to the case $\Qcal_\lambda=0$. We have first that \begin{equation} \label{QID} \tr(\Qcal_\lambda) = \tr \Big( (\AAA-\lambda\,\III)*(\AAA-\lambda\,\III) \Big) = \sigma(\AAA-\lambda\,\III) \end{equation} Denoting the 3 real solutions of the characteristic equation (\ref{CharDet}) by $\lambda$, $\mu$, $\nu$, so that \begin{eqnarray} \tr\,\AAA &=& \lambda + \mu + \nu \\ \sigma(\AAA) &=& \lambda (\mu + \nu) + \mu\nu \end{eqnarray} we then have \begin{equation} \label{SigmaID} \sigma(\AAA-\lambda\,\III) = \sigma(\AAA) - 2\lambda \, \tr\,\AAA + 3\lambda^2 = (\lambda - \mu) (\lambda - \nu) \end{equation} But by (\ref{QSq}) and (\ref{NormZero}), $\Qcal_\lambda=0$ if and only if $\tr(\Qcal_\lambda)=0$. Using (\ref{QID}) and (\ref{SigmaID}), we therefore see that $\Qcal_\lambda=0$ if and only if $\lambda$ is a solution of (\ref{CharDet}) of multiplicity greater than 1. We will return to this case below.

Putting this all together, if there are no repeated solutions of the characteristic equation (\ref{CharDet}), then the eigenmatrix problem leads to the decomposition \begin{equation} \label{Decomp} \AAA = \sum_{i=1}^3 \lambda_i \Pcal_{\lambda_i} \end{equation} in terms of orthogonal primitive idempotents, which expresses each Jordan matrix $\AAA$ as a sum of squares of quaternionic columns. / \footnote{ / 1) We emphasize that the components of the eigenmatrices $\Pcal_{\lambda_i}$ need not lie in the same quaternionic subalgebra, and that $\AAA$ is octonionic. Nonetheless, it is remarkable that $\AAA$ admits a decomposition in terms of matrices which are, individually, quaternionic.

We now return to the case $\Qcal_\lambda=0$, corresponding to repeated eigenvalues. If $\lambda$ is a solution of the characteristic equation (\ref{CharDet}) of multiplicity 3, then $\tr\,\AAA=3\lambda$ and $\sigma(\AAA)=3\lambda^2$. As shown in [arXiv:math.RA/9807126] in a different context, or using an argument along the lines of Footnote ??, this forces $\AAA=\lambda\,\III$, which has a trivial decomposition into orthonormal primitive idempotents. We are left with the case of multiplicity 2, corresponding to $\AAA\ne\lambda\,\III$ and $\Qcal_\lambda=0$.

Since $\Qcal_\lambda=0$, $\AAA-\lambda\,\III$ is (up to normalization) in the Cayley plane, and we have \begin{equation} \label{IISq} \AAA - \lambda\,\III = \pm ww^\dagger \end{equation} with the components of $w$ in some quaternionic subalgebra of $\OO$. While $ww^\dagger$ is indeed an eigenmatrix of $\AAA$, it has eigenvalue $\mu=\tr(\AAA)-2\lambda\ne\lambda$. However, it is straightforward to construct a vector $v$ orthogonal to $w$ in a suitable sense. For instance, if \begin{equation} w = \begin{pmatrix}x\\ y\\ r\\\end{pmatrix} \end{equation} with $r\in\RR$, then choosing \begin{equation} v = \begin{pmatrix}|y|^2\\ -y\bar{x}\\ 0\\\end{pmatrix} \end{equation} leads to \begin{equation} \label{VWperp} vv^\dagger \circ ww^\dagger = 0 \end{equation} and only minor modifications are required to adapt this example to the general case. But (\ref{IISq}) now implies that \begin{equation} \AAA \circ vv^\dagger = \lambda \, vv^\dagger \end{equation} so that we have constructed an eigenmatrix of $\AAA$ with eigenvalue $\lambda$.

We can now perturb $\AAA$ slightly by adding $\epsilon\,vv^\dagger$, thus changing the eigenvalue of $vv^\dagger$ by $\epsilon$. The resulting matrix will have 3 unequal eigenvalues, and hence admit a decomposition (\ref{Decomp}) in terms of orthogonal primitive idempotents. But these idempotents will also be eigenmatrices of $\AAA$, and hence yield an orthogonal primitive idempotent decomposition of $\AAA$.  2) In summary, decompositions analogous to (\ref{Decomp}) can also be found when there is a repeated eigenvalue, but the terms corresponding to the repeated eigenvalue can not be written in terms of the projections $\Pcal_\lambda$, and of course the decomposition of the corresponding eigenspace is not unique. 3)