Without being able to solve (some version of) the characteristic equation in the $3\times3$ case, it is not possible in general to determine all the (non-real) eigenvalues of a given Hermitian octonionic matrix. It is therefore instructive to consider several explicit examples.

Example 1

Consider the matrix \begin{equation} \BBB = \begin{pmatrix}   p &   iq & kqs \\ \noalign{\smallskip} -iq &   p & jq\\ \noalign{\smallskip} -kqs & -jq & p \\ \end{pmatrix} \end{equation} where \begin{equation} s=\cos\theta+k\ell\sin\theta \label{Seq2} \end{equation} Note that $\BBB$ is quaternionic if $\theta=0$.

The real eigenvalues of $\BBB$, and corresponding orthonormal bases of eigenvectors, were given in [arXiv:math.RA/9807133] . But $\BBB$ also admits eigenvectors with eigenvalues which are not real. For instance: \begin{align} \lambda_{\hat u} = p \pm q\bar{s}: \quad& \hat u_\pm = \begin{pmatrix}i\\ 0\\ j\\\end{pmatrix} S_\pm \\ \noalign{\smallskip} \lambda_{\hat v} = p \pm q\bar{s}: \quad& \hat v_\pm = \begin{pmatrix}j\\ 2ks\\ i\\\end{pmatrix} S_\pm \\ \noalign{\smallskip} \lambda_{\hat w} = p \mp 2q\bar{s}: \quad& \hat w_\pm = \begin{pmatrix} j\\ -ks\\  i\\\end{pmatrix} S_\pm \end{align} where \begin{equation} S_\pm = \begin{cases}-k\ell\\   1\\\end{cases} \end{equation} These eigenvectors and eigenvalues reduce to the ones given in [arXiv:math.RA/9807133] when $\theta\to0$. Somewhat surprisingly, these eigenvectors (when normalized) yield a decomposition of the form (\ref{DecompGen}). Remarkably, they also yield a decomposition of the form \begin{equation} \BBB = \sum_{\alpha=1}^3 \left( v_\alpha \lambda_\alpha \right) v_\alpha^\dagger \label{DecompSp} \end{equation}

Example 2

A related example is given by the matrix \begin{equation} \hat{\BBB} = \begin{pmatrix}   p &   qi & {q\over2}ks \\ \noalign{\smallskip} -qi &   p & {q\over2}j\\ \noalign{\smallskip} -{q\over2}ks & -{q\over2}j p \\ \end{pmatrix} \end{equation} with $s$ again given by (\ref{Seq2}). We choose $\theta$ such that \begin{equation} s= {\sqrt{5}\over3} - {2\over3} k\ell \end{equation} resulting in \begin{equation} \hat{\BBB} = \begin{pmatrix}   p &   qi & {q\over6}(\sqrt{5}k+2l) \\ \noalign{\smallskip} -qi &   p & {q\over2}j\\ \noalign{\smallskip} -{q\over6}(\sqrt{5}k+2l) & -{q\over2}j & p \\ \end{pmatrix} \end{equation} The two families of real eigenvalues of $\hat{\BBB}$ turn out to be $\{p\pm q, p\mp{q\over2}(1+\sqrt{3}), p\mp{q\over2}(1-{\sqrt{3}\over2})\}$. Some eigenvectors for $\hat{\BBB}$ corresponding to eigenvalues which are not real are \begin{align} \lambda_{u_1} = (p+{\sqrt{5}\over2}q) - {q\over2}k\ell: \quad& u_1 = \begin{pmatrix}3k\\ \sqrt{5}\,j-2\,i\ell\\ 1+\sqrt{5}\,k\ell\\\end{pmatrix} \\ \noalign{\smallskip} \lambda_{u_2} = (p+{\sqrt{5}\over2}q) + {q\over2}k\ell: \quad& u_2 = \begin{pmatrix}\sqrt{5}\,k+2\ell\\ 3j\\ \sqrt{5}-k\ell\\\end{pmatrix} \\ \noalign{\smallskip} \lambda_{v_1} = (p-{\sqrt{5}\over3}q) + {2q\over3}k\ell: \quad& v_1 = \begin{pmatrix}\sqrt{5}\,j-2\,i\ell\\ 3k\\ 0\\\end{pmatrix} \\ \noalign{\smallskip} \lambda_{v_2} = (p-{\sqrt{5}\over3}q) - {2q\over3}k\ell: \quad& v_2 = \begin{pmatrix}3j\\ \sqrt{5}\,k+2\ell\\ 0\\\end{pmatrix} \\ \noalign{\smallskip} \lambda_{w_1} = (p-{\sqrt{5}\over6}q) - {q\over6}k\ell: \quad& w_1 = \begin{pmatrix}3k\\ \sqrt{5}\,j-2\,i\ell\\ -7-\sqrt{5}\,k\ell\\\end{pmatrix} \\ \noalign{\smallskip} \lambda_{w_2} = (p-{\sqrt{5}\over6}q) + {q\over6}k\ell: \quad& w_2 = \begin{pmatrix}\sqrt{5}\,k+2l\\ 3j\\ -3\sqrt{5}-3\,k\ell\\\end{pmatrix} \\ \noalign{\smallskip} \end{align} However, we have been unable to find any decompositions of $\hat{\BBB}$ involving these vectors. It is intriguing that, for instance, $v_1$ is orthogonal to both $u_1$ and $w_1$ (in the sense of (\ref{Ortho}), but that $u_1$ and $w_1$ are not orthogonal. In fact, we have shown using \textsl{Mathematica} that there is no eigenvector triple containing $w_1$ which is orthogonal in the sense of (\ref{Ortho}) Unless $w_1$ is special in some as yet to be determined sense, we are forced to conclude that neither (\ref{Ortho}) nor (\ref{DecompGen}) are generally true for eigenvectors whose eigenvalues are not real. It is curious, however, that the sum of the squares (outer products) of all six of these (normalized) vectors is indeed (twice) the identity!

Example 3

In all of the examples considered so far, the eigenvalues have been in the complex subalgebra of $\OO$ determined by the associator $[a,b,c]$ (with $a$, $b$, $c$ as in (\ref{Three}). We now give an example for which this is not the case.

Consider \begin{equation} \CCC = \begin{pmatrix}   p & iq & -q(j-i\ell-j\ell) \\ \noalign{\smallskip} -iq & p &  q(1+k+l) \\ \noalign{\smallskip} q(j-i\ell-j\ell) & -q(1-k-l) p \\ \end{pmatrix} \end{equation} which admits an eigenvector \begin{equation} v = \begin{pmatrix}j\\ l\\ 0\\\end{pmatrix} \end{equation} with eigenvalue \begin{equation} \lambda_v = p + q \, lk \end{equation} However, the associator takes the form \begin{equation} {[a,b,c] \over q^3} = [i,(j-i\ell-j\ell),(1+k+l)] = 2 (l-k) \end{equation}

Further discussion of these examples appears in [arXiv:math/0010255] .