The eigenvalue problem as usually stated is to find vectors $v\ne0$ and numbers $\lambda$ that satisfy \begin{equation} \AA v = \lambda v \label{wrong} \end{equation} for a given square matrix $\AA$, which we will assume to be complex and Hermitian $(\AA^\dagger=\AA)$. The basic properties of the eigenvalue problem for such matrices are well-understood:
Property 3 follows from the previous properties using the additional fact that there are exactly the right number of independent eigenvectors to form a basis, which we will not prove. 1) Property 4 is the statement that $\AA$ can be expanded as \begin{equation} \AA = \sum_{m=1}^n \lambda_m v_m v_m^\dagger \end{equation} where $\{v_m\}$ is an orthonormal basis of eigenvectors corresponding to the eigenvalues $\lambda_m$. This expansion follows from the other properties by checking that \begin{equation} \left(\sum_{m=1}^n \lambda_m v_m v_m^\dagger\right) v_k = \lambda_k v_k \end{equation}
Which of these properties, if any, hold over the other division algebras?
Over the quaternions, there are Hermitian matrices which admit non-real eigenvalues. For instance, we have \begin{equation} \begin{pmatrix}0& -i\cr i& 0\cr\end{pmatrix} \begin{pmatrix}1\cr k\cr\end{pmatrix} = \begin{pmatrix}j\cr i\cr\end{pmatrix} = j \begin{pmatrix}1\cr k\end{pmatrix} \end{equation} What went wrong? The proof of Property 1 uses commutativity to move the eigenvalue $\lambda$ around; this is no longer valid. Is there a way around this?
A bit of thought reveals that (\ref{wrong}) is no longer the only eigenvalue equation. The right eigenvalue problem turns out to be, well, the right eigenvalue problem, that is \begin{equation} \AA v = v\lambda \label{right} \end{equation} If the quaternionic matrix $\AA$ is Hermitian, then a careful computation shows that \begin{equation} \bar\lambda (v^\dagger v) = (\bar\lambda v^\dagger) v = (A v)^\dagger v = (v^\dagger A) v = v^\dagger (A v) = v^\dagger (v \lambda) = (v^\dagger v) \lambda \end{equation} which uses associativity, but not commutativity. Since $v^\dagger v\in\RR$, we can still conclude that $\lambda\in\RR$. Similarly, orthogonality follows from \begin{align} \lambda_1 (v_1^\dagger v_2) &= (\lambda_1 v_1^\dagger) v_2 = (A v_1)^\dagger v_2 = (v_1^\dagger A) v_2 \nonumber\\ &= v_1^\dagger (A v_2) = v_1^\dagger (v_2 \lambda_2) = (v_1^\dagger v_2) \lambda_2 \end{align} (and the fact that $\lambda_m\in\RR$). Properties 1–4 therefore hold over the quaternions, so long as the eigenvalues are written on the right, as in (\ref{right}). What happens over the octonions?
The use of associativity in the last two derivations leads one to suspect that something will go wrong. It does; even the right eigenvalues of octonionic Hermitian matrices need not be real. For instance, we have \begin{equation} \begin{pmatrix}0& -i\cr i & 0\cr\end{pmatrix} \begin{pmatrix}j\cr \ell\cr\end{pmatrix} = \begin{pmatrix}-i\ell\cr k\cr\end{pmatrix} = \begin{pmatrix}j\cr \ell\end{pmatrix} k\ell \end{equation} Nevertheless, it turns out that there is a sense in which all of Properties 1–4 hold over the octonions, at least for $2\times2$ and $3\times3$ octonionic Hermitian matrices [arXiv:math.RA/9807126] . We discuss each of these cases in turn.