The eigenvalue problem as usually stated is to find solutions $\lambda$, $v$ (with $v\ne0$) to the equation \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:

1. The eigenvalues of complex Hermitian matrices are real.
2. Eigenvectors of a complex Hermitian matrix corresponding to different eigenvalues are orthogonal.
3. The eigenvectors of any complex Hermitian matrix (can be chosen to) form an orthonormal basis.
4. Any complex Hermitian matrix admits a decomposition in terms of an orthonormal basis of eigenvectors.

Property 1 follows by computing \begin{equation} \bar\lambda v^\dagger v = (\AA v)^\dagger v = v^\dagger \AA v = \lambda v^\dagger v \end{equation} so that if $v\ne0$ we have $v^\dagger v\ne0$, which forces $\bar\lambda=\lambda$. Property 2 follows by considering eigenvectors $v_m$ with eigenvalues $\lambda_m$. By Property 1, $\lambda_m\in\RR$. Then \begin{equation} \lambda_1 v_1^\dagger v_2 = (\AA v_1)^\dagger v_2 = v_1^\dagger \AA v_2 = \lambda_2 v_1^\dagger v_2 \end{equation} Then if $\lambda_1\ne\lambda_2$ we must have $v_1^\dagger v_2=0$, which defines orthogonality.

Property 3 follows from the previous properties using the additional fact that there are exactly the right number of 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 $\{v_m;  m=1,…,n\}$ is an orthonormal basis of eigenvectors corresponding to eigenvalues $\lambda_m$. This follows from the other properties by checking \begin{equation} \sum_{m=1}^n \lambda_m v_m v_m^\dagger 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 The proof of Property 1 uses commutativity to move the eigenvalue $\lambda$ around; this is no longer valid.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 \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. What about 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 [ 23 ]. We will return to this topic in a later chapter.