We consider the eigenvector problem $\AA v = v\lambda$ over $\OO$, and look (only) for solutions with real eigenvalues, that is, where $\lambda\in\RR$.

There are no surprises in this case, or rather the only surprise is that we can not show that $\lambda$ is real, but must assume this property separately. First of all, there is only one independent octonion in $\AA$; the components of $\AA$ live in a complex subalgebra $\CC\subset\OO$. We can write \begin{equation} \AA = \begin{pmatrix}p & \bar{a} \\ a & m\\\end{pmatrix} \label{mat2} \end{equation} with $a\in\CC$, $p,m\in\RR$, and we will assume $a\ne0$. Setting \begin{equation} v = \begin{pmatrix}x\\ y\\\end{pmatrix} \end{equation} with $x,y\in\OO$ brings the eigenvalue equation to the form \begin{align} px + \bar{a}y &= \lambda x \label{eigen2a}\\ ax + my &= \lambda y \label{eigen2b} \end{align} either of which suffices to show that $a,x,y$ associate, and therefore lie in some quaternionic subalgebra $\HH\subset\OO$. So at first sight, the eigenvalue problem for $2\times2$ octonionic Hermitian matrices with real eigenvalues reduces to the quaternionic case.

Not so fast. Each eigenvector must indeed be quaternionic in the sense above, but different eigenvectors can, together with $\AA$, determine different quaternionic subalgebras.

Let's try again. Since $\AA$ is complex, we can start by solving the complex eigenvalue problem, which allows us to write \begin{equation} \AA = \lambda_1 v_1 v_1^\dagger + \lambda_2 v_2 v_2^\dagger \label{Cdecomp} \end{equation} where the $\lambda_m$ are real, and where the components of both $\AA$ and the $v_m$ all lie in $\CC$. If we now set \begin{equation} w_m = v_m \xi_m \label{ovec2} \end{equation} for any octonions $\xi_m\in\OO$, it's clear that $w_m$ is still an eigenvector of $\AA$ with eigenvalue $\lambda_m$; there are no associativity issues here, since $\AA$ and (a single) $\xi_m$ determine (at most) a quaternionic subalgebra of $\OO$. Furthermore, if we normalize the $\xi_m$ by setting \begin{equation} |\xi_m| = 1 \end{equation} then we can replace $v_m$ by $w_m$ in (\ref{Cdecomp}) without changing anything.

We claim that all eigenvectors $v$ of $\AA$ have the form (\ref{ovec2}). One way to see this is to note that, over $\HH$, we can make at least one component of $v$ real by right-multiplying by a suitable (quaternionic) phase, which does not otherwise affect the eigenvalue equation. But the eigenvalue equation then forces the other component to lie in $\CC$, not merely in $\HH$. Thus, the result of multiplying $v$ by this phase is one of the complex eigenvectors of $\AA$. Reversing this process establishes the claim.

It is instructive to work out these properties using explicit components. From (\ref{eigen2a}) and (\ref{eigen2b}) we obtain \begin{equation} (\lambda-p)(\lambda-m)x = \bar{a}(\lambda-m)y = \bar{a}(ax) = |a|^2 x \label{eigen2det} \end{equation} where we have used the fact that $p,m,\lambda\in\RR$, so that \begin{equation} \left((\lambda-p)(\lambda-m)-|a|^2\right) x = 0 \end{equation} Assuming $x\ne0$ (if not, start over with $y$), we recover the usual characteristic equation \begin{equation} \det(\AA-\lambda\II) = 0 \label{char2} \end{equation} for the eigenvalues of $\AA$. The rest is easy; at least one of (\ref{eigen2a}) and (\ref{eigen2b}) determines $y$ in terms of $x$. None of these manipulations uses either commutativity or associativity, although alternativity is required in the last equality of (\ref{eigen2det}).

Putting this all together, the eigenvectors of the $2\times2$ octonionic Hermitian matrix (\ref{mat2}) with eigenvalue $\lambda$ can be given in either of the forms \begin{equation} v = \begin{pmatrix}|a|^2\\ a(\lambda-p)\\\end{pmatrix} \xi \qquad\qquad v = \begin{pmatrix}\bar{a}(\lambda-m)\\ |a|^2\\\end{pmatrix} \xi \end{equation} where of course $\lambda$ must solve (\ref{char2}).

There is however one last surprise. Associativity hasn't been a problem yet, since a single eigenvector involves only $a$ and $\xi$, and hence has components that live in a quaternionic subalgebra of $\OO$. So if $v_m$ are the complex eigenvectors of $\AA$ corresponding to eigenvalues $\lambda_m$, with $m=1,2$, then of course \begin{equation} v_1^\dagger v_2 = 0 \end{equation} that is, $v_1$ is orthogonal to $v_2$. However, with $w_m$ as in (\ref{ovec2}), we have \begin{equation} w_1^\dagger w_2 = (v_1\xi_1)^\dagger (v_2\xi_2) = (\bar\xi_1v_1^\dagger) (v_2\xi_2) \end{equation} which is not necessarily zero, since we can not in general move the parentheses. A simple example can be constructed using \begin{equation} v_1 = \begin{pmatrix}1\\ i\\\end{pmatrix} \qquad\qquad v_2 = \begin{pmatrix}1\\ -i\\\end{pmatrix} \end{equation} and setting $\xi_1=j$, $\xi_2=\ell$. So what does orthogonality mean?

The answer is both simple and elegant, and can be motivated by looking again at (\ref{Cdecomp}), in which we can use either $v_m$ or $w_m$ (assuming that $|\xi_m|=1$). Why? Because $\xi_m$ cancels out in the square! This suggests that the correct notion of orthogonality is given by \begin{equation} v\perp w \Longleftrightarrow (vv^\dagger)w = 0 \label{Ortho} \end{equation} and it is easy to see that this notion of orthogonality does indeed hold between any two eigenvectors of $\AA$ with different eigenvalues. In the associative case, we can move the parentheses in (\ref{Ortho}), but not in general.