The general $2\times2$ octonionic Hermitian matrix can be written \begin{equation} \AA = \begin{pmatrix}p& a\\ \noalign{\smallskip} \bar{a}& m\\\end{pmatrix} \end{equation} with $p,m\in\RR$ and $a\in\OO$, and satisfies its characteristic equation \begin{equation} \AA^2 - (\tr \AA) \, \AA + (\det \AA) \, \II = 0 \end{equation} where $\tr \AA$ denotes the trace of $\AA$, and where there is no difficulty with commutativity and associativity in defining the determinant of $\AA$ as usual via \begin{equation} \det \AA = pm - |a|^2 \end{equation} since the components of $\AA$ lie in a complex subalgebra $\CC\subset\OO$. If $a=0$ the eigenvalue problem is trivial, so we assume $a\ne0$. We also set \begin{equation} v = \begin{pmatrix}x\\ y\\\end{pmatrix} \label{vform} \end{equation} with $x,y\in\OO$.

Left Eigenvalue Problem

As pointed out in § 13.1, even quaternionic Hermitian matrices can admit left eigenvalues which are not real, as is shown by the following example: \begin{equation} \begin{pmatrix}1&-i\\ \noalign{\smallskip} i&  1\\\end{pmatrix} \begin{pmatrix}1\\ \noalign{\smallskip} k\\\end{pmatrix} = \begin{pmatrix}1+j\\ \noalign{\smallskip} k+i\\\end{pmatrix} = (1+j) \begin{pmatrix}1\\ \noalign{\smallskip} k\\\end{pmatrix} \label{Unex} \end{equation}

Direct computation allows us to determine which Hermitian matrices $\AA$ admit left eigenvalues which are not real. Inserting (\ref{vform}) into the left eigenvalue equation (\ref{wrong}) leads to \begin{equation} (\lambda-p) x = a y \qquad\qquad (\lambda-m) y = \bar{a} x \end{equation} which in turn leads to \begin{equation} {\bar{a} \Bigl( (\lambda-p) x\Bigr) \over |a|^2} = y = {(\bar\lambda-m) (\bar{a} x) \over |\lambda-m|^2} \end{equation} Assuming without loss of generality that $x\ne0$ and taking the norm of both sides yields \begin{equation} |a|^2 = |\lambda-p| |\lambda-m| \label{aNorm} \end{equation} resulting in \begin{equation} {\bar{a} \Bigl( (\lambda-p) x\Bigr) \over |\lambda-p|} = {(\bar\lambda-m) (\bar{a} x) \over |\lambda-m|} \label{EqLeft} \end{equation} This equation splits into two independent parts, the terms (in the numerator) which involve the imaginary part of $\lambda$, which is nonzero by assumption, 1) and those which don't. Looking first at the latter leads to \begin{equation} p = m \end{equation} which in turn reduces (\ref{EqLeft}) to \begin{equation} \bar{a} (\lambda x) = \bar\lambda (\bar{a} x) \end{equation} which forces $a$ to be purely imaginary (and orthogonal to $\lambda$), but which puts no conditions on $x$.

Denoting the $2\times2$ identity matrix by $\II$, setting \begin{equation} \JJ(r) = \begin{pmatrix}0& -r\\ \noalign{\smallskip} r&   0\\\end{pmatrix} \end{equation} and noting that $r$ is a pure imaginary unit octonion if and only if $r^2=-1$, we have

Lemma: The set of $2\times2$ Hermitian matrices $\AA$ for which non-real left eigenvalues exist is \begin{equation} \Amat = \{ \AA = p \, \II + q \, \JJ(r) ; \quad p,q\in\RR, \, q\ne0, \, r^2=-1 \} \label{Adef} \end{equation}

The set $\Amat$ has some remarkable properties, which will be further discussed below. Without loss of generality, we can take $r=i$, so that $\AA$ takes the form \begin{equation} \AA = \begin{pmatrix}p&-iq\\ \noalign{\smallskip} iq&  p\\\end{pmatrix} \label{Aform} \end{equation} Let us find the general solution of the left eigenvalue problem for these matrices. Taking $\AA$ as in (\ref{Aform}) and $v$ as in (\ref{vform}), the left eigenvalue equation becomes \begin{align} {\lambda - p \over q} \> x &= -iy \label{Leq1}\\ {\lambda - p \over q} \> y &= ix \label{Leq2} \end{align} Taking the norm of both sides immediately yields \begin{equation} |x|^2 = |y|^2 \label{xyNorm} \end{equation} and we can normalize both of these to 1 without loss of generality. We thus obtain \begin{align} {\lambda - p \over q} &= -(iy)\bar{x} = (ix)\bar{y} \nonumber\\ &= -[i,y,\bar{x}] - i(y\bar{x}) = [i,x,\bar{y}] + i(x\bar{y}) \end{align} But since \begin{equation} [z,y,\bar{x}] = -[z,y,x] = [z,x,y] = -[z,x,\bar{y}] \end{equation} for any $z$, the two associators are equal, and we are left with \begin{equation} x \cdot y = x\bar{y}+y\bar{x} = 0 \label{xyDot} \end{equation} Thus, $x$ and $y$ correspond to orthonormal vectors in $\OO$ thought of as $\RR^8$. This argument is fully reversible; any suitably normalized $x$ and $y$ which are orthogonal yield an eigenvector of $\AA$. We have therefore shown that all matrices in $\Amat$ have the same left eigenvectors:

Lemma: The set of left eigenvectors for any matrix $\AA\in\Amat$ is given by \begin{equation} \Vvec = \left\{ \begin{pmatrix}x\\ y\\\end{pmatrix}{:}  |x|^2=|y|^2 ;  x \cdot y =0 \right\} \end{equation}

Given $x$ and $y$, the left eigenvalue is given in each case by either (\ref{Leq1}) or (\ref{Leq2}). Furthermore, left multiplication by an arbitrary octonion preserves the set $\Vvec$, so that matrices in $\Amat$ have the property that left multiplication of left eigenvectors yields another left eigenvector (albeit with a different eigenvalue). 2) It follows from (\ref{Leq1}) or (\ref{Leq2}) and (\ref{xyNorm}) that \begin{equation} |\lambda - p| = q \label{pqeq} \end{equation} Inserting (\ref{pqeq}) into either (\ref{Leq1}) or (\ref{Leq2}), multiplying both sides by $i$, and using the identities \begin{align} a \cdot (xb) &= b \cdot (\bar{x}a) \\ (ax) \cdot (bx) &= |x|^2 \> a \cdot b \\ \end{align} for any $a,b,x\in\OO$ then shows that (\ref{xyDot}) forces $\lambda\cdot i=0$, or more generally \begin{equation} \lambda\cdot a = 0 \label{ldot} \end{equation} However, (\ref{pqeq}) and (\ref{ldot}) are the only restrictions on $\lambda$, in the sense that (\ref{Leq1}) or (\ref{Leq2}) can be used to construct eigenvectors having any eigenvalue satisfying these two conditions.

Right Eigenvalue Problem

As discussed in § 13.1, the right eigenvalues of quaternionic Hermitian matrices must be real, which is a strong argument in favor of putting the eigenvalues on the right. However, as also pointed out § 13.1, there do exist octonionic Hermitian matrices which admit right eigenvalues which are not real, as is shown by the following example: \begin{equation} \begin{pmatrix}1&-i\\ \noalign{\smallskip} i&  1\\\end{pmatrix} \begin{pmatrix}j\\ \noalign{\smallskip} \ell\\\end{pmatrix} = \begin{pmatrix}j-i\ell\\ \noalign{\smallskip} \ell+k\\\end{pmatrix} = \begin{pmatrix}j\\ \noalign{\smallskip} \ell\\\end{pmatrix} (1+k\ell) \label{SpecEx} \end{equation}

Proceeding as we did for left eigenvectors, we can determine which matrices $\AA$ admit right eigenvalues which are not real. Inserting (\ref{vform}) into the right eigenvalue equation leads to \begin{equation} x (\lambda-p) = a y \qquad\qquad y (\lambda-m) = \bar{a} x \label{Right} \end{equation} which in turn leads to \begin{equation} {\bar{a} \Bigl(x (\lambda-p)\Bigr) \over |a|^2} = y = {(\bar{a} x) (\bar\lambda-m) \over |\lambda-m|^2} \end{equation} Taking the norm of both sides (and assuming $x\ne0$) again yields (\ref{aNorm}), resulting in \begin{equation}{\bar{a} \Bigl(x (\lambda-p)\Bigr) \over |\lambda-p|} = {(\bar{a} x) (\bar\lambda-m) \over |\lambda-m|} \label{EqRight} \end{equation} Just as for the left eigenvector problem, this equation splits into two independent parts, the terms (in the numerator) which involve the imaginary part of $\lambda$, which is nonzero by assumption, and those which don't. Looking first at the latter again forces $p=m$, which in turn forces $|y|=|x|$. The remaining condition is now \begin{equation} \bar{a} (x \lambda) = (\bar{a} x) \bar\lambda \end{equation} so that $\bar{a}$, $\Im(\lambda)$, and $x$ antiassociate. In particular, this forces both $a$ and $x$ to be pure imaginary, as well as \begin{align} \lambda \cdot a &= 0 \label{laz} \\ \lambda\cdot x &= 0 = a\cdot x \label{Xdot} \end{align} with corresponding identities also holding for $y$. 3) But (\ref{laz}) and (\ref{Xdot}) are conditions on $\lambda$ and $v$, not on $\AA$. We conclude that the necessary and sufficient condition for matrices to admit non-real right eigenvalues real is that $\AA\in\Amat$:

Lemma: The set of $2\times2$ Hermitian matrices $\AA$ for which non-real right eigenvalues is $\Amat$ as defined in (\ref{Adef})

Thus, all $2\times2$ Hermitian matrices which admit non-real right eigenvalues also admit non-real left eigenvalues, and vice versa!

Corollary: A $2\times2$ octonionic Hermitian matrix admits right eigenvalues which are not real if and only if it admits left eigenvalues which are not real.

Turning to the eigenvectors, inserting $p=m$ into (\ref{Right}) leads to \begin{equation} \bar{x} (ay) = \bar{y} (\bar{a}x) \end{equation} and inserting the conditions on $a$, $x$, and $y$ now leads to \begin{equation} x \cdot y = 0 \end{equation} just as for left eigenvectors. All right eigenvectors with non-real eigenvalues are hence in $\Vvec$, although the converse is false (since right eigenvectors have no real part). Furthermore, not all of the remaining elements of $\Vvec$ will be eigenvectors for any given matrix $\AA$ (since right eigenvectors have no “quaternionic” part).

Putting all of this together, typical solutions of the (right) eigenvalue problem for $\AA$ as in (\ref{Aform}) can thus be written as \begin{align} v &= n \begin{pmatrix}j\\ k\bar{s}\\\end{pmatrix} \qquad \lambda_v = p+q\bar{s} \\ \noalign{\smallskip} w &= n \begin{pmatrix}k\bar{s}\\ j\\\end{pmatrix} \qquad \lambda_w = p-q\bar{s} \label{GenEx} \end{align} where $p,q,n\in\RR$ and where \begin{equation} s=\cos\theta+k\ell\sin\theta \label{Seq} \end{equation} The example given in (\ref{SpecEx}) is a special case of the first of (\ref{GenEx}) with $p=q=n=1$ and $\theta=\pi/2$.

Further Properties

We list some further properties of the $2\times2$ non-real eigenvalue problem without proof. For further details, see [arXiv:math/0006069] .

• The eigenvalues of the right eigenvalue problem satisfy the characteristic equation \begin{equation} \lambda^2 - \lambda (\tr \AA) + (\det \AA) = [\bar{a},x,y] \, {(\lambda-p) \over |y|^2} = [a,y,x] \, {(\lambda-m) \over |x|^2} \label{CharII} \end{equation}

If the associator $[a,x,y]$ vanishes, then $\lambda$ satisfies the ordinary characteristic equation, and hence is real (since $\AA$ is complex Hermitian). Otherwise, comparing real and imaginary parts of the last two terms in (\ref{CharII}) provides an alternate derivation of $|y| = |x|$, and we recover $p=m$ as expected. Furthermore, since the left-hand-side of (\ref{CharII}) lies in a complex subalgebra of $\OO$, so does the right-hand-side, and it is then straightforward to solve for $\lambda$ by considering its real and imaginary parts. The generalized characteristic equation (\ref{CharII}) then yields the following equation for $\lambda$ \begin{equation} \Bigl(\Re(\lambda)\Bigr)^2 - \Re(\lambda) (\tr \AA) + (\det \AA) = \Bigl(\Im(\lambda)\Bigr)^2 < 0 \label{CharIIa} \end{equation} together with the requirement that \begin{equation} {[\bar{a},x,y] \over |x| |y|} = 2 \, \Im(\lambda) \label{ALT} \end{equation} The explicit form of the eigenvalues given in (\ref{GenEx}) and (\ref{Seq}) verifies that there are no further restrictions on $\lambda$ other than (\ref{laz}) and (\ref{CharIIa}). Furthermore, having shown in the previous subsection that $a$ and $x$ (and therefore also $y$) are pure imaginary, (\ref{ALT}) yields an alternate derivation that $\lambda$ is orthogonal to $a$, which is (\ref{laz}) as well as to $x$ and $y$, which is (\ref{Xdot}).

• If $\AA v=\lambda_v v$ with $v^\dagger v=1$ and $[a,x,y]=0$, then \begin{equation} \AA = \lambda_v vv^\dagger + \lambda_w ww^\dagger \end{equation} where \begin{equation} w=\begin{pmatrix}0& 1\\ 1& 0\\\end{pmatrix} v = \begin{pmatrix}y\\ x\\\end{pmatrix} \in\Vvec \label{Flip} \end{equation} and where $\lambda_v=\lambda$ is obtained by solving (\ref{Leq1}) or (\ref{Leq2}), and $\lambda_w$ is obtained from $\lambda_v$ by interchanging $x$ and $y$.

We illustrate this result by returning to the example (\ref{Unex}), for which we obtain the decomposition \begin{equation} \begin{pmatrix}1&-i\\ \noalign{\smallskip} i&  1\\\end{pmatrix} = {(1+j)\over2} \begin{pmatrix}1\\ k\\\end{pmatrix} \begin{pmatrix}1\\ k\\\end{pmatrix} ^\dagger + {(1-j)\over2} \begin{pmatrix}k\\ 1\\\end{pmatrix} \begin{pmatrix}k\\ 1\\\end{pmatrix} ^\dagger \end{equation} where the factor of $2$ is due to the normalization of the eigenvectors.

The above construction fails if $[i,x,y]\ne0$.

• Any $\AA\in\Amat$ with (normalized) $v\in\Vvec$ such that $\AA v=v\lambda_v$ can be expanded as \begin{equation} \AA = \lambda_v \> (v v^\dagger) + \lambda_w \> (w w^\dagger) \label{Surprise} \end{equation} where $w$ is defined by (\ref{Flip}) and satisfies $\AA w=w\lambda_w$.

As before, we can assume without loss of generality that $\AA$ is given by (\ref{Aform}) and that $v$ and $w$ are given by (\ref{GenEx}). Returning to our example (\ref{SpecEx}) yields the explicit decomposition \begin{equation} \begin{pmatrix}1&-i\\ \noalign{\smallskip} i&  1\\\end{pmatrix} = {(1+k\ell)\over2} \left( \begin{pmatrix}j\\ \ell\\\end{pmatrix} \begin{pmatrix}j\\ \ell\\\end{pmatrix} ^\dagger \right) + {(1-k\ell)\over2} \left( \begin{pmatrix}\ell\\ j\\\end{pmatrix} \begin{pmatrix}\ell\\ j\\\end{pmatrix} ^\dagger \right) \end{equation}

While it is true that \begin{equation} (v v^\dagger) \, w = v \, (v^\dagger w) \label{VAssoc} \end{equation} for any $v$, $w$ related by (\ref{Flip}) (but not necessarily in $\Vvec$), the decomposition (\ref{Surprise}) is surprising because the eigenvalues $\lambda_v$, $\lambda_w$ do not commute or associate with the remaining terms. Specifically, although (\ref{VAssoc}) is zero here, we have \begin{equation} \Big(\lambda_v (v v^\dagger) \Big) \, w \ne 0 \end{equation}

• Any $\AA\in\Amat$ with (normalized) $v\in\Vvec$ such that $\AA v=v\lambda_v$ can be expanded as \begin{equation} \AA = (v \lambda_v) \, v^\dagger + (w \lambda_w) \, w^\dagger \label{SurpriseII} \end{equation} where $w$ is defined by (\ref{Flip}) and satisfies $\AA w=w\lambda_w$.

The decomposition (\ref{SurpriseII}) is less surprising than (\ref{Surprise}) when one realises that orthogonality in the form \begin{equation} \Big( (v \lambda) \, v^\dagger \Big) \, w = (v \lambda) \, (v^\dagger w) = 0 \label{NewOrtho} \end{equation} holds for any $\lambda\in\OO$ and $v,w\in\Vvec$ satisfying (\ref{Flip}).