The group $F_4$ is the automorphism group of both the Jordan and Freudenthal products, that is, if $\XXX,\YYY\in\HHH$, and $\phi\in F_4$, then \begin{align} \phi(\XXX\circ\YYY) &= \phi(\XXX)\circ\YYY \\ \phi(\XXX*\YYY) &= \phi(\XXX)*\YYY \\ \end{align} Loosely speaking, $F_4$ allows us to change basis in $\HHH$, without affecting normalization or inner products. We have seen in § 13.6 that there is indeed an eigenvalue problem for any Jordan matrix $\XXX\in\HHH$, with exactly three real eigenvalues (counting multiplicity). We therefore expect to be able to find a basis in which $\XXX$ is diagonal. We show here that this expectation is correct, and any Jordan matrix can be diagonalized using $F_4$ transformations.

We start with a Jordan matrix in the form (\ref{Three}), and show how to diagonalize it using nested $F_4$ transformations. As discussed in [Harvey] , a set of generators for $F_4$ can be obtained by considering its $SO(9)$ subgroups, which in turn can be generated by $2\times2$ tracefree, Hermitian, octonionic matrices.

Just as for the traditional diagonalization procedure, it is first necessary to solve the characteristic equation for the eigenvalues. Let $\lambda$ be a solution of (\ref{Char}), and let $vv^\dagger\ne0$ be a solution of (\ref{JEigen}) with eigenvalue $\lambda$. 1) We assume further that the phase in $v$ is chosen such that \begin{equation} v = \begin{pmatrix}x\\ y\\ r\\\end{pmatrix} \end{equation} where $x,y\in\OO$ and $r\in\RR$. Define \begin{align} \MMM_1 &= \frac{1}{N_1} \begin{pmatrix} -r& 0& x\\ 0& N_1& 0\\ \bar{x}& 0& r\\ \end{pmatrix} \\ \MMM_2 &= \frac{1}{N_2} \begin{pmatrix} N_2& 0& 0\\ 0& -N_1& y\\ 0& \bar{y}& N_1\\ \end{pmatrix} \end{align} where the normalization constants are given by $N_1^2=|x|^2+r^2$ and $N_2^2=N_1^2+|y|^2\equiv v^\dagger v\ne0$. (If $N_1=0$, then $\AAA$ is already block diagonal.) It is straightforward to check that \begin{equation} \MMM_2 \MMM_1 v = \begin{pmatrix}0\\ 0\\ 1\\\end{pmatrix} \end{equation} and, since everything so far is quaternionic, this implies \begin{equation} \MMM_2 \MMM_1 vv^\dagger \MMM_1 \MMM_2 = \begin{pmatrix}0&0&0\\ 0&0&0\\0&0&1\\\end{pmatrix} =: \EEE_3 \end{equation} since there are no associativity problems.

But conjugation by each of the $\MMM_i$ is an $F_4$ transformation (which is well-defined since each $\MMM_i$ separately has components which lie in a complex subalgebra of $\OO$); this is precisely the form of the generators referred to earlier. Furthermore, $F_4$ is the automorphism group of the Jordan product (\ref{JProd}). Thus, since \begin{equation} (\AAA-\lambda\,vv^\dagger) \circ vv^\dagger = 0 \end{equation} then after applying the (nested!) $F_4$ transformation above, we obtain \begin{equation} \Bigg( \MMM_2 \Big(\MMM_1 (\AAA-\lambda\,\III) \MMM_1\Big) \MMM_2 \Bigg) \circ \EEE_3 = 0 \end{equation} which in turn forces \begin{equation} \MMM_2 (\MMM_1 \AAA \MMM_1) \MMM_2 = \begin{pmatrix}\XX& 0\\ 0& \lambda\\\end{pmatrix} \end{equation} where \begin{equation} \XX = \begin{pmatrix}s& z\\ \bar{z}& t\\\end{pmatrix} \end{equation} is a $2\times2$ octonionic Hermitian matrix (with $z\in\OO$ and $s,t\in\RR$).

The final step amounts to the diagonalization of $\XX$, which is easy. Let $\mu$ be any eigenvalue of $\XX$ (which in fact means that it is another solution of (\ref{Char})) and set \begin{equation} \MMM_3 = \frac{1}{N_3} \begin{pmatrix} \mu-t& 0& 0\\ 0& t-\mu& z\\ 0& \bar{z}& N_3\\ \end{pmatrix} \end{equation} where $N_3=(\mu-t)^2+|z|^2$. (If $N_3=0$, $\XX$ is already diagonal.) This finally results in \begin{equation} \MMM_3 \Big(\MMM_2 \left(\MMM_1 \AAA \MMM_1\right) \MMM_2\Big) \MMM_3 = \begin{pmatrix}\mu& 0& 0\\ 0& \tr(\XX)-\mu& 0\\ 0& 0& \lambda\\\end{pmatrix} \end{equation} and we have succeeded in diagonalizing $\AAA$ using $F_4$ as claimed.