I consider $1,i,j,k,l,m,n,o$ the standard basis of the (complexified if you like) octonions ($\mathbb{O}$ for the octonions).
Let $a = x\_1.1 +\ldots + x\_8.o$, $b = x\_9.1+ \ldots + x\_{16}.o$ and $c = x\_{17}.1+ \ldots +x\_{24}.o$, where $x\_1, \ldots, x\_{24}$ are indeterminates over the base field (say $\mathbb{C}$).


I denote by $L\_a$ the $8 \times 8$ matrix which represents the left multiplication by $a$ in $\mathbb{O} \simeq \mathbb{C}^8$ and $R\_a$ the $8 \times 8$ matrix which represents the right multiplication by $a$. Similar notations for $b$ and $c$. I would like to compute the characteristic polynomial of the symmetric matrix:
$$ S = \dfrac{1}{2}(R\_a L\_b L\_c + {}^{t}(R\_a L\_b L\_c)),$$
where ${}^{t} X$ is the transpose of $X$.


I have tried with Macaulay2 and this computation seems to be far beyond the possibilities my machine (which is supposed to be a quite powerful portable workstation).


A simple reforrmulation of the eigenvalue problem in a well-chosen basis (namely let $\mathbb{H}$ be the quaternionic subalgebra generated by $b$ and $c$, split $\mathbb{O}$ as $\mathbb{H} \bigoplus \mathbb{H}.e$, where $e$ is orthogonal to $\mathbb{H}$ and take a basis adapted to this decomposition) shows that:
$$ (T - \mathrm{Re}((bc)\overline{a}))^4 \ \textrm{divides} \ det(S- T.id),$$
where $\mathrm{Re}(z)$ is the real part of $z \in \mathbb{O}$.


I put $f(T) = \dfrac{det(S- T.id)}{(T - \mathrm{Re}((bc)\overline{a})^4}$. A vast number of computations over finite fields and specializing the $x\_i$ to random values suggests that $f(T)$ is in fact a square, say $f(T) = g(T)^2$, where $g$ is a quadratic polynomial in $T$.


I would like to get a closed expression of $g(T)$. May it be a clean formula involving $a,b$ and $c$ or a dirty "in coordinates" polynomial. I would really appreciate any suggestion. I would also be interested in a theoretical argument which shows that $f(T)$ is indeed a square.


Thanks a lot!



Characteristic polynomial of an $8 \times 8$ symmetric matrix with indeterminate entries related to octonionic multiplication
Let $a,b,c\in\mathbb{O}$ be octonions and consider the linear map $L:\mathbb{O}\to\mathbb{O}$ defined by
$$
L(x) = (b(cx))a = R\_aL\_bL\_c(x).
$$
One desires a formula for the characteristic polynomial of $S$, the *symmetric* part of $L$, i.e.,
$$
S(x) = \tfrac12\bigl(R\_aL\_bL\_c + {}^t(R\_aL\_bL\_c) \bigr).
$$
(I note that the OP seems to have inadvertently omitted the factor of $\tfrac12$ in the formula for $S$; this becomes apparent when one compares the claimed formula for the characteristic polynomial of $S$ when $a = b = c = 1$.)
This is equivalent to knowing the symmetric functions of the eigenvalues of the quadratic form
$$
Q(x) = L(x)\cdot x = (b(cx))a\cdot x
$$
relative to the quadratic form $Q\_0(x) = x\cdot x$.


First, note that, if any of $a$, $b$ or $c$ vanishes, then, of course,
$L$ and $Q$ vanish identically, and all of the eigenvalues of $S$ are equal to zero. Thus, we can assume that none of $a$, $b$, or $c$ vanishes. Then, dividing by
$|abc|\not=0$, we can assume that $|a| = |b| = |c| = 1$.


In this case, since $L$, being a product of elements of $\mathrm{SO}(8)$, belongs to $\mathrm{SO}(8)$, it follows that $L$ is conjugate in $\mathrm{SO}(8)$ to an element of the maximal torus ${\mathrm{SO}(2)}^4$, i.e., a blocked diagonal matrix
where the diagonal elements are the $2$-by-$2$ rotation matrices $R(\theta\_i)$ for $1\le i\le 4$. The matrix $S$ (the symmetric part of $L$) is then diagonal with double eigenvalues $\cos(\theta\_i)$, and hence the characteristic polynomial of $S$ is
$$
p(t) = (t-\cos\theta\_1)^2(t-\cos\theta\_2)^2(t-\cos\theta\_3)^2(t-\cos\theta\_4)^2.
$$
When $p(t)= t^8 + r\_1\,t^7 + r\_2\,t^6 + \cdots + r\_8$, we have $p(t) = q(t)^2$ where
$$
q(t) = t^4 + \tfrac12\,r\_1\,t^3 + \tfrac18(4r\_2-{r\_1}^2)\,t^2 + \cdots
$$
(I leave it to the interested reader to work out the formulae for the $t$ and constant coefficients of $q$ as polynomials in $r\_1,r\_2,r\_3,r\_4$).


Now, by a theorem of Dickson, $b$ and $c$ lie in a quaternion subalgebra $\mathbb{A}\subset\mathbb{O}$. Also, we can write $a = \cos\theta\,a\_0 + \sin\theta\, u$ where $a\_0\in\mathbb{A}$ is a unit vector, $u\in\mathbb{A}^\perp$ is a unit octonion, and $0\le \theta\le \tfrac12\pi$. [This expression for $a$ will be unique relative to $\mathbb{A}$ if $0<\theta<\tfrac12\pi$.]


In what follows, it will be useful to remember that elements of $\mathbb{O}$ satisfy $xy\cdot z = x\cdot z\bar y = y\cdot \bar x z$. Recalling that $\mathrm{Re}(x) = x\cdot\mathbf{1}$ (where $\mathbf{1}\in\mathbb{O}$ is the multiplicative unit), these identities imply that $\mathrm{Re}(xy) = \mathrm{Re}(yx)$ and that $\mathrm{Re}\bigl(a(bc)\bigr) = \mathrm{Re}\bigl((ab)c\bigr)$, so that $\mathrm{Re}(abc)$ is unambiguous, even though $\mathbb{O}$ is not associative. We also have
$$
\mathrm{Re}(abc) = \mathrm{Re}(bca) = \mathrm{Re}(cab) 
= \mathrm{Re}(\bar c\,\bar b\,\bar a),
$$
but note that $\mathrm{Re}(abc)\not=\mathrm{Re}(bac)$ in general.


While $\mathbb{O} = \mathbb{A}\oplus \mathbb{A} u$
(note the orthogonal direct sum) is not associative,
we have the product formula of Cayley and Dickson:
$$
(a+b\,u)(c+d\,u) = \bigl( ac - \bar d b\bigr) + (da + b\bar c)\,u
$$
for all $a,b,c,d\in\mathbb{A}$.


Writing $x\in\mathbb{O}$ as $x = x\_0 + x\_1\,u$ where $x\_i\in\mathbb{A}$
and using the Cayley-Dickson formula several times, we have
$$
\begin{aligned}
Q(x) 
&= (b(cx))a\cdot x = b(cx)\cdot x\bar a \\
&= b(c(x\_0+x\_1\,u))\cdot (x\_0+x\_1\,u)(\cos\theta\,\overline{a\_0} -\sin\theta\,u)\\
&= \bigl(bcx\_0 + (x\_1cb)\,u\bigr)\cdot\bigl((\cos\theta\,x\_0\overline{a\_0}+\sin\theta\,x\_1) + (\cos\theta\,x\_1a\_0-\sin\theta\,x\_0)\,u\bigr)\\
&=\cos\theta\,bcx\_0\cdot x\_0\overline{a\_0}
+ \sin\theta(bcx\_0\cdot x\_1-x\_1cb\cdot x\_0) + \cos\theta\,x\_1cb\cdot x\_1a\_0\\
&= \cos\theta\,bcx\_0\cdot x\_0\overline{a\_0}
+ \sin\theta(bcx\_0-x\_0\bar b\bar c)\cdot x\_1 
+ \cos\theta\,x\_1\cdot x\_1a\_0\bar b\bar c\\
&= \cos\theta\,bcx\_0a\_0\cdot x\_0
+ \sin\theta(bcx\_0-x\_0\bar b\bar c)\cdot x\_1 
+ x\_1{\cdot}x\_1\,\mathrm{Re}(\cos\theta\,a\_0\bar b\bar c)
\end{aligned}
$$
(Note that I have used $x\_1\cdot x\_1a\_0\bar b\bar c = \overline{x\_1}x\_1\cdot a\_0\bar b\bar c = |x\_1|^2 \mathrm{Re}(a\_0\bar b\bar c)$.


This expression can be further simplified. Since $b$ and $c$ are unit vectors in
$\mathbb{A}$, we can write $bc = w$, which implies that $b = w\bar c$ and hence that $\bar b\bar c = c\bar w\bar c$. Making the substitution $x\_0 = y\_0\bar c$
and $x\_1 = y\_1\bar c$ then yields $Q\_0(x) = |x\_0|^2+|x\_1|^2 = |y\_0|^2+|y\_1|^2$ while setting $v = \bar c a\_0 c$ yields
$$
\begin{aligned}
Q(x) &= \cos\theta\,wy\_0{\bar c}a\_0 \cdot y\_0\bar c
+ \sin\theta\,(wy\_0-y\_0\bar w)\cdot y\_1 
+ |y\_1|^2\,\mathrm{Re}(\cos\theta\,a\_0\bar b\bar c)\\
& = \cos\theta\,wy\_0v \cdot y\_0
+ \sin\theta\,(wy\_0-y\_0\bar w)\cdot y\_1 
+ |y\_1|^2\,\mathrm{Re}(\cos\theta\,v\bar w)
\end{aligned}
$$
Using this reduced form, it is straightforward to compute that the characteristic polynomial of $S$ is
$$
\bigl(t-\mathrm{Re}(\cos\theta\,v\bar w)\bigr)^4
\bigl(t^2-2\cos\theta\,\mathrm{Re}(v)\mathrm{Re}(w)\,t
+ \cos^2\theta\,\mathrm{Re}(v)^2w\bar w - |\mathrm{Im}(w)|^2\bigr)^2,
$$
which can also be written as
$$
\bigl(t-\mathrm{Re}(\cos\theta\,v\bar w)\bigr)^4
\bigl((t-\cos\theta\,\mathrm{Re}(v)\mathrm{Re}(w))^2
- |\mathrm{Im}(w)|^2(1-\cos^2\theta\,\mathrm{Re}(v)^2)\bigr)^2,
$$
so that its roots are $t = \mathrm{Re}(\cos\theta\,v\bar w)$ with multiplicity $4$
and
$$
t = \cos\theta\,\mathrm{Re}(v)\mathrm{Re}(w)
\pm |\mathrm{Im}(w)|\bigl(1-\cos^2\theta\,\mathrm{Re}(v)^2)^{1/2},
$$
each with multiplicity $2$.


Finally, tracing back through the definitions and normalizations, we see that the characteristic polynomial of $S$ can be written in the form
$$
\bigl(t-\mathrm{Re}(a\bar b\bar c)\bigr)^4
\bigl(\bigl(t-\mathrm{Re}(a)\,\mathrm{Re}(bc)\bigr)^2-\mathrm{Im}(a)^2\,\mathrm{Im}(bc)^2\bigr)^2.
$$


**Note on a computation:** Since the OP asked, here is a bit of
detail of the computation of the characteristic polynomial.


We can choose a basis of $\mathbb{A}\simeq\mathbb{H}$ as $(\bf{1},\bf{i},\bf{j},\bf{k})$
in such a way that $w = w\_0\,{\bf{1}}+w\_1\,\bf{i}$ while $v = v\_0\,{\bf{1}}+v\_1\,{\bf{i}}+v\_2\,{\bf{j}}+v\_3\,{\bf{k}}$. Now write out $y\_0$
in the orthonormal basis $(\bf{1},\bf{i},\bf{j},\bf{k})$ and $y\_1$
in the orthonormal basis $(-\bf{i},\bf{1},\bf{j},\bf{k})$.


Set $\lambda\_1 = c(v\_{{0}}w\_{{0}}{-}v\_{{1}}w\_{{1}})$ and $\lambda\_2=c(v\_{{0}}w\_{{0}}{+}v\_{{1}}w\_{{1}})$ where $c = \cos\theta$ and $s=\sin\theta$
(to make the matrix below more readable). Then the matrix of
the quadratic form $Q$ in this basis is
$$
\left( 
\begin{array}{cccccccc} 
\lambda\_1&0&-cw\_{{1}}v\_{{3}}&cw\_{{1}}v\_{{2}}&sw\_{{1}}&0&0&0\\
0&\lambda\_1&-cw\_{{1}}v\_{{2}}&-cw\_{{1}}v\_{{3}}&0&sw\_{{1}}&0&0\\
-cw\_{{1}}v\_{{3}}&-cw\_{{1}}v\_{{2}}&\lambda\_2&0&0&0&0&0\\
cw\_{{1}}v\_{{2}}&-cw\_{{1}}v\_{{3}}&0&\lambda\_2&0&0&0&0\\ 
sw\_{{1}}&0&0&0&\lambda\_2&0&0&0\\ 
0&sw\_{{1}}&0&0&0&\lambda\_2&0&0\\ 
0&0&0&0&0&0&\lambda\_2&0\\ 
0&0&0&0&0&0&0&\lambda\_2
\end{array} 
\right)
$$
Now tell MAPLE to compute the characteristic polynomial of this matrix and tell it to factor the result. (I confess that it's a bit surprising that $\lambda\_2$ turns out to be a root of multiplicity $4$ instead of just $2$. I don't have a 'theoretical' understanding of this. Looking at the matrix, you can see that by choosing the basis of $\mathbb{A}$ a bit more carefully, you could arrange that $v\_2=0$, and that makes it clear that the characteristic polynomial will be a square, but we knew that already.)


