An annihilating polynomial: $M(M^2 + 4)(M^2 + (2 - 2c)) = 0$

Theorem OctonionD8.flow_annihilating. Let $c, s \in \mathbb{R}$ be arbitrary real numbers with the single hypothesis

Then the real $8\times 8$ matrix $F = F_{c,s}$ (defined below, rows and columns indexed by $\{0,1,\dots,7\}$) satisfies the matrix identity

where $I_8$ is the $8\times 8$ identity matrix, products are ordinary matrix products (in exactly this left-to-right order), and $0$ is the $8\times 8$ zero matrix. No other hypotheses are made on $c, s$.

The bilinear product on $\mathbb{R}^8$. Let $e_0,\dots,e_7$ be the standard basis of $\mathbb{R}^8$. A product $p \cdot q$ on $\mathbb{R}^8$ is defined bilinearly by

with integer structure constants $T(i,j,k)$ (so $e_i \cdot e_j = \sum_k T(i,j,k)\, e_k$) given by the following case analysis, checked in this order:

1. if $i = 0$: $T = 1$ when $k = j$, else $0$ (so $e_0\cdot e_j = e_j$);
2. else if $j = 0$: $T = 1$ when $k = i$, else $0$ (so $e_i \cdot e_0 = e_i$);
3. else if $i = j$ (both nonzero): $T = -1$ when $k = 0$, else $0$ (so $e_i\cdot e_i = -e_0$ for $i=1,\dots,7$);
4. else if $k = 0$: $T = 0$;
5. otherwise ($i, j, k$ all nonzero, $i \ne j$) the indices are mapped to points of $\mathbb{Z}/7$ by $\varphi(i) = (i + 6) \bmod 7$, i.e. $\varphi(i) = i-1$ for $i = 1,\dots,7$. The "lines" are the seven 3-element sets $L_\ell = \{\ell,\ \ell+1,\ \ell+3\} \subseteq \mathbb{Z}/7$ for $\ell \in \mathbb{Z}/7$ (arithmetic mod 7). Then $T(i,j,k) = +1$ if there is an $\ell$ with $L_\ell = \{\varphi(i),\varphi(j),\varphi(k)\}$ (as sets) and the ordered pair $(\varphi(i),\varphi(j))$ is one of $(\ell,\ell+1)$, $(\ell+1,\ell+3)$, $(\ell+3,\ell)$; otherwise $T(i,j,k) = -1$ if there is an $\ell$ with $L_\ell = \{\varphi(i),\varphi(j),\varphi(k)\}$; otherwise $T(i,j,k) = 0$.

Translated back to indices (point $p$ corresponds to index $p+1$), the lines are the index triples $\{a, a+1, a+3\}$ with indices taken cyclically in $\{1,\dots,7\}$ (e.g. $\{1,2,4\}, \{2,3,5\}, \dots, \{7,1,3\}$), and for distinct nonzero $i,j$: $e_i\cdot e_j = +e_k$ if $(i,j,k)$ is one of the cyclic orders $(a, a+1, a+3)$, $(a+1, a+3, a)$, $(a+3, a, a+1)$ of such a triple, and $e_i \cdot e_j = -e_k$ if $(i,j)$ is in the reverse order, $k$ being the third index of the unique line through $i$ and $j$. (If $k$ coincides with $i$ or $j$, the set $\{\varphi(i),\varphi(j),\varphi(k)\}$ has fewer than 3 elements and cannot be a line, so $T = 0$.)

The matrix $F_{c,s}$. For $a, b \in \mathbb{R}^8$, $R_a$ is the $8\times 8$ matrix with entries $(R_a)_{k j} = (e_j \cdot a)_k$, so $R_a x = x \cdot a$ (right multiplication by $a$); and $L_b$ is the matrix with $(L_b)_{kj} = (b\cdot e_j)_k$, so $L_b x = b \cdot x$ (left multiplication by $b$). Then

where $e_1$ and $e_2$ are the basis vectors with index $1$ and $2$ (the imaginary units corresponding to Fano points $0$ and $1$), and $c\,e_1 + s\,e_2$ is the vector with $c$ in coordinate $1$, $s$ in coordinate $2$ and $0$ elsewhere.

Edge cases. The hypothesis $c^2+s^2=1$ is satisfiable (e.g. $c = 1, s = 0$), so the statement is not vacuous; it forces $c \in [-1,1]$. At $c = 1$ (hence $s = 0$) the last factor is $F^2$ and the claim reads $F^3(F^2 + 4I_8) = 0$; at $c = -1$ (hence $s = 0$) the last factor is $F^2 + 4I_8$ and the claim reads $F(F^2+4I_8)^2 = 0$. The sign of $s$ is unrestricted. The statement asserts only that this particular degree-5 polynomial in $F$ annihilates $F$; it says nothing about minimality or about the eigenvalues individually.