I am working over $\mathbb{R}$ (though most of the story goes over any field). I am looking for linear spaces of matrices such that the restriction of the determinant to this spaces can be written (non-trivially) as the power of another polynomial. Let me give some examples where such a phenomenon appears, before asking my question in more details.

let $n = 2m$ be an even integer. Then, the restriction of the determinant to $\bigwedge^2 \mathbb{R}^n \subset \mathrm{End}(\mathbb{R}^n)$ can be written as the square of a non-zero polynomial : the pfaffian.

let $n = 2m$ be again even. Let $\mathcal{H}_{m}(\mathbb{C})$ be the set of $m \times m$ Hermitian matrices with complex coefficients. Using the matrix representation of $i$ (square root of $-1$) as: $$ i = \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix},$$ one can embed $\mathcal{H}_{m}(\mathbb{C})$ as a sub-algebra of $\mathrm{S}^{2} \mathbb{R}^{n} \subset \mathrm{End}(\mathbb{R}^{n})$. Then, it is easily checked that the following equality holds on $\mathcal{H}_{m}(\mathbb{C})$:

$$ \mathrm{det}_{\mathrm{End}(\mathbb{R}^{n})} = \left( \mathrm{det}_{\mathcal{H}_{m}(\mathbb{C})} \right)^2.$$

- let $n = 4m$. Let $\mathcal{H}_{m}(\mathbb{H})$ be the set of $m \times m$ Hermitian matrices with quaternionic coefficients. Using the $4 \times 4$ matrix representation of $i,j,k$ (square roots of $-1$ in $\mathbb{H}$), One can embed $\mathcal{H}_{m}(\mathbb{C})$ as a sub-algebra of $\mathrm{S}^{2} \mathbb{R}^{n} \subset \mathrm{End}(\mathbb{R}^{n})$. Then, it is again easily checked that the following equality holds on $\mathcal{H}_{m}(\mathbb{H})$:

$$ \mathrm{det}_{\mathrm{End}(\mathbb{R}^{n})} = \left( \mathrm{det}_{\mathcal{H}_{m}(\mathbb{H})} \right)^4.$$

**Question** : Is there a similar story for $\mathcal{H}_{3}(\mathbb{O})$?

More precisely, one can define a good notion of determinant for Hermitian $3 \times 3$ matrices with octonionic coefficients. I was wondering if there is an embedding $\mathcal{H}_{3}(\mathbb{O}) \hookrightarrow \mathrm{S}^{2} \mathbb{R}^{24} \subset \mathrm{End}(\mathbb{R}^{24})$, such that the following hold on $\mathcal{H}_{3}(\mathbb{O})$:

$$ \mathrm{det}_{\mathrm{End}(\mathbb{R}^{24})} = \left( \mathrm{det}_{\mathcal{H}_{3}(\mathbb{O})} \right)^8?$$

The algebra $\mathcal{H}_{3}(\mathbb{O})$ being non-associative, it can not be embedded **as an algebra** into a matrix algebra. On the other hand, what I am asking is (probably) considerably weaker : I just want an embedding of vector spaces $\mathcal{H}_{3}(\mathbb{O}) \hookrightarrow \mathrm{End}(\mathbb{R}^{24})$ such that the restriction of $\mathrm{det}_{\mathrm{End}(\mathbb{R}^{24})}$ to $\mathcal{H}_{3}(\mathbb{O})$ is the $8$-th power of $\mathrm{det}_{\mathcal{H}_{3}(\mathbb{O})}$.

Of course, since $\mathcal{H}_{3}(\mathbb{O})$ can not be embedded into $\mathrm{End}(\mathbb{R}^{24})$ as an algebra, it will probably be harder to check such an equality of determinants on $\mathcal{H}_{3}(\mathbb{O})$. Indeed, we can't use the representative of the Hermitian comatrix in $\mathrm{End}(\mathbb{R}^{24})$ and just multiply it with the orginal matrix to get $\mathrm{det}_{\mathcal{H}_{3}(\mathbb{O})}.Id_{24}$. But I thought that it was perhaps possible to find another trick for $\mathcal{H}_{3}(\mathbb{O})$?

Thanks for your help.