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I have what looks like the first half of an answer, but for some strange reason I can't see how to finish the job.

Let $A$ be an $n \times n$ real matrix with eigenvalues $x_1,...,x_n$. On the one hand, the adjoint operator $ad(A): B \mapsto [A,B]$, acting from the space $C(A)^\perp$ of symmetric matrices orthogonal to centraliser of A to the space of skew-symmetric matrices, has determinant $\prod_{1 \leq i < j \leq n} (x_i - x_j)$ (as can be seen by diagonalising $A$, and after fixing some sign conventions).

On the other hand, generically the centraliser $C(A)$ is an $n$-dimensional space spanned by $1, A, \ldots, A^{n-1}$, and the determinant of this basis is $\det(x_j^{i-1})_{1 \leq i < j \leq n}$ (again up to some normalisations).

So presumably there must be some special property of the basis $1,A,\ldots,A^{n-1}$ of the kernel of the adjoint operator $ad(A)$ which would connect the two quantities and finish the job, but I can't see it yet, though it looks very close; I have a vague feeling one wants to work somehow in the non-commutative polynomial ring $M_n({\bf R})(X)$ formed by adjoining an non-commutative indeterminate X to the matrix ring $M_n({\bf R})$, and then specialise X to A, but my algebra is not good enough to push this through immediately. The fact that $\det(T^*) = \det(T)$ for any linear transformation T also seems relevant somehow.

1

I have what looks like the first half of an answer, but for some strange reason I can't see how to finish the job.

Let $A$ be an $n \times n$ real matrix with eigenvalues $x_1,...,x_n$. On the one hand, the adjoint operator $ad(A): B \mapsto [A,B]$, acting from the space $C(A)^\perp$ of symmetric matrices orthogonal to centraliser of A to the space of skew-symmetric matrices, has determinant $\prod_{1 \leq i < j \leq n} (x_i - x_j)$ (as can be seen by diagonalising $A$, and after fixing some sign conventions).

On the other hand, generically the centraliser $C(A)$ is an $n$-dimensional space spanned by $1, A, \ldots, A^{n-1}$, and the determinant of this basis is $\det(x_j^{i-1})_{1 \leq i < j \leq n}$ (again up to some normalisations).

So presumably there must be some special property of the basis $1,A,\ldots,A^{n-1}$ of the kernel of the adjoint operator $ad(A)$ which would connect the two quantities and finish the job, but I can't see it yet, though it looks very close; I have a vague feeling one wants to work somehow in the polynomial ring $M_n({\bf R})(X)$ formed by adjoining an indeterminate X to the matrix ring $M_n({\bf R})$, and then specialise X to A, but my algebra is not good enough to push this through immediately.