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Let $A \in \mathrm{M}_n(F)$ be a matrix over a field $F$. Consider its adjoint representation $\mathrm{ad}_A \in \mathrm{End}(\mathrm{M}_n(F))$, defined by $$ \mathrm{ad}_A(X) = [A, X] = AX - XA. $$ I am interested in understanding if there is a clear relationship between the characteristic polynomials of $A$ and $\mathrm{ad}_A$.

For example, in the $2 \times 2$ case, the characteristic polynomial of $\mathrm{ad}_A$ is $$ x^2 \left(x^2 - \mathrm{tr}(A)^2 + 4 \det(A)\right). $$ Note that the coefficients here involve $\mathrm{tr}(A)$ and $\det(A)$, which are also coefficients of the characteristic polynomial of $A$.

In general, if the characteristic polynomial of $A$ is given by $\prod_i (x - \lambda_i)$, then the characteristic polynomial of $\mathrm{ad}_A$ is of the form $\prod_{i,j} (x - (\lambda_i - \lambda_j))$. However, I’m curious if there is a more direct or intuitive relationship between the characteristic polynomials of $A$ and $\mathrm{ad}_A$, particularly one that resembles the connection observed in the $2 \times 2$ case.

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    $\begingroup$ Let me mention that it is always possible to recover the characteristic polynomial of $\DeclareMathOperator\ad{ad}\ad_A$ from that of $A$. The key point is that $\prod_{i,j}(t-(\lambda_i-\lambda_j))\in\mathbb Z[t,\lambda_1,\dots,\lambda_n]$ is a symmetric $\mathbb Z[t]$-polynomial in variables $\lambda_1,\dots,\lambda_n$. $\endgroup$
    – Z. M
    Commented Nov 14 at 15:16
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    $\begingroup$ More formally, the recoverability is in fact true for any commutative rings $F$ (not necessarily fields). It suffices to establish the universal case $F=\mathbb Z[a_{11},a_{12},\dots,a_{1n},a_{21},\dots,a_{nn}]$ with $A=(a_{ij})$. But then it suffices to establish the result after pass to its fraction field, and moreover, the splitting field of the characteristic polynomial of $A$. $\endgroup$
    – Z. M
    Commented Nov 14 at 15:20

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