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Nov 14 at 15:20 comment added Z. M 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 $A$.
Nov 14 at 15:16 comment added Z. M 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 $\lambda_1,\dots,\lambda_n$.
Nov 14 at 13:15 history asked darko CC BY-SA 4.0