For $n \geq 2$ a fixed integer, how to prove that there is no polynomial $P\in \mathbb{C}[X]$ such that for every matrix $A \in \mathcal{M}_{n}(\mathbb{C})$ : $P(A)$ is diagonalisable? Edit : $ P \neq 0$

  • $\begingroup$ Guessing here: perhaps explicitly work out $P(A)$ for every nontrivial Jordan block matrix $A$ (that is, constant on the diagonal and $1$ on the first superdiagonal)? $\endgroup$ – Greg Martin Jan 13 at 20:03
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    $\begingroup$ also needs the polynomials nonconstant $\endgroup$ – Will Jagy Jan 13 at 20:45

You need $p$ to be non-constant, otherwise $p=1$ is a counterexample.

If $p$ is non-constant, then $p'(z)\ne0$ for some $z$. Let $J$ be the nilpotent Jordan block of size $n$ and $A=zI_n+J$, i.e. let $A$ be the $n\times n$ Jordan block for the eigenvalue $z$. Then $$ p(A)=p(z)I_n+p'(z)J+\frac{p''(z)}{2!}J^2+\cdots+\frac{p^{(n-1)}(z)}{(n-1)!}J^{n-1} $$ is not diagonalisable because all eigenvalues of $p(A)$ are equal to $p(z)$ but $p(A)\ne p(z)I_n$.


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