# Polynomial making every matrix diagonalisable

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 : $$P \neq 0$$

• 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)? – Greg Martin Jan 13 at 20:03
• also needs the polynomials nonconstant – 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 $$p(A)\ne p(z)I_n$$.