$A,B \in M(n,\mathbb C)$ with $AB=BA$ , then does there exist $P \in M(n,\mathbb C)$ and $f(X) , g(X) \in \mathbb C[X]$ such that $A=f(P), B=g(P)$?

Question

Let $A,B \in M(n,\mathbb C)$ ; if $\exists P \in M(n, \mathbb C)$ and $f(X) , g(X) \in \mathbb C[X]$ such that $A=f(P) , B=g(P)$ then $AB=BA$ . Now suppose $AB=BA$ ; then does there exist $P \in M(n,\mathbb C)$ and $f(X) , g(X) \in \mathbb C[X]$ such that $A=f(P)$ and $ B=g(P)$ ? If this is not true in general , then what if we assume that both $A,B$ are diagonalizable ? Is it true then ?

Answer 1

Certainly, this is true in the case that $A$ and $B$ are diagonalizable: if $A$ and $B$ are diagonlizable and commute, then they are simultaneously diagonalizable. Suppose then, without loss of generality, that $A$ and $B$ are diagonal. We have $$ A = \pmatrix{a_1\\&\ddots\\&&a_n} \quad B = \pmatrix{b_1\\&\ddots\\&&b_n} $$ If we take $$ P = \pmatrix{1 \\ & \ddots \\ & &n} $$ Then it suffices to find interpolating polynomials $f,g$ such that $f(j) = a_j$ and $g(j) = b_j$ for $j = 1,\dots,n$.

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I don't think this is true in general. I believe the following is a counterexample: $$ A = \pmatrix{0&1\\&0\\&&1\\&&&1}, \quad B = \pmatrix{1\\&1\\&&0&1\\&&&0} $$