# If $A\in\mathbb C^{n\times n}\!$, $\,f\in\mathbb C[t]$ and $f(A)$ diagonalizable, and $f'(A)$ is invertible, then $A$ is diagonalizble.

Suppose that $$A$$ is a square complex matrix and $$f$$ is a polynomial in $$\mathbb C[t]$$ such that $$f(A)$$ is diagonalizable. If $$f'(A)$$ is invertible, where $$f'$$ is the derivative of $$f$$, prove that $$A$$ is diagonalizable in $$\mathbb C$$.

My attempt:

Since $$f(A)$$ is diagonalizable, then the minimal polynomial of $$f(A)$$ has no multiple roots. Say $$g(t)=\prod_{i=1}^n (t-r_i)\in\mathbb C[t]$$, with $$r_i$$ distinct. And $$g(f(A))=0$$. In order to prove $$A$$ is diagonalizable in $$\mathbb C$$, we need to find a polynomial $$p(t)\in\mathbb C[t]$$ with no multiple roots such that $$A$$ is annihilated by $$p(t)$$. We know that $$p(t)|g(f(t))$$. Now the problem is how to use the fact that $$f'(A)$$ is invertible?

Let $$g(x)$$ be the minimal polynomial of $$f(A)$$. Clearly, $$g$$ has only simple roots, since $$f(A)$$ is diagonalisable. Hence $$g$$ and $$g'$$ do not have any common roots, and since the eigenvalues of $$f(A)$$ are roots of $$g$$, no root of $$g'$$ is an eigenvalue of the matrix $$f(A)$$, and therefore the matrix $$g'\big(f(A)\big)$$ is invertible.

Also, as $$f'(A)$$ is also invertible, then $$(g\circ f)'(A)=g'\big(f(A)\big)f'(A)$$ is also invertible. Thus the roots of $$(g\circ f)'(x)$$ are not eigenvalues of $$A$$.

Meanwhile $$(g\circ f)(A)=0$$, and hence the minimal polynomial $$h(x)$$ of $$A$$ divides $$g\circ f$$. If $$h$$ had a multiple root, say $$x=\lambda$$, then $$(x-\lambda)^2$$ would divide $$h$$ and $$g\circ f$$, and $$x-\lambda$$ would divide $$(g\circ f)'=(g'\circ f)f'$$.

In particular, $$(g'\circ f)(\lambda)f'(\lambda)=0$$. But in such case $$g'\big(f(A)\big)f'(A)$$ would not be invertible.

Hints:

1) $$g’(f(A))$$ is invertible.

2) if $$B$$ is a matrix, and $$p$$ is a polynomial such that $$p(B)=0$$ and $$p’(B)$$ is invertible, then $$B$$ is diagonalizable.