A is invertible and $A^d$ is diagonalizable for some $d\geq 1$, then show that A is diagonalizable.

Question

Let A be an invertible matrix with complex entries and $d\geq 1$. Prove that if $A^d$ is diagonalizable then A is diagonalizable.

Also, What happens if we don't assume that A is invertible?

Ok, so A invertible and on $\mathbb{C}$ so, all the eigenvalues are distinct plus A invertible will imply $A^d$ will be invertible.

Say, $AC = I$

and , there exists P invertible and B diagonal such that , $PA^d P^{-1} = B$

then , $A = A^d C^{d-1} = P^{-1}BP C^{d-1} $

$\implies PAP^{-1} = BPC^{d-1}P^{-1}$

If I can say , $PC^{d-1}P^{-1}$ is diagonal I am done? Is it visible? Or is this a wrong approach?

Another thing I was thinking that maybe can solve this by some kind of induction and $A^d$ diagonalizable will imply $A^d A^{-1} = A^{d-1}$ also diagonalizable and so on, and the key point will be A is invertible.

Also, for the second part of the question, if A is not invertible then I think A will not be diagonalizable but I need better reasoning.

Edit : I just realize showing A is similar to $A^d$ will also suffice.

Answer 1

Take$$A=\begin{bmatrix}0&1\\0&0\end{bmatrix}.$$Then $A$ is not invertible, $A^2=0$, and $A$ is not diagonalizable.

However, if $A$ is invertible, its Jordan normal $J$ form will not have zeros on the main diagonal and $J^d$ is not diagonalizable unless it is already a diagonal matrix, in which case $A$ will be diagonalizable.

Answer 2

I think this works.

Since $A^d$ is diagonalizable, there exists a polynomial with only roots of multiplicity $1$ $\mu=\prod (X-\alpha_i)$ such that $\mu(A^d)=0$.

Then if you consider $\nu = \prod (X^d-\alpha_i)$, then $\nu(A)=0$. But if you choose a $d$-root $\beta_i$ of $\alpha_i$, you can write $\nu= \prod_i \prod_{k=0,..,d-1} (X-\zeta^k \beta_i)$ where $\zeta= \exp(2i\pi /n)$. Since $A^d$ is invertible, none of the $\alpha_i$ is zero, and thus $\beta_i \neq 0$. So the roots of $\nu$ are of multiplicity $1$ and we conclude that $A$ is diagonalizable.