Infinite number of square roots of a complex matrix

Question

Given A a diagonalizable matrix of $\mathcal{M}_{n}(\mathbb{C})$ which has at least one eigenvalue (other than zero) with multiplicity $\geq$ 2. Can we prove that there exists an infinite number of diagonalizable square roots of A (i.e. diagonalizable matrices $M$ such that $M^2=A$) ? Or has anyone a counter-example? I have easily shown that for some cases the number of such matrices is not finite, but I do not know where to start for this proof.

Answer 1

Note that $\pmatrix{1&z\\ 0&-1}^2=I_2$ for every scalar $z$. So, if $A=P\operatorname{diag}(\lambda_1,\lambda_1,\lambda_3,\lambda_4,\ldots,\lambda_n)P^{-1}$, then $$ B=P\left[\sqrt{\lambda_1}\pmatrix{1&z\\ 0&-1}\oplus\operatorname{diag}(\sqrt{\lambda_3},\ldots,\sqrt{\lambda_n})\right]P^{-1} $$ is a square root of $A$, where $\sqrt{\lambda_i}$ is any square root of $\lambda_i$. As $z$ is arbitrary, $B$ assumes infinitely many different values.