Diagonalization of a complex matrix

Question

I'm trying to prove this: Let $A$ be a complex matrix. If $A^2$ is diagonalizable and $A$ is invertible then $A$ is diagonalizable.

So, if $\lambda$ is an eigenvalue of $A$ then $\lambda²$ is an eigenvalue of $A^2$. I tried proving that $\dim (Ker (A-\lambda Id)) = \dim (Ker (A^2-\lambda^2 Id))$ but I couldn't get anywhere with that.

Thanks for helping me out.

Answer 1

Hint: think about minimal polynomials. $A^2 - \lambda^2 I = (A - \lambda I)(A + \lambda I)$.

Answer 2

Suppose that $A^2$ is diagonalizable. It follows that $A^2$ has a minimal polynomial of the form $$ m_{A^2}(x) = (x - \lambda_1)(x-\lambda_2)\cdots (x - \lambda_k) $$ Where $\lambda_i \neq 0$ for each $i$ and $\lambda_i\neq\lambda_j$ when $i \neq j$. We note that $A^2$ must satisfy $m(A^2)=0$, and that $(A^2 - \lambda_iI) = (A-I\sqrt\lambda_i)(A+I\sqrt \lambda_i)$. It follows that $$ m_A^2(A^2) = (A-I\sqrt\lambda_1)(A+I\sqrt{\lambda_1})\cdots(A-I\sqrt\lambda_k)(A+I\sqrt{\lambda_k}) =0 $$ So, defining $$ p(x) = (x-\sqrt\lambda_1)(x+\sqrt{\lambda_1})\cdots(x-\sqrt\lambda_k)(x+\sqrt\lambda_k) $$ we note that $A$ satisfies $p(A) = 0$. It follows that the minimal polynomial $m_{A}(x)$ of $A$ divides $p$. It therefore follows that $m_A(x)$ has no repeated factors. Thus, we conclude that $A$ is diagonalizable.