On the centralizer

Suppose that $A \in \mathcal{M}_n(\mathbb{C})$ has this property that if $\lambda$ is an eigenvalue of $A$ then $-\lambda$ is not an eigenvalue of $A$ . Show that $AX=XA$ if and only if $A^2X=XA^2$ for any $X \in \mathcal{M}_n(\mathbb{C})$ . In other words the centralizer of $A$ equals the centralizer of $A^2$ .

Solution

It is clear that $AX=XA$ implies $A^2X=XA^2$ for any $X \in \mathcal{M}_n(\mathbb{C})$ . Now suppose that $A^2X=XA^2$ for some $X \in \mathcal{M}_n(\mathbb{C})$ and set $Y=AX-XA$ . We want to prove that $Y=0$ . We have

$\begin{align*}AY+YA &=A(AX-XA)+(AX-XA)A\\ &=A^2X-XA^2\\ &=0 \end{align*}$

and so $AY=-YA$ . It now follows that $A^kY=(-1)^kYA^k$ for any integer $k \geq 0$ and thus for any $\lambda \in \mathbb{C}$ and any integer $m \geq 0$ we have

$\begin{align*}(A+\lambda \mathbb{I})^mY &=\sum_{k=0}^m\binom{m}{k}\lambda^{m-k}A^kY\\ &=\sum_{k=0}^m\binom{m}{k}(-1)^k\lambda^{m-k}YA^k \\ &=(-1)^mY\sum_{k=0}^m\binom{m}{k}(-\lambda)^{m-k}A^k \\ &=(-1)^mY(A-\lambda \mathbb{I}_{n})^m \quad (*) \end{align*}$

where $\mathbb{I}$ is the identity matrix. Now let $v$ be a generalized eigenvector corresponding to an eigenvalue $\lambda$ of $A$ . Then $(A-\lambda \mathbb{I}_{n})^mv=0$ for some integer $m$ and thus, by $(*)$ we have $(A+\lambda \mathbb{I}_n)^mYv=0$ . Therefore, since we are assuming that $-\lambda$ is not an eigenvalue of $A$ , we must have $Yv=0$ . So, since every element of $\mathbb{C}^n$ is a linear combination of some generalized eigenvectors of $A$ , we get $Yu=0$ for all $u \in \mathbb{C}^n$ , i.e. $Y=0$ and hence $AX=XA$ .