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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.


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.

The exercise can also be found here.

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