A sufficient condition under which the centralizer of a matrix A equals the centralizer of A^2

Posted: December 21, 2019 in Elementary Algebra; Problems & Solutions, Linear Algebra
Tags: , , ,

Problem. Suppose that A \in 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 M_n(\mathbb{C}). In other words, the cenralizer of A equals the centralizer of A^2.

Solution. It is clear that AX=XA implies A^2X=XA^2 for any X \in M_n(\mathbb{C}). Now suppose that A^2X=XA^2 for some X \in M_n(\mathbb{C}) and put Y:=AX-XA. We want to prove that Y=0. We have

AY+YA=A(AX-XA)+(AX-XA)A=A^2X-XA^2=0

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

\displaystyle \begin{aligned}(A+\lambda 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\end{aligned}

=(-1)^mY(A-\lambda I)^m, \ \ \ \ \ \ \ \ \ \ (*)

where I is the identity matrix. Now let v be a generalized eigenvector corresponding to an eigenvalue \lambda of A. Then (A-\lambda I)^mv=0 for some integer m and thus, by (*), we have (A+\lambda I)^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 eigenectors of A, we get Yu=0 for all u \in \mathbb{C}^n, i.e. Y=0 and hence AX=XA. \ \Box

Example. The converse of the result given in the above problem is trivially false. For example, consider

A=\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix} \in M_2(\mathbb{C}).

Then A^2=A and so AX=XA if and only if A^2X=XA^2 but 0=-0 is an eigenvalue of A.

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