## 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
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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 $A.$