If $A$ is a real $n\times n$ matrix such that $AB=BA$ for any real skew-symmetric $n\times n$ matrix $B$, then $A=kI_n$ for some $k\in\mathbb{R}$.

Question

My attempt was to use the fact that $A-A^\top$ is skew-symmetric. Then $A(A-A^\top)=(A-A^\top)A$ implies $AA^\top=A^\top A$, but this does not imply $A=kI$ even for $n=2$. It seems that $A-A^\top$ is not enough. I need to find another skew-symmetric matrix that helps me out. Is it possible to construct one from $A-A^\top$? Any other approach?

Answer 1

Consider the matrices $M_{ij}$ for $1\leq i \neq j \leq n$ such that $$(M_{ij})_{kl} = \begin{cases}1 & i=k,j=l\\ -1 & i=l,j=k\\ 0 & \textrm{otherwise} \end{cases}$$

These are skew-symmetric. What happens if $AM_{ij} = M_{ij}A$ for each $i,j$?

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The above hint is partially incomplete. I will complete the answer.

What is $(AM_{ij})_{ij}$? Observe that $$ (AM_{ij})_{ij} = \sum_{m=1}^{n} A_{im}(M_{ij})_{mj} $$ Only one term $m=i$ will be nonzero. So you just get $A_{ii}$ out of this. On the other hand, $$ (M_{ij}A)_{ij} = \sum_{m=1}^n (M_{ij})_{im}A_{mj} $$ Again, only the term $m=j$ will survive. This leads to $A_{jj}$. Thus, $M_{ij}A = AM_{ij}$ implies that $A_{ii} = A_{jj}$ for all $i \neq j$. The diagonal entries of $A$ are all the same. What about the non-diagonal entries?

Simple : look at $(A M_{ij})_{ii}$ now. That is $$ (AM_{ij})_{ii} = \sum_{m=1}^n A_{im}(M_{ij})_{mi} $$ again, because $m=j$ is the only surviving term, this evaluates to $-A_{ij}$. On the other hand,$$ (M_{ij}A)_{ii} = \sum_{m=1}^n (M_{ij})_{im}A_{mi} $$ with $m=j$ surviving, we get $A_{ji}$. Thus, $A_{ji} = -A_{ij}$ for all $i\ \neq j$.

We have proven that $A = kI + A'$ where $k$ is a(the only) diagonal entry of $A$, and $A'$ is skew symmetric. I leave as an exercise using the transpose carefully, that $A' = 0$ in fact, so that $A=kI$ as desired.