Could the product of a skew-symmetric matrix and an invertible matrix be nilpotent? Suppose that $A$ is a $d\times d$ skew-symmetric matrix, $B$ is a $d\times d$ invertible matrix and $AB$ is a nilpotent matrix. The unique example I could find is $A=O$, the null matrix.
My question: Does there exist another pair of matrices $A,B$ such that $A$ is skew-symmetric, $B$ is invertible and $AB$ is nilpotent?
I guess $A=O$ is the only possibility. Since $AB$ is nilpotent, its Jordan canonical form is \begin{bmatrix} 
   S_1 & 0 & \ldots & 0 \\ 
   0 & S_2 & \ldots & 0 \\
   \vdots & \vdots & \ddots & \vdots \\
   0 & 0 & \ldots & S_r 
\end{bmatrix}
where each of the blocks $ S_{1},S_{2},\dots ,S_{r}$ is a shift matrix (possibly of different sizes). And the column transformations can not change this matrix to be skew-symmetric.
 A: In fact, for any matrix $A$, skew-symmetric or otherwise, there is some invertible matrix $B$ such that $AB$ is nilpotent if and only if $A=0$ or $A$ is not invertible.
First and foremost, it's an elementary fact about column-equivalence that, for any two matrices $X,Y\in\Bbb F^{n\times m}$, the following are equivalent:

*

*there is some $Z\in GL(m,\Bbb F)$ such that $X=YZ$;

*$\operatorname{col}X=\operatorname{col}Y$.

As a side note, the various Gaussian-like algorithms can even provide such a $Z$ explicitly, given $X$ and $Y$.
Secondly, notice that a vector subspace $V\subseteq \Bbb F^d$ is the column space of some nilpotent matrix if and only if $V=0$ or $V\ne \Bbb F^d$. The "only if" is obvious, because nilpotent endomorphisms on a non-zero vector space cannot be surjective. For the "if" part, consider a basis $v_1,\cdots,v_d$ such that $v_1,\cdots, v_k$ is a basis of $V$. Then, consider the endomorphism $N$ such that $Nv_1=0$, $Nv_j=v_{j-1}$ for $1<j\le k+1$, and $Nv_j=0$ for $j>k+1$.
Second side note about effective computability: given $A$ you can find explicitly a basis that extends some basis of $\operatorname{col}A$, and then you can find explicitly this endomorphism.
The two facts together prove the result. Notice that skew-symmetric matrices in characteristic $\ne2$ have automatically $\det A=0$ if $d$ is odd.
