Let $A, B$ be skew-symmetric matrices such that $AB = -BA$. Show that $AB = 0$ 
Let $A, B$ be skew-symmetric matrices such that $AB = -BA$. Show that $AB = 0$.


I know that $AB$ is skew-symmetric,because $$(AB)^t=B^tA^t=BA=-AB$$
but I don't know how show that $AB=0$.
 A: Let
$$ A = \begin{pmatrix}
 0 & 0 & 0 & -1 \\
 0 & 0 & -1 & 0 \\
 0 & 1 & 0 & 0 \\
 1 & 0 & 0 & 0
\end{pmatrix} 
\quad\text{and}\quad
B = \begin{pmatrix}
 0 & -1 & 0 & 0 \\
 1 & 0 & 0 & 0 \\
 0 & 0 & 0 & -1 \\
 0 & 0 & 1 & 0
\end{pmatrix}.$$
Then
$$ AB = -BA = \begin{pmatrix}
 0 & 0 & -1 & 0 \\
 0 & 0 & 0 & 1 \\
 1 & 0 & 0 & 0 \\
 0 & -1 & 0 & 0
\end{pmatrix} $$
but this is obviously non-zero.
Remark. This example comes from the matrix representation of the quaternion, see the Wikipedia article for instance.

Addendum. In general, let $F$ be a field whose characteristic is not equal to 2, and let $A$ and $B$ be $n\times n$ skew-symmetric matrices over $F$ such that $AB=-BA$. Then we claim the following:


*

*If $n \leq 3$, then either $A = \mathbf{0}$ or $B = \mathbf{0}$.

*If $n \geq 4$, it is possible to have $AB \neq \mathbf{0}$.


When $n \geq 4$, the assertion is obvious by the first part of this answer. Also, the cases of $n=1, 2$ are easy to tackle by a direct computation. So we move on to the case of $n = 3$.
Suppose that $n = 3$. Then $A$ and $B$ takes the form
$$ A = \begin{pmatrix} 0 & a_1 & a_2 \\ -a_1 & 0 & a_3 \\ -a_2 & -a_3 & 0 \end{pmatrix} \quad\text{and}\quad B = \begin{pmatrix} 0 & b_1 & b_2 \\ -b_1 & 0 & b_3 \\ -b_2 & -b_3 & 0 \end{pmatrix}. $$
Then a direct computation tells that $AB = -BA$ is equivalent to
$$ a_i b_j + a_j b_i = 0, \qquad \forall i, j \in \{1, 2, 3 \}. \tag{*} $$
We claim that this implies either $A=\mathbf{0}$ or $B=\mathbf{0}$, hence the conclusion still holds. To this end, we assume $A \neq \mathbf{0}$ without losing the generality. Then there exists $i_0 \in \{1,2,3\}$ such that $a_{i_0} \neq 0$. Then by $\text{(*)}$ with $i = j = i_0$, we have $b_{i_0} = 0$. Now plugging $i = i_0$ to $\text{(*)}$ shows that $b_j = 0$ for any $j$, and therefore $B=\mathbf{0}$ as desired.
