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

Let

$$ A = begin{pmatrix} 0 & 0 & 0 & -1 \ 0 & 0 & -1 & 0 \ 0 & 1 & 0 & 0 \ 1 & 0 & 0 & 0 end{pmatrix} quadtext{and}quad B = begin{pmatrix} 0 & -1 & 0 & 0 \ 1 & 0 & 0 & 0 \ 0 & 0 & 0 & -1 \ 0 & 0 & 1 & 0 end{pmatrix}.$$

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.

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

1. If $n leq 3$, then either $A = mathbf{0}$ or $B = mathbf{0}$.
2. 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

$$ A = begin{pmatrix} 0 & a_1 & a_2 \ -a_1 & 0 & a_3 \ -a_2 & -a_3 & 0 end{pmatrix} quadtext{and}quad B = begin{pmatrix} 0 & b_1 & b_2 \ -b_1 & 0 & b_3 \ -b_2 & -b_3 & 0 end{pmatrix}. $$

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.