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

Mathematics Asked by Carlos Andres Henao Acevedo on December 8, 2020

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 $$AB=0$$.

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.

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.

Correct answer by Sangchul Lee on December 8, 2020

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