# Product of skew symmetric matrices

Let $$A,B$$ be skew symmetric 3-dimensional real non-zero matrices. Because dimension is odd they have non-trivial one-dimensional kernels.

• Is it true that $$AB$$ is nilpotent iff $$\text{ker}(A)$$ $$\perp$$ $$\text{ker}(B)$$? How to prove it?

The example illustrating one direction of the implication:

$$\begin{bmatrix} 0 & 1 & 2 \\ -1 & 0 & 4 \\ -2 & -4 & 0 \end{bmatrix}\begin{bmatrix} 0 & -2 & 3 \\ 2 & 0 & -1 \\ -3 & 1 & 0 \end{bmatrix}=\begin{bmatrix} -4 & 2 & -1 \\ -12 & 6 & -3 \\ -8 & 4 & -2 \end{bmatrix}=C$$

we have here $$C^2=0$$.

• $0$ is skew-symmetric and its kernel is $3$-dimensional. – logarithm May 6 at 14:55
• @logarithm Ok. Exclude this trivial case. – Widawensen May 6 at 14:56
• Interesting question. Where does it come from? – user1551 May 6 at 17:28
• @user1551 I invented it myself, making calculations with skew-symmetric matrices when I noticed that in some cases all eigenvalues of the result are 0. I have been thinking also about the generalization of this for higher dimensions but as kernels in this case can be more dimensional so they are harder for conceptualization. – Widawensen May 7 at 7:45
• @user1551 I have read in Om.'s answer that also you presented your answer (but it was cancelled from unknown reasons), I wonder what it were, I assume that as usual it was very interesting. It was also based on calculations of eigenvalues or some other method? – Widawensen May 7 at 8:01

As user1551 mentioned in his answer (deleted at the time of writing), every real $$3 \times 3$$ skew-symmetric matrix is a cross product matrix. That is, there exist two non-zero vectors $$u$$ and $$v$$ such that $$Ax=u\times x$$ and $$Bx=v\times x$$ for every $$x\in\mathbb R^3$$.
By the vector triple-product formula, we have $$ABx = [vu^T - (u^T v)I]x$$ so that $$AB = vu^T - (u^T v)I$$. Since $$AB$$ is a rank-1 update of a scalar matrix, we easily find that $$AB$$ has eigenvalues $$\{0,-u^Tv,-u^Tv\}$$.
If $$AB$$ is nilpotent, then $$AB$$ must have $$0$$ as its only (repeated) eigenvalue. This occurs if and only if $$u^Tv = 0$$, which is to say that $$u \perp v$$. Of course, $$u$$ spans the kernel of $$A$$, and $$v$$ spans the kernel of $$B$$.
We conclude that your statement is true: $$AB$$ is nilpotent if and only if $$A$$ and $$B$$ have orthogonal kernels.
• Very nice. Your proof also shows that $AB$ is nilpotent if and only if its trace is zero. – user1551 May 6 at 17:24
• @Widawensen you flatter me. User1551's answer reminded me that we could frame things in terms of cross-products and made it clear that consecutive cross-products were making things unclear, hence the need for the triple-product formula. Once I used that formula to get a neat form for $AB$, the path to the solution was clear. – Omnomnomnom May 7 at 14:29