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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 $C^2=0$.

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  • $\begingroup$ $0$ is skew-symmetric and its kernel is $3$-dimensional. $\endgroup$ – logarithm May 6 at 14:55
  • $\begingroup$ @logarithm Ok. Exclude this trivial case. $\endgroup$ – Widawensen May 6 at 14:56
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    $\begingroup$ Interesting question. Where does it come from? $\endgroup$ – user1551 May 6 at 17:28
  • $\begingroup$ @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. $\endgroup$ – Widawensen May 7 at 7:45
  • $\begingroup$ @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? $\endgroup$ – Widawensen May 7 at 8:01
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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.

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    $\begingroup$ Very nice. Your proof also shows that $AB$ is nilpotent if and only if its trace is zero. $\endgroup$ – user1551 May 6 at 17:24
  • $\begingroup$ Excellent answer, I have not expected that it could be so short, the clue it seems is to use the appropriate formulas at the right moment, but to guess what formulas is some kind of art. $\endgroup$ – Widawensen May 7 at 7:54
  • $\begingroup$ @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. $\endgroup$ – Omnomnomnom May 7 at 14:29
  • $\begingroup$ @Omnomnomnom ok. so user1551 is at some degree co-author of the way which led to the solution :) $\endgroup$ – Widawensen May 7 at 15:35

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