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Suppose $A$ and $B$ are nonzero, commuting, skew-symmetric, nilpotent matrices in $M_4(k)$, $k$ a field (char $k\ne 2$). Must $A=\lambda B$ for some $\lambda\in k$? I have shown that this is true for $3\times 3$ matrices, and I believe it should also be true for $4\times 4$ matrices.

Thanks in advance to anyone willing to help me with this fairly dry question.

Jared
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  • how dou you show that its true for $3\times 3$ – M.H Mar 08 '13 at 18:07
  • Brute calculation. It turns out that skew-symmetric nilpotent 3x3 matrices satisfy the equation a^2+b^2+c^2=0, where a,b, and c are the off diagonal elements. If you write two such matrices, and force them to commute, the equations show that (a,b,c) and (a',b',c') must be scalar multiples. I'm hoping to find a more elegant way, because the same method for 4x4 matrices is ugly. – Jared Mar 08 '13 at 18:10
  • You have restrictions that basically constrain the $3 \times 3$ case down to a single dimension: nilpotent, skew, and real (since you did not say skew-Hermitian). I believe if you allow complex, it will not be true for $3\times 3$ matrices. And neither true as you believe for higher dimensions, consider same spectrum but differing eigenvalues. – adam W Mar 08 '13 at 18:14
  • @adamW: There cannot be differing eigenvalues, the matrices are nilpotent. – Jim Mar 08 '13 at 18:21
  • @Jim oh... was thinking zero determinant. nilpotent then gives all eigenvalues as zero. – adam W Mar 08 '13 at 18:25

2 Answers2

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It's not true. Consider $$ A = \left[ \begin {array}{cccc} 0&0&1&-i\\ 0&0&i&1\\ -1&-i&0&0\\ i&-1&0&0 \end {array} \right],\ B = \left[ \begin {array}{cccc} 0&1&-i&0\\ -1&0&0&i\\ i&0&0&1\\ 0&-i&-1&0\end {array} \right]$$ where $i$ is a square root of $-1$.

EDIT: Or, a bit more generally, with the same $A$, $$ B = \left[ \begin{array}{cccc} 0 & a & b & c \\ -a & 0 & c & -b \\ -b & -c & 0 & a \\ -c & b & -a & 0 \end{array}\right]$$ where $a^2 + b^2 + c^2 = 0$ and $a$, $b$, $c$ are not all $0$.

EDIT: Also, try $$ A = \left[ \begin{array}{cccc} 0 & -a & b & c \\ a & 0 & -c & b \\ -b & c & 0 & a \\ -c & -b & -a & 0 \end{array}\right],\ B = \left[ \begin{array}{cccc} 0 & a & b & c \\ -a & 0 & c & -b \\ -b & -c & 0 & a \\ -c & b & -a & 0 \end{array}\right]$$ where again $a^2 + b^2 + c^2 = 0$ and $a$, $b$, $c$ are not all $0$. In particular we get counterexamples over every field of nonzero characteristic (and every field of Stufe $\le 2$).

Basically I found these examples by starting with a suitable $A$ and a general form for $B$ and (with Maple's help) solving the equations $A B - B A = 0$ and $B^2 = 0$.

Robert Israel
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  • Wonderful. Thank you. Any insight you could share on how you found these matrices would be appreciated. – Jared Mar 08 '13 at 22:38
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If the matrices are real and skew-symmetric (i.e. $A^T=-A$), then they are normal, hence diagonalizable. If they're additionally nilpotent, they must both be zero.

So, this holds in real matrices of any size, but it seems rather anticlimactic :/

rschwieb
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    To add some drama--symmetric but in the complex, then the normal condition is out. But I think that would make it an even more "dry" question (though I did not find it very dry to begin with). – adam W Mar 08 '13 at 18:52