Commuting Skew-symmetric Nilpotent 4x4 Matrices 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.
 A: 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 $B^2 = 0$.
A: 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 :/
