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Let $A$ be a nilpotent square matrix, $J$ be the antidiagonal matrix with 1's on the secondary diagonal (i.e. $J^{2}=E$) and let $B=J A J.$ Suppose we conjugate the matrices $A,B$ by a matrix $C:$ $A_1=C A C^{-1}, B_1=C B C^{-1}.$

Question 1. Suppose we know the matrix $A_1$ only (but don't know matrices $A$ and $C.$) Is there any trick (i.e. a $C$-invariant construction) to find the matrix $B_1$?

Question 2. Is it possible (knowing the matrix $A_1$ only) to find the conjugated commutator $C[A,B]C^{-1}$?

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Where does the come from? – Igor Rivin Sep 23 '11 at 14:36
@Igor. It is an original problem and it come from nowhere – Melania Sep 24 '11 at 6:32
You can reformulate the question in terms of $J$, $A_1$, $B_1$ and $C$ only, and the input of $A$ seems irrelevant. Let $J_1=CJC^{-1}$, then $B_1=J_1A_1J_1$. So the possibilities for $B_1$ are related to the adjoint orbit of $J$ under the action of an appropriate general linear group. Maybe there is a nice description of this? Similarly, your second question is to find $[A_1,B_1]$, which again is determined up to conjugation by something in the adjoint orbit of $J$. – David Hill Sep 26 '11 at 20:07
@David. Thanks for ansver. Unfortunatelly I dont know any fine description of the class of matrices. – Melania Sep 28 '11 at 17:13

I don't think that there can be a positive answer to either of the two questions. The matrix $J$ is equivalent to a fixed decomposition of your intial vector space into two linear subspaces of the same dimension (in even dimensions) or of dimensions differing by one (in odd dimensions). The two subspaces are just the $\pm 1$-eigenspaces of $J$. Hence $J_1=CJC^{-1}$ represents any other decomposition of the initial space into a direct sum of this type, and if you don't know $C$, you don't know what the two spaces are. In particular, if $A_1$ respects this decomposition, you have $J_1A_1J_1=A_1$ while if $A_1$ exchanges the two spaces, then $J_1A_1J_1=-A_1$ (an both behaviors are possible for nilpotent matrices without problems). Thus there cannot be a uniform way to obtain $B_1$ from $A_1$ without knowing where the eigenspaces of $J_1$ are, which means knowing $C$ (to some extent). As similar line of argument applies to question 2, since the behavior of $[A,B]$ with respect to the decomposition can be taylored in differnt ways.

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