Now I have a matrix equation $AX-XA=0$, where $A=A^T$ is real symmetric and $X=-X^T$ is unknown and skew-symmetric. I have transformed the equation into the following, $$(I_n\otimes S-S\otimes I_n)~\mathrm{vec}X=0$$ According to relevant theory of Sylvester equation, if $X$ is general, this doesn't admit a unique solution. But is there any non-trivial skew-symmetric solution? What's the dimension of solution space? Or, how to solve it?
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Related. – Git Gud Apr 29 '14 at 01:10
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Going back to the original matrix equation, it asks which matrices $X$ commute with a given matrix $A$. While the solutions $X$ form a subspace of matrices (of whatever size $A$ is), its unclear what you mean by asking "the dimension of null space". Of course the equation itself is homogeneous. – hardmath Apr 29 '14 at 01:11
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@hardmath Yes. And null space here is exactly the space of solution. – zhangwfjh Apr 29 '14 at 01:24
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@PedroTamaroff Thanks. But my field is real. And is it possible there being a unique solution when $X$ is skew-symmetric? – zhangwfjh Apr 29 '14 at 01:25
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If by "unique solution" you mean "only the trivial solution" $X=0$, then no, that is not the case for any matrix $A$. – hardmath Apr 29 '14 at 01:32
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@hardmath Oh, sorry. I mean uniqueness of nontrivial solution. – zhangwfjh Apr 29 '14 at 01:33
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1But it's a homogeneous problem. There's a whole vector space of solutions having positive dimension. – hardmath Apr 29 '14 at 01:34
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@hardmath Yes, up to a scalar. – zhangwfjh Apr 29 '14 at 01:37
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If you are suggesting the dimension might be one, this would only be true if $A$ were a $1\times 1$ matrix. – hardmath Apr 29 '14 at 02:00
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@hardmath Probably not. If $A$ is a real, $X\equiv 0$ is the unique solution, hence the dimension of solution space is zero. – zhangwfjh Apr 29 '14 at 02:12
2 Answers
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Let $A$ be real symmetric, and thus orthogonally diagonalizable:
$$ D = QAQ^T $$
If $D$ has any two diagonal entries equal, then a nontrivial skew-symmetric $Y$ will commute with $D$. Set $X = Q^T Y Q$ and you will have a skew-symmetric $X$ that commutes with $A$.
Without loss of generality we may assume the first two diagonal entries of $D$ are equal:
$$ D = \begin{pmatrix} rI & 0 \\ 0 & D' \end{pmatrix} $$
where block diagonal $rI$ is of dimension at least $2\times 2$. Then take compatible block diagonal matrix:
$$ Y = \begin{pmatrix} B & 0 \\ 0 & 0 \end{pmatrix} $$
$Y$ will now commute with $D$ for any block $B$, and in particular if $B$ is skew-symmetric, so too will $Y$ be skew-symmetric.
The rest of the verification (that $X = Q^T Y Q$ is skew-symmetric and commutes with $A$) is routine.
hardmath
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HINT: Diagonalize $A$. Typically you'll get $X = 0$ as the solution
S.B.
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I'm not sure what you mean by "typically". $X=0$ is always a solution, but there are nontrivial solutions as well. – hardmath Apr 29 '14 at 01:33
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@hardmath I mean that for generic $A$ (e.g., drawn randomly) you only get $X=0$. – S.B. Apr 29 '14 at 01:35
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@user1551 As I said for a generic $A$ the only answer is $X=0$. If I wanted to explain more about how the solutions look like depending on the eigenvalues of $A$, I would have given the full answer. – S.B. Apr 29 '14 at 10:45