On existence and uniqueness of non-trivial solution of matrix equation $AX=XA$

Question

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?

Answer 1

HINT: Diagonalize $A$. Typically you'll get $X = 0$ as the solution

Answer 2

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.