I'm looking to solve some matrix equations. One of the equations involves a commutator, so my question is as follows: let $A$ be a skew-self-adjoint, traceless matrix, does the equation $[X,Y] = A$ always have a self-adjoint solution? For every size of matrices.

I hope that this is a well-known fact. Perhaps it is related to the fact that the traceless skew-adjoint matrices are $\mathfrak{s}\mathfrak{u}_n$.

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This may be relevant: math.stackexchange.com/a/181516/27978 – copper.hat Dec 5 '12 at 17:12
Thanks copper.hat, but I think I need a bit more. Let $H$ be the space (over $\mathbb{R}$) of self-adjoint matrices and $S$ the space of skew-adjoints. Then $[H,H] \subset S$ and it is rather easy to show that the image spans $S$. But I want only one commutator and not a sum of them. – shamovic Dec 5 '12 at 18:11
You are right, my optimism was unfounded :-(. – copper.hat Dec 5 '12 at 18:17
But the geometric proof of this fact is really nice. So I've learned something. Thank you :) – shamovic Dec 5 '12 at 18:20

Yes, self adjoint solutions always exist.

Since $A$ is skew Hermitian and traceless, it is unitarily similar to some skew Hermitian matrix with zero diagonal (the proof is deferred to the end of this answer). So, WLOG, we may assume that $A$ has a zero diagonal. We now take $X$ as an arbitrary diagonal matrix $\mathrm{diag}(x_1,\ldots,x_n)$ with real and distinct diagonal entries. The equation $A=XY-YX$ then boils down to $(x_i-x_j)y_{ij} = a_{ij}$, which is solvable as $y_{ij}=a_{ij}/(x_i-x_j)$.

Finally, we show that $A$ is unitarily similar to a skew Hermitian matrix with zero diagonal. Firstly, unitarily diagonalize $A$ to a purely imaginary diagonal matrix $D$. Let $U$ be a unitary matrix with $\frac{1}{\sqrt{n}}(1,1,\ldots,1)$ as its last row. Then the $(n,n)$-th entry of $\tilde{A}=UDU^\ast$ is a multiple of the sum of all entries of $D$, which is zero by assumption. Do the similar for the leading principal $(n-1)\times(n-1)$ submatrix of $\tilde{A}$, and continue in this manner recursively, we get a skew Hermitian matrix with zero diagonal.

Edit: On a second thought, actually every real (resp. complex) matrix with zero trace is similar to a matrix with zero diagonal. Hence the above idea of directly solving $Y$ can be employed to show that every real (resp. complex) matrix with trace zero is the commutator of two real (resp. complex) matrices. Consequently, we have an elementary proof that the set of matrix commutators over $\mathbb{R}$ or $\mathbb{C}$ form a matrix subspace (which is the space of all matrices with zero trace).

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Thank you very much! The crucial step of showing that every traceless skew-hermitian is unitarily equivalent to one with zeroes on the diagonal was precisely what I was missing. Thank you again! – shamovic Dec 6 '12 at 11:48

For real matrices (where the trace condition is superfluous), I think I can prove it. In fact, I think I can prove more: For a generic matrix self-adjoint matrix $X$, the equation $[X,Y] = A$ is solvable for all $A$. I think you're asking about complex matrices, so I don't think this quite answers your question, but maybe it'll give some insight.

Let $S$ denote all the self-adjoint matrices. It is well known that the dimension of $S$ is $\frac{1}{2}(n^2 + n)$. Likewise, let $T$ denote the skew-self-adjoint matrices. Then the dimension of $T$ is $\frac{1}{2}(n^2-n)$.

Now, let $X\in S$ where the center of $X$, denoted $Z(X)$ has dimension $\leq n$. For example, if all of $X$s eigenspaces are one dimensional, then this holds. (This is a generic condition on $X$).

Proof: Since $P\cdot Z(X)\cdot P^{-1} = Z(PXP^{-1})$ we may assume wlog that $X$ is diagonal. If the eigenspaces are one dimensional (i.e., the diagonal entries of $X$ are all distinct), then the only things it commutes with are other diagonal matrices, so $n$ dimensions worth.

Now, define $f_X:S\rightarrow T$ by $f_X(Y) = [X,Y]$. This is a linear map from $S$ into $T$. It's kernel consist of $Z(X)\cap S$, so has dimension at most $n$. Then, by counting dimensions we have $$\dim (f_X(S)) = \dim S - \dim\ker f_X \geq \frac{1}{2}(n^2+n) - n = \frac{1}{2}(n^2-n).$$

On the other hand, since $\dim(T) = \frac{1}{2}(n^2-n)$, we have $$\dim f_X(S) \leq \frac{1}{2}(n^2-n).$$

Hence, the inequality are equalities, so $f_X$ is surjective.

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Thanks, Jason! This argument is true for symmetric matrices. The problem is that neither $S$, nor $T$ are vector spaces over $\mathbb{C}$. They are however vector spaces over $\mathbb{R}$. But isn't the $\mathbb{R}$-linear map, given by multiplication by $i$ is an isomorphism from $S$ to $T$ as vector spaces over $\mathbb{R}$? – shamovic Dec 6 '12 at 6:13
You're right that multiplication by $i$ is a (real) isomorphism from $S$ to $T$ when everything is complex, so this argument is doomed to work in that case. On the other hand, I proved something significantly stronger in the real case - you only need a single generic $X$ to make all the skew-self adjoint things (in the real case). I was hoping something could use this as a spring board to show that, in the complex case, if you allow all Xs, that should give all the skew-self adjoint traceless things. – Jason DeVito Dec 6 '12 at 13:35
Yes it does, indeed, show that a single $X$, with distinct eigenvalues will do. I though along the same lines and also through factoring the commutator through the tensor product, but I saw user1551's post. Shows you that knowing high-tech stuff can get in the way sometimes :) – shamovic Dec 6 '12 at 14:49