Describe the space of solutions of the matrix equation $UV^T=VU^T$

I would like to describe the space of solutions to the following matrix equation. Here $U$ and $V$ are two unknown real matrices with $n$ lines and $p$ columns (i.e. they belong to $M_{n,p}(\mathbb{R})$) :

$$U V^T - V U^T = 0$$

Note : The result is a square matrix with $n$ lines and $n$ columns.

What I found up to now :

• if $p=1$ then the equations are equivalent with the statement $U$ and $V$ are colinear.
• The system can be rewritten under the following form : Let $X$ be the following matrix with $2p$ lines and $n$ columns (the unknown) : $$X = \left[ \begin{matrix}U^T\\V^T\end{matrix}\right]$$ Let $\Omega$ be the following square matrix with $2p$ lines and $2p$ columns : $$\Omega = \left[ \begin{matrix}0_{p \times p} & I_{p \times p}\\-I_{p \times p} & 0_{p \times p}\end{matrix}\right]$$ Then my equations can be re-written like this : $$X^T \Omega X = 0$$

This formal similarity with isotropic spaces in symplectic geometry makes me think that there is some litterature on the subject (the difference with symplectic geometry is that we do not have a "form" since the result is not a scalar when $n \neq 1$). I however cannot find anything relevant yet. Any hint or partial answer is most welcome.

Edit : Trying to generalize the special $p=1$ case to general $p$, I found the following sufficient condition : If there exists a symmetric $p\times p$ matrix $P$ such that $U=VP$ then $(U,V)$ is a solution to the equation. Is this condition sufficient ? Can it be re-written so that it highlights the symmetric role of $U$ and $V$ ?

• This looks a lot like a sylvester equation Jun 25, 2015 at 15:50
• Not really. In my equation, BOTH $U$ and $V$ are unknowns, whereas in a Sylverster equation $AX + XB = C$, there is only a single unknown $X$. Aditionnaly, Sylvester equations are linear, whereas mine is non-linear. Jun 25, 2015 at 15:55
• I didn't notice that both were unknown here, my mistake Jun 25, 2015 at 15:58
• No problem, this was indeed not clear. I'll edit the question. Jun 25, 2015 at 15:58
• Here's an observation (might not really be helpful): if we take $X$ to be $$X = \pmatrix{U\\V}$$ and take $\Omega$ to be the $2n \times 2n$ matrix of the same form as the matrix you presented, then we once again can write our equality as $X^T\Omega X = 0$. If we restrict our $U$ and $V$ so that the columns of $X$ are orthonormal, then finding $X$ corresponds to finding an $n \times n$ zero principal sub-matrix of $M\Omega M^T$ for some orthogonal matrix $M$. Jun 25, 2015 at 16:07

We assume $n=p$. Then $Z=\{U,V\in M_n(\mathbb{C})|UV^T=VU^T\}$ is an algebraic variety; let $\Delta_n=dim(Z)$, that is the maximal dimension of its connected components. Note that we work over the complex numbers.

Proposition: $\Delta_n=(3n^2+n)/2$.

Proof: i) Let $U\in M_n$ be a fixed matrix. According to Horn, Sergeichuk, ArXiv 0709.2473, lemma 2, $U$ is congruent to $diag(B,J_{r_1},\cdots,J_{r_s})$ where $J_k$ is the nilpotent Jordan block of dimension $k$ and $B$ is invertible. We give a proof when $U=diag(B,J),B\in M_p,J\in M_q,p+q=n$; indeed, the general case works in a same way and gives the same result. Putting $V^T=\begin{pmatrix}P&Q\\R&S\end{pmatrix}$, we obtain $BP,JS$ symmetric and $Q=B^{-1}(JR)^T$. $P$ depends on $p(p+1)/2$ parameters, $S$ depends on $q(q+1)/2$ parameters and $(Q,R)$ depends on $pq$ parameters. Finally $V^T$ or $V$ depends on $n(n+1)/2$ independent parameters that is a function of $n$ only.

ii) The maximal dimension with respect to $U$ is obtained for $p=n$ and therefore $\Delta_n=n^2+n(n+1)/2$.

EDIT . Let $\Delta_{n,p}=dim(\{U,V\in M_{n,p}(\mathbb{C})|UV^T=VU^T\})$. Note that "$UV^T$ symmetric" is equivalent to $n(n-1)/2$ relations; thus $\Delta_{n,p}\geq 2np-n(n-1)/2=(4np-n^2+n)/2$. Numerical calculations seem to "show" that follows: if $p\leq n$, then $\Delta_{n,p}=np+p(p+1)/2$ and if $p\geq n$, then $\Delta_{n,p}=(4np-n^2+n)/2$.

• Extremely interesting ! My intuition is that this number becomes $np + p(p+1)/2$ when $p$ and $n$ are not constrained to be equal. I think it can be proven (I did notmanage it yet though) that $(UV^T = VU^T) \Leftrightarrow ( \exists R$ rotation, $D,D'$ diagonal such that $URD + VRD' = 0)$. This stems from the observation that $UV^T=VU^T$ for $U=VP$ and $P$ symmetric. Jun 27, 2015 at 7:57

Let $$XY^T$$ be a rank decomposition of $$UV^T$$. Then $$XY^T=YX^T$$. Therefore $$X$$ and $$Y$$ are two full-rank matrices having the same range. Hence $$Y=X\Sigma$$ for some invertible matrix $$\Sigma$$, and from $$XY^T=YX^T$$, we obtain $$X\Sigma^TX^T=X\Sigma X^T$$. Thus $$\Sigma$$ is symmetric.

Now let $$r=\operatorname{rank}(X)=\operatorname{rank}(Y)$$ and $$U=XA^T$$ and $$V^T=BY^T$$, where $$A$$ and $$B$$ are $$p\times r$$. From $$XA^TBY^T=UV^T=XY^T$$, we get $$A^TB=I_r$$. Thus $$V=YB^T=X\Sigma B^T=XA^TB\Sigma B^T=U(B\Sigma B^T)$$.

It follows that the general solution to the equation $$UV^T=VU^T$$ is given by $$V=US$$, where $$U$$ arbitrary and $$S$$ is an arbitrary symmetric matrix.

• Congrats for digging up old questions like this! What do you call a "rank decomposition"? Can you give a set of defining properties? Aug 23 at 9:36
• @G.Fougeron See Wikipedia. Aug 23 at 12:23