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I encountered the following matrix equation for $A, N, Q \in \mathbb{R}^{n \times n}$ and $A^T=-A$ and $N^T=-N$

$$[X,A]+N^TXN+Q = 0$$

where $Q$ is symmetric, positive definite. My final goal is to understand under which conditions this equation has a unique symmetric, positive definite solution $X$.

In this case, we have that $$\langle [X,A]y,y\rangle=-\langle Qy,y\rangle-\langle XNy,Ny \rangle \le 0\ ,$$ so that $2 \Re\left(\langle Ay,Xy\rangle \right)=\langle Ay,Xy \rangle + \overline{\langle Ay,Xy \rangle} \le 0$ for all $y \in \mathbb{C}^n.$

Thus, if $X$ is positive and symmetric, then we must have $\Re\left(\langle AXy,y\rangle \right) \ge 0$ as well.

As a first step, I would like to know if this matrix equation has ever been studied before or if anybody sees some valuable properties of this equation.

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    $\begingroup$ What is the difference between the $*$ and $T$ superscripts (when working over $\mathbb R$)? $\endgroup$
    – Ilya Bogdanov
    Dec 27, 2016 at 13:06
  • $\begingroup$ there is no difference $\endgroup$
    – Joppi
    Dec 27, 2016 at 13:07
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    $\begingroup$ Suppose that a solution $X$ exists. Taking the trace you get $tr(Q)=0$ and hence $Q=0$; similarly $tr(N^TXN)=0$ and hence $N^TXN=0$. Thus, you need to search for $X$ satisfying $XA-AX=0$ and $XN=0$. In any case, these equations are homogenous, so that if there is a solution, then there will be many solutions. $\endgroup$
    – Andreas Thom
    Dec 29, 2016 at 11:22
  • $\begingroup$ If you are interested in existence of symmetric solution to general matrix polynomial equations, maybe arxiv.org/abs/1608.04204 is of interest to you. $\endgroup$
    – Andreas Thom
    Dec 29, 2016 at 11:35
  • $\begingroup$ @AndreasThom How do you get $\operatorname{Tr} Q = 0$? $\endgroup$
    – Federico Poloni
    Dec 29, 2016 at 13:45

2 Answers 2

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Since you look for literature pointers: similar equations appear in the study of stochastic linear-quadratic control problems. See for instance the work of Tobias Damm; for instance this paper (Damm, Mena, Stillfjord, Numerical Solution of the Finite Horizon Stochastic Linear Quadratic Control Problem), and in particular Theorem 2.1.

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We have the following linear matrix equation in $\mathrm X$

$$\mathrm X \mathrm A - \mathrm A \mathrm X + \mathrm N^{\top} \mathrm X \mathrm N + \mathrm Q = \mathrm O_n$$

where all the matrices are real and $n \times n$. Matrices $\mathrm A$ and $\mathrm N$ are skew-symmetric. Multiplying both sides by $-1$, we rewrite the matrix equation as follows

$$\mathrm A \mathrm X - \mathrm X \mathrm A - \mathrm N^{\top} \mathrm X \mathrm N = \mathrm Q$$

Exploiting the skew-symmetry of $\mathrm A$ and $\mathrm N$, we obtain

$$\mathrm A \mathrm X + \mathrm X \mathrm A^{\top} - \mathrm N \mathrm X \mathrm N^{\top} = \mathrm Q$$

Vectorizing both sides, we obtain a linear system of $n^2$ equations in $n^2$ unknowns

$$\underbrace{\left( (\mathrm I_n \otimes \mathrm A) + (\mathrm A \otimes \mathrm I_n) - (\mathrm N \otimes \mathrm N) \right)}_{=: \mathrm M} \, \mbox{vec} (\mathrm X) = \mbox{vec} (\mathrm Q)$$

If $\mathrm M$ has full rank, then the linear matrix equation has only one solution, which may or may not be positive definite. If $\mathrm M$ does not have full rank and the linear system is

  • consistent, then the linear matrix equation has infinitely many solutions, and we can look for one that is positive definite. This is the most interesting case.
  • inconsistent, then the linear matrix equation has no solution.

Let $\mathrm C_i \in \mathbb R^{n \times n}$ be the "un-vectorization" of the $i$-th row of matrix $\mathrm M$. We then have $n^2$ equations

$$\langle \mathrm C_i, \mathrm X \rangle = \langle \mathrm e_i, \mbox{vec} (\mathrm Q) \rangle$$

each defining an affine space in $\mathbb R^{n \times n}$. Relaxing the positive definiteness constraint $\mathrm X \succ \mathrm O_n$ and imposing the positive semidefiniteness constraint $\mathrm X \succeq \mathrm O_n$ instead, we have the intersection of $n^2$ affine spaces with the positive semidefinite cone, i.e., we have a (closed) spectrahedron.

  • If this spectrahedron has no interior (i.e., if it is "flat"), then there is no positive definite solution to the given linear matrix equation.

  • If this spectrahedron has an interior, then there are infinitely many positive definite solutions to the given linear matrix equation. Note that there can be no unique positive definite solution.

We can attempt to find one of the positive definite solutions by solving the following semidefinite program (SDP)

$$\begin{array}{ll} \text{minimize} & \langle \mathrm O_n, \mathrm X \rangle\\ \text{subject to} & \langle \mathrm C_1, \mathrm X \rangle \,\,= q_{11}\\ & \langle \mathrm C_2, \mathrm X \rangle \,\,= q_{21}\\ & \quad\quad\quad \vdots\\ & \langle \mathrm C_{n^2}, \mathrm X \rangle = q_{nn}\\ & \mathrm X \succeq \mathrm O_n\end{array}$$

where $q_{ij}$ is the $(i,j)$-th entry of $\mathrm Q$. The objective function is zero because we would like to find a solution in the interior of the spectrahedron, not on its boundary (where it would only be positive semidefinite). If the SDP is infeasible, then there are neither positive semidefinite nor positive definite solutions.

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