Sylvester equation

Question

Solving the Sylvester equation $$A\cdot X+X\cdot B = C,$$ where $A$ and $B$ are similar. I am aware that the solution is non-unique, however I have the information that all entries in $X$ are positive. With this added condition of positive entries, can I say that $X$ is unique? Is it possible to calculate the minimum non-negative solution?

Answer 1

By vectorizing: $$(I\otimes A' + B'\otimes I)\times vec(X) = vec(C)$$ Thus all solutions to this equation are given by $vec(X) = X_0 + N\theta$, where $X_0$ is any solution, $N$ is the right null matrix of $I\otimes A' + B'\otimes I$ and $\theta$ is any vector of appropriate size. If $X_0$ has positive entries and $I\otimes A' + B'\otimes I$ has nonempty kernel, then one can always find a vector $\theta$, such that $X_0 + N\theta$ is also a solution and has positive entries.

Thus, even if solution must have positive entries, the solution need not be unique.

If there exists a positive solution $X_0$, then you can always find $\theta$ that solves $$\min_\theta \|X_0 + N\theta\|^2_2\\ X_0 + N\theta \geq 0$$ i.e. minimal nonnegative solution. This solution is unique since $\|X_0 + N\theta\|^2_2 = X_0'X_0 + 2X_0'N\theta + \theta'(N'N)\theta$, and $N'N$ is positive definite.

Answer 2

What do you mean by "where $A,B$ are similar" ($A=PBP^{-1}$ ?). Why did you say "I am aware that the solution is non-unique" ? If $A,B$ are generic matrices (for example, if they are randomly chosen), then the solution is unique. Indeed, let $spectrum(A)=(\lambda_i)_i,spectrum(B)=(\mu_i)_i$; the eigenvalues of the linear function $f:X\rightarrow AX+XB$ are the $(\lambda_i+\mu_j)_{i,j}$ and, in general, $0\notin spectrum(f)$, that is, $f$ is a bijection. One has the same result if $A,B$ are assumed to be similar in the above sense.

Conclusion. If $C$ is also random, then the solution in $X$ has no chance of being positive.

Remark. Of course, the conclusion is totally different if $A$ and $-B$ are similar.