Solving underdetermined Lyapunov equation?

Question

I'm solving the following for $X$ with $A,B$ singular positive semidefinite matrices.

$$AX + XA = B$$

Because $A$, $B$ are singular, standard Lyapunov solver fails

However, if I heuristically skip dividing by 0 elements in Matlab implementation of the Lyapunov solver, I get the same answer as the least squares solution to $(A\otimes I + I\otimes A)\operatorname{vec}X=\operatorname{vec}B$

Does this appear in the literature, or does someone see a way to prove that this works?

(originally posted on math.SE)

Answer 1

Since $A$ is symmetric, it has an eigendecomposition $A = QDQ^*$ with $Q$ orthogonal. Then $$ M = A \otimes I + I\otimes A = (Q\otimes Q)(D\otimes I + I \otimes D)(Q\otimes Q)^* $$ is an eigendecomposition and also a singular value decomposition. The standard formula for the pseudoinverse in terms of the SVD gives $$ M^+ = (A \otimes I + I\otimes A)^+ = (Q\otimes Q)(D\otimes I + I \otimes D)^+(Q\otimes Q)^*, $$ where the pseudoinverse in the middle is a diagonal matrix with elements $\frac{1}{d_i+d_j}$ if $d_i+d_j\neq 0$ or $0$ otherwise.

Your implementation computes the minimum-norm least-squares solution $\operatorname{vec}X = M^+\operatorname{vec} B$ with this formula, if I am not mistaken.

Note also that in your case the explicit inverse IT can be avoided, since T ($Q$ above) is orthogonal.