How could I extend this result to a case where the matrices were not of full rank?

Question

I'm reading this paper by Bhatia, Jain and Lim and on page 6 theorem 2, they state

$$ \sqrt{\operatorname{tr}(A^{1/2}BA^{1/2})} = \max_{X>0} \{ |\operatorname{tr} X|: A\geq XB^{-1}X^{*} \} $$ where $A$ and $B$ are positive definite matrices.

Is there a way to relax this in the case that $A$ and $B$ are instead PSD and not of full rank? I don't think I can carry out the similar steps as they do in the proof since $B$ may not necessarily be invertible now.

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Someone told me that this is doable since $A$ is PD on the same subspace as $B$ because if $A$ is positive where $B$ is zero, then the scale of $X$ in that direction can be infinite. If $B$ is positive where $A$ is zero, then $X$ is zero in that direction, so the max is over the subspace.

I don't quite follow this -- can someone explain why this is the case? Also doesn't it matter that $X$ can be infinite in some direction?

Answer 1

The natural way to relax the condition $A \geq XB^{-1}X^*$ to the case when $B$ is not full-rank is transforming it to $$ \begin{bmatrix} A & X\\ X^* & B \end{bmatrix} \geq 0, $$ which is equivalent by standard results on Schur complements (when $A \geq 0$). So this change would give you a candidate statement that is valid also for PSD matrices. Then proving it is another question --- but it looks like the kind of thing that can be proved using a continuity argument.