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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.

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.

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?

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

$$ F(A,B) = \max_{X>0} \{ |\operatorname{tr} X|: A\geq XB^{-1}X^{*} \} $$ 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.

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.

I'm reading this paper by Bhatia et al. and on page 6 theorem 2, they state

$$ F(A,B) = \max_{X>0} \{ |\operatorname{tr} X|: A\geq XB^{-1}X^{*} \} $$ 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.

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

$$ F(A,B) = \max_{X>0} \{ |\operatorname{tr} X|: A\geq XB^{-1}X^{*} \} $$ 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.