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Let $A, B_i$ be positive semidefinite real matrices. Let $\dagger$ stand for the Moore-Penrose generalized inverse. I managed to prove that if $\operatorname{Ran}B_1\subseteq\operatorname{Ker}B_2$ then $$\operatorname{trace}\left((A + B_1 + B_2)^\dagger B_1 \right) \leq\operatorname{trace}\left(( A + B_1)^\dagger B_1\right) $$

Does it still hold without this assumption?

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Yes. In general, if $X,Y$ are positive semidefinite and $P=XX^\dagger$ denotes the orthogonal projection onto $\operatorname{ran}(X)$, then $P(X+Y)^\dagger P\preceq X^\dagger$. This can be easily proved by using Schur complements.

Now, if you put $X=A+B_1$ and $Y=B_2$, you get $$ B_1^{1/2}P(A+B_1+B_2)^\dagger PB_1^{1/2}\preceq B_1^{1/2}(A+B_1)^\dagger B_1^{1/2}. $$ Since $B_1^{1/2}P=PB_1^{1/2}=B_1^{1/2}$, the result follows.

Edit. If $X=0$, the inequality $P(X+Y)^\dagger P\preceq X^\dagger$ simply means $0\preceq0$. Suppose $X$ is PSD but nonzero. Since $\operatorname{ran}(X)\subseteq\operatorname{ran}(X+Y)$, by a change of orthonormal basis, we may assume that $$ X=\pmatrix{X_1&0&0\\ 0&0&0\\ 0&0&0},\ P=\pmatrix{I&0&0\\ 0&0&0\\ 0&0&0},\ Y=\pmatrix{H&R&0\\ R^T&S&0\\ 0&0&0},\ X+Y=\pmatrix{X_1+H&R&0\\ R^T&S&0\\ 0&0&0} $$ where $X_1$ and $Z:=\pmatrix{X_1+H&R\\ R^T&S}$ are the matrix representations of $X|_{\operatorname{ran}(X)}$ and $(X+Y)|_{\operatorname{ran}(X+Y)}$ respectively and they are positive definite.

As $Z\succ0$, we must have $S\succ0$. Yet $Y\succeq0$ by assumption. Therefore the Schur complement $H-RS^{-1}R^T$ must be $\succeq0$. It follows that $X_1+H-RS^{-1}R^T\succeq X_1\succ0$ and in turn $0\prec X_1^{-1}\preceq(X_1+H-RS^{-1}R^T)^{-1}$. But this means $P(X+Y)^\dagger P\preceq X^\dagger$, because $$ X^\dagger=\pmatrix{X_1^{-1}&0&0\\ 0&0&0\\ 0&0&0},\ (X+Y)^\dagger=\pmatrix{(X_1+H-RS^{-1}R^T)^{-1}&\ast&0\\ \ast&\ast&0\\ 0&0&0}, $$

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  • $\begingroup$ Could you please expand a bit about the Schur complements demonstration? Or provide some reference. Many thanks. $\endgroup$
    – Manuel
    Oct 19, 2018 at 20:54
  • $\begingroup$ @Manuel See my edit. $\endgroup$
    – user1551
    Oct 20, 2018 at 5:40
  • $\begingroup$ I have started a bounty as an appreciation of this answer. I also belive it should receive more attention. Thanks a lot. I would accept in a few days. $\endgroup$
    – Manuel
    Oct 22, 2018 at 13:17
  • $\begingroup$ @Manuel Thanks for your bounty! $\endgroup$
    – user1551
    Oct 23, 2018 at 18:05
  • $\begingroup$ Thank you for a very useful well explained answer. $\endgroup$
    – Manuel
    Oct 23, 2018 at 18:25

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