# Trace inequality for a product of p.s.d. matrices and their pseudo inverse.

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?

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 $$X^\dagger=\pmatrix{X_1^{-1}&0&0\\ 0&0&0\\ 0&0&0},\ (X+Y)^\dagger=\pmatrix{(X_1+H-RS^{-1}R^T)^{-1}&\ast&0\\ \ast&\ast&0\\ 0&0&0},$$

• Could you please expand a bit about the Schur complements demonstration? Or provide some reference. Many thanks. – Manuel Oct 19 '18 at 20:54
• @Manuel See my edit. – user1551 Oct 20 '18 at 5:40
• 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. – Manuel Oct 22 '18 at 13:17
• @Manuel Thanks for your bounty! – user1551 Oct 23 '18 at 18:05
• Thank you for a very useful well explained answer. – Manuel Oct 23 '18 at 18:25