I have some questions about Moore–Penrose inverse.

1. Let $$A, P\in \mathbb{R}^{d\times d}$$. Suppose $$A$$ is positive definite and $$P$$ is a projection matrix with $$P^2=P, P^\top=P$$. I try to prove that $$A^{-1}-(PAP)^{-}$$ is semi-positive definite. Here $$B^-$$ denotes the Moore–Penrose inverse of $$B$$. (I can prove it using some statistical methods. But how to verify it directly.)

2. Suppose $$A_n\to A$$ with $$A_n$$'s and $$A$$ being semi-positive definite. Do we have $$A_n^- \to A^-$$? (It holds when $$A$$ is positive definite.)

3. As pointing out by user1551, 2 is not true in general. Actually, I try to answer the following question. Suppose $$A_n\to A$$ with $$A_n$$'s and $$A$$ being positive definite, do we have $$(PA_nP)^- \to (PAP)^-$$? Here $$P$$ is a projection matrix with $$P^2=P$$ and $$P^\top=P$$.

Thanks.

• If $P$ is a projection and $P^2 = I$, then $P = I$. Did you perhaps mean $P^2 = P$? Commented Mar 5, 2021 at 17:04
• The answer to the second question is definitely negative. E.g. if $\{A_n\}$ is a sequence of positive scalars that converges to $A=0$, then $A_n^-\to+\infty\ne0=A^-$. Commented Mar 5, 2021 at 17:15
• Yes, I mean $P^2=P$. Thanks for pointing out the typo. Commented Mar 5, 2021 at 17:30
• Thanks for the answer to the second question. Commented Mar 5, 2021 at 17:31

1. Yes. By a change of orthonormal basis, we may assume that $$P$$ and $$A$$ are partitioned as $$P=\pmatrix{I&0\\ 0&0}\text{ and }A=\pmatrix{X&Y\\ Y^T&Z}.$$ Let $$S$$ be the Schur complement of $$X$$ in $$A$$ (i.e. $$S=Z-Y^TX^{-1}Y$$). Then \begin{aligned} A^{-1} &=\pmatrix{X^{-1}+X^{-1}YS^{-1}Y^TX^{-1}&-X^{-1}YS^{-1}\\ -S^{-1}Y^TX^{-1}&S^{-1}}\\ &=\pmatrix{X^{-1}&0\\ 0&0}+\pmatrix{X^{-1}YS^{-1}Y^TX^{-1}&-X^{-1}YS^{-1}\\ -S^{-1}Y^TX^{-1}&S^{-1}}\\ &=(PAP)^-+\pmatrix{X^{-1}YS^{-1/2}\\ -S^{-1/2}}\pmatrix{S^{-1/2}Y^TX^{-1}&-S^{-1/2}}\\ &\succeq(PAP)^-. \end{aligned}
2. No. E.g. when $$\{A_n\}_{n\in\mathbb N}$$ is a sequence of positive real numbers that converges to $$A=0$$, we have $$A_n^-\to+\infty\ne0=A^-$$ as $$n\to\infty$$.
3. Yes. Since $$A_n\to A$$, we have $$PA_nP\to PAP$$. As $$P$$ is an orthogonal projection and both $$A_n$$ and $$A$$ are positive definite, the range of $$P$$ is an invariant subspace as well as the column spaces of both $$PA_nP$$ and $$PAP$$, and the latter two matrices are nonsingular on the range of $$P$$. It follows that the inverse of $$PA_nP$$ on the range of $$P$$ converges to the inverse of $$PAP$$ on the range of $$P$$. Hence $$(PA_nP)^-\to(PAP)^-$$. In terms of block matrices, we mean the assumption $$A_n=\pmatrix{X_n&Y_n\\ Y_n^T&Z_n}\to A=\pmatrix{X&Y\\ Y^T&Z}$$ implies that $$X_n\to X$$ and in turn $$(PA_nP)^-=\pmatrix{X_n^{-1}&0\\ 0&0}\to (PAP)^-=\pmatrix{X^{-1}&0\\ 0&0}$$ when $$n\to\infty$$.
• Is Statement 2 true provided $A_n$ and $A$ have the same rank? Commented Mar 15, 2021 at 19:36
• Statement 2 will be true if $A_n$ and $A$ have the same rank, I use the Davis-Kahan theorem to prove the convergence of the eigenvectors. Commented Mar 16, 2021 at 1:05