Questions about Moore–Penrose inverse I have some questions about Moore–Penrose inverse.

*

*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.)


*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.)


*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.
 A: *

*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}


*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 $n\to\infty$.


*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 $n\to\infty$.
