Showing $A-(I-B)A(I-B)'$ is positive semidefinite

Question

I need to show that for two specific matrices $A$ and $B$, where $A$ is symmetric and positive (semi)definite and $B$ is a projection, the difference $$A-(I-B)A(I-B)^\prime$$ is positive semidefinite. How do I start this?

Answer 1

This is not always true. E.g. $$ A=\pmatrix{1&1\\ 1&2},\ B=\pmatrix{0&0\\ 0&1},\ A-(I-B)A(I-B)=\pmatrix{0&1\\ 1&2}\not\geq0. $$ In general, suppose $A$ is positive semidefinite and $B$ is an orthogonal projection. Every vector can be written as a sum of the form $u+v$, where $u\in\operatorname{ran}(B)$ and $v\in\ker(B)=\operatorname{ran}(B)^\perp$. Thus $$ (u+v)^\ast\left[A-(I-B)A(I-B)\right](u+v)=u^\ast Au+2\Re(v^\ast Au).\tag{1} $$ If $v^\ast Au\ne0$, then $(e^{i\theta}v)^\ast Au<0$ for an appropriate $\theta$. So, if we replace $v$ by $ke^{i\theta}v$ for a sufficiently large $k>0$, the RHS of $(1)$ can be made negative.

Therefore, $A-(I-B)A(I-B)$ is positive semidefinite if and only if $v^\ast Au$ is identically zero, i.e., if and only if $\operatorname{ran}(B)$ and $\ker(B)$ are invariant subspaces of $A$.