Does $(I-BB^\dagger)C=0$ hold for a symmetric positive semidefinite matrix $G = \begin{pmatrix} A & B \\ B^T &C \end{pmatrix}$ with $C \preceq B$?

Question

Given a symmetric positive semidefinite matrix

$$ G = \begin{pmatrix} A & B \\ B^T &C \end{pmatrix}$$

where $A$, $B$ and $C$ are not invertible, and $C\preceq B$, does the following equality hold?

$$ \left( I - B B^\dagger \right) C = 0 $$

Or, equivalently, does the following hold?

$$ C = B B^\dagger C$$

where $B^\dagger$ is the Moore-Penrose pseudoinverse of $B$.

I already know from here (Theorem 4.3) that we have

$$(I-AA^\dagger)B=0 \text{ and } (I-CC^\dagger)B^T=0$$ (we have it even without $C \preceq B$).

Edit : user 1551 proposed a direct counter-example to my proposition, I realized that it missed an assumption from my application: $C \preceq B$.

Answer 1

Since $G\succeq0$, if $Cx=0$, then $$ x^TBABx+2tx^TB^2x =\pmatrix{x^TB&tx^T}\pmatrix{A&B\\ B&C}\pmatrix{Bx\\ tx}\ge0 $$ for every real number $t$. Therefore $\|Bx\|_2^2=x^TB^2x$ must be zero and in turn $Bx=0$. Hence $\ker(C)\subseteq\ker(B)$.

On the other hand, if $Bx=0$, then $-x^TCx=x^T(B-C)x\ge0$. Hence $Cx=0$ and $\ker(B)\subseteq\ker(C)$.

Thus $\ker(C)=\ker(B)$ and the orthogonal projections onto them coincide. That is, $I-CC^+=I-BB^+$. It follows that $(I-BB^+)C=(I-CC^+)C=0$.