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

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: $$C \preceq B$$.

• Consider $B=0\ne C$. Commented Feb 14, 2022 at 8:28
• @user1551 Thank you very much for your answer. Actually, you made me realize that I forgot an assumption in addition to the symmetric positive semidefiniteness of G, I have $C \preceq B$.
– Niz
Commented Feb 14, 2022 at 8:58
• It yields $(I-BB^\dagger)C = (I-(2C)(2C)^\dagger)C = (C-CC^\dagger C) = 0$
– Niz
Commented Feb 14, 2022 at 9:12
• May I ask you details about the implications : 1) $G \succeq 0$ and $C \preceq B \Rightarrow \mathcal{C}(B) = \mathcal{C}(C)$. 2) $\mathcal{C}(B) = \mathcal{C}(C) \Rightarrow BB^\dagger = CC^\dagger$ Thank you for your help
– Niz
Commented Feb 14, 2022 at 9:48
• @user1551 is the implication 2) related to the fact that $BB^\dagger$ is a projector onto the column space of $B$, hence, if the column space of $B$ and $C$ are the same then the projectors are the same i.e. $BB^\dagger = CC^\dagger$ ?
– Niz
Commented Feb 14, 2022 at 10:37

## 1 Answer

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 $$(I-BB^+)C=(I-CC^+)C=0$$.

• Perfect ! Thank you very much !
– Niz
Commented Feb 14, 2022 at 16:40