Prove that the given block matrix is positive semi-definite

Question

How do I show $M = \begin{bmatrix} A & B \\ B^T & C \end{bmatrix} \succeq 0$ i.e. $M$ is positive semi-definite (PSD) given that $A$ is PSD and for some $\Lambda = \text{diag}(\lambda_1, \lambda_2, \cdots, \lambda_p), \lambda_i \in [0,1]$ $$ B = B_1 (I-\Lambda) + B_2 \Lambda\\ C = (I-\Lambda)C_1(I-\Lambda) + (I-\Lambda)C_2\Lambda + \Lambda C_2^T (I-\Lambda) + \Lambda C_3 \Lambda, $$ where $B_1, B_2, C_1, C_2, C_3$ are all PSD. I know about the Schur complement, but I'm not sure how to take the inverse of $C$. Is there any other way I can approach to solve this problem?

Thanks for the help.

Answer 1

Consider $K=\left(\begin{matrix} A & 0 & B_2 C_1 & B_1\\ 0 & \Lambda C_3 \Lambda & (I-\Lambda)C_2 C_1 & \Lambda\\ C_1^TB_2^T & C_1^TC_2^T(I-\Lambda) & -C_1^{-1} & 0\\ B_1^T & \Lambda & 0 & -I\end{matrix}\right).$

Then $M \approx K/L$ is Schur complement of $K$ w/respect to the right lower block $L$.

Answer 2

Isn't that obvious? Unless $B=0$ or $B,C$ are related to $A$ in some way, you can never prove that $M$ is necessarily PSD, because it is not.

When $B\ne0$, since neither $B$ nor $C$ depend on $A$, if $C$ is PSD, when $A$ approaches zero, $M$ will eventually become indefinite; if $C$ is not PSD, then $M$ is not PSD in the first place.