Q:

Prove that the given block matrix is positive semi-definite

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. 

A:

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.