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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.

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  • $\begingroup$ Try constructing a PSD block matrix $K$, such that $M=K/L$... $\endgroup$ – DVD Jul 8 '15 at 5:17
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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.

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  • $\begingroup$ To add, concretely, just consider the (indefinite) matrix $\begin{bmatrix}1 & 10 \\ 10 & 1\end{bmatrix}$, with eigenvalues $-9$ and $11$. $\endgroup$ – Nick Alger Dec 22 '15 at 0:45
  • $\begingroup$ I guess my basics in matrix is not very strong. What if the elements of $B$ and $C$ are picked from $A$ ? $\endgroup$ – jMathew Dec 22 '15 at 1:37
  • $\begingroup$ @jMathew That's why I said $B$ or $C$ should be somehow related to $A$. Otherwise you can always construct an example of $M$ that is not PSD (provided that $B\ne0$). Without knowing the relations between $A,B$ and $C$, whether $M$ is PSD is really case-dependent. $\endgroup$ – user1551 Dec 22 '15 at 1:50
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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 $L$.

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  • $\begingroup$ I'm not sure I understand.Did you mean to show that since $K$ is PSD and $M$ is the Schur complement of $K$, $M$ is also PSD. How can I show that $K$ is PSD, also what did you mean by $M\approx K/L$ $\endgroup$ – jMathew Jul 9 '15 at 3:37

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