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Let $M$ to be a real-valued symmetric and positive-definite (PD) matrix (also sparse and banded if it helps)

$$ M= \begin{bmatrix} A & B\\ B^T & D \end{bmatrix} $$

Under what conditions the Schur complement of $M$ ( $S=D-B^T A^{-1} B$) is PD?

As far as I found, it holds if $M$ and $A$ are both PD. If this is true, how can say if $A$ is PD?

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  • $\begingroup$ There are many ways in which you might show theoretically that the matrix $A$ in your problem is PD. Computationally, you might verify that $A$ is PD by computing a Cholesky factorization or looking at $A$'s eigenvalues. Without more context, it's impossible to provide a more specific answer. Please tell us more about your problem. $\endgroup$ Jan 24, 2018 at 13:10
  • $\begingroup$ I don't quite understand what you're asking. Are you trying to prove that $M$ is PD implies that $S$ is PD? $\endgroup$ Jan 24, 2018 at 13:11
  • $\begingroup$ I'm trying to find a way to prove that $S$ is PD. $\endgroup$
    – Mat123
    Jan 24, 2018 at 13:17
  • $\begingroup$ @Mat123 trying to prove that $S$ is PD... using the fact that $M$ is PD? $\endgroup$ Jan 24, 2018 at 13:19

2 Answers 2

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Whatever it is that you're trying to do, it's helpful to understand the relationship between the Schur complement and the matrix $M$. Notably, we have (using block-matrix multiplication) $$ \pmatrix{I & 0\\-B^TA^{-1}&I}\pmatrix{A & B\\B^T & D}\pmatrix{I & 0\\-B^TA^{-1}&I}^T = \\ \pmatrix{A & 0\\0&D - B^TA^{-1}B} $$ Also, note that if $M$ is PD, then $A$ (which is a principal submatrix of $M$) must also be PD.

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(In this answer, "$A \geq 0$" means $A$ is PSD and "$A > 0$" means $A$ is PD)

Let $f(u, v) = [u^T \ v^T]M[u^T \ v^T]^T$, where $M=\begin{bmatrix} A & B\\ B^T & D \end{bmatrix}.$ If $A>0$, one can show that $\inf_{u}f(u,v)=v^TSv$, where $S=D-B^T A^{-1} B.$

Theorem: if $A > 0$, then $M \geq 0$ iff $S \geq 0$.

Proof: If $S \geq 0$, then $$f(u,v) \geq \inf_{u}f(u,v)=v^TSv \geq 0\Rightarrow f(u,v) \geq 0$$ for all $u, v$, hence, $M \geq 0.$

If $M \geq 0$, then $$f(u,v) \geq 0 \Rightarrow \inf_{u}f(u,v) \geq 0 \Rightarrow v^TSv \geq 0$$ for all $v$, hence, $S \geq 0.$

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