4

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?

Rodrigo de Azevedo
  • 23,223
Mat123
  • 321
  • 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. – Brian Borchers Jan 24 '18 at 13:10
  • I don't quite understand what you're asking. Are you trying to prove that $M$ is PD implies that $S$ is PD? – Ben Grossmann Jan 24 '18 at 13:11
  • I'm trying to find a way to prove that $S$ is PD. – Mat123 Jan 24 '18 at 13:17
  • @Mat123 trying to prove that $S$ is PD... using the fact that $M$ is PD? – Ben Grossmann Jan 24 '18 at 13:19

2 Answers2

6

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.

Ben Grossmann
  • 234,171
  • 12
  • 184
  • 355
5

(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, $S \geq 0.$

durdi
  • 783
  • 5
  • 15