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Let $A=\begin{bmatrix} A_{11} & A_{21}^T\\ A_{21} & A_{22} \end{bmatrix}\in \mathbb{R}^{nxn}$ , which is a symmetric positive definite matrix, and $A_{11} \in \mathbb{R}^{pxp}$ which is invertible.

(a) Prove the Schur complement

$S=A_{22}-A_{21}A_{11}^{-1}A_{21}^T$

satisfies $\kappa_2(S) \leq \kappa_{2}(A)$

(b) Prove that $\lVert A_{21}A_{11}^{-1} \rVert_2 \leq \kappa_2(A)^{1/2}$

where $\kappa_2$ is the condition number with matrix 2 norm

For problem (a) I have already done it.

I can't figure out how to solve problem (b) though.

I know that in this case $\kappa_2(A)^{1/2}=\frac{\lambda_{max}(A)}{\lambda_{min}(A)}$, and I tried everything in my head but still couldn't get it. Am I missing something?

Archer914
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  • It might be helpful to rewrite the inequality as $$ |A_{21}A_{11}^{-1}|2^2 \leq \kappa_2(A) \implies\ \lambda{\max}(A_{21}A_{11}^{-2}A_{21}^T) \leq \frac{\lambda_{\max}(A)}{\lambda_{\min}(A)} $$ – Ben Grossmann Oct 12 '21 at 13:44
  • Also, if you could briefly explain how you did part a, that might be helpful – Ben Grossmann Oct 12 '21 at 14:01
  • I am using the method – Archer914 Oct 13 '21 at 06:21
  • I can understand the inequality you gave me, but do I need another decomposition for A or something to solve the problem? – Archer914 Oct 13 '21 at 06:23
  • I'm not really sure. Another potentially helpful observation, however, is that the condition that $A$ is positive definite is necessary for this part (but wasn't necessary for part a). Otherwise, the matrix $$ A = \pmatrix{t & 1\ 1 & t} $$ serves as a counterexample. As $t \to \infty$, $\kappa(A) \to 1$ but $|A_{21}A_{11}^{-1}|_2 = t \to \infty$. – Ben Grossmann Oct 13 '21 at 12:19
  • Is there a chance that you copied the question incorrectly? – Ben Grossmann Oct 13 '21 at 12:22
  • No, I've checked it, nothing wrong. – Archer914 Oct 13 '21 at 12:28

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