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If symmetric matrix $A\geq0$, $P>0$, can $APA\leq \lambda_{max}^2(A) P$ always hold?

Notation:

$\lambda_{max}(A)$ means matrix $A$'s largest eigenvalue.

$A\geq0$ means matrix A is a positive semi-definite matrix.

$P>0$ means matrix P is a positive definite matrix.

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  • $\begingroup$ What do you mean by "can we get"? Do you mean "is it ever the case that [inequality]" or do you mean "is it always the case that [inequality]"? $\endgroup$ – Omnomnomnom Mar 31 '17 at 1:10
  • $\begingroup$ I revised the title for clarity $\endgroup$ – wayne Mar 31 '17 at 1:11
  • $\begingroup$ If that's the case, we can simply take $A = P = I$ ($I$ being the identity matrix). $\endgroup$ – Omnomnomnom Mar 31 '17 at 1:13
  • $\begingroup$ it's too special, can other positive matrix rather than $I$ hold? $\endgroup$ – wayne Mar 31 '17 at 1:14
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No, it is not always, the case. For example $$A=\left(\begin{array}{cc}1&1\\1&1\end{array}\right), P=\left(\begin{array}{cc}100&0\\0&1\end{array}\right)$$

Then $$APA = \left(\begin{array}{cc}101&101\\101&101\end{array}\right)\not\le 4P$$

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  • $\begingroup$ Thanks for your answer. A should be positive semi-definite. However, in your case, A is not. $\endgroup$ – wayne Mar 31 '17 at 4:46
  • $\begingroup$ Is it correct now? $\endgroup$ – Quang Hoang Mar 31 '17 at 5:11
  • $\begingroup$ Yes. You are right. Thanks $\endgroup$ – wayne Mar 31 '17 at 5:18

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