1
$\begingroup$

Consider a symmetric positive definite matrix $P$ and arbitrary matrix $R<-I$. Does the following inequality hold? $$ P < RPR^T $$

If yes, provide some references. If no, guide me under what conditions on matrix $R$ the aforementioned inequality holds.

Thanks

$\endgroup$
  • $\begingroup$ by $R\lt I$ do you mean $R-I$ is negative definite? $\endgroup$ – Ellya May 4 '14 at 20:31
1
$\begingroup$

It does not hold in general. For a counterexample, consider $$ R=R^T=-\pmatrix{4&1\\ 1&4},\quad D=\pmatrix{1\\ &16},\quad P=D^2,\quad x=\pmatrix{-4\\ 1}. $$ The spectrum of $R$ is $\{-5,-3\}$ (with eigenvectors $(1,\pm1)^T$), so that $R\prec -I$. However, we have $$ Dx=\pmatrix{-4\\ 16}\ \text{ and }\ DRx=D\pmatrix{15\\ 0}=\pmatrix{15\\ 0}, $$ so that $x^TPx=x^TD^2x=\|Dx\|^2=272>225=\|DRx\|^2=x^TR^TPRx=x^TRPR^Tx$. Therefore $P\not\preceq RPR^T$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.