6

Suppose that $A$ and $B$ are positive definite, symmetric, $n\times n$ real matrices such that $A-B$ is positive semidefinite. Write $A\succeq B$. When is it true that $A^2\succeq B^2$?

I know given the conditions, $A^2\succeq B^2$ is not true in general. On the other hand, let $u_1\geq\ldots\geq u_n\geq 0$ and $v_1\geq\ \ldots \geq v_n\geq 0$ be the eigenvalues of $A$ and $B$. By the Min-Max Principle, we have $$ u_i\geq v_i,\quad i=1,\ldots,n. $$ Then, $u_1^2\geq\ldots\geq u_n^2$ and $v_1^2\geq\ldots\geq v_n^2$ are the eigenvalues of $A^2$ and $B^2$ and satisfy $$ u_i^2\geq v_i^2,\quad i=1,\ldots,n.\tag{$*$} $$ I used to think that ($*$) would be enough to give $A^2\succeq B^2$ but as the counter-example linked above shows, it clearly doesn't. So I'm wondering how we can strengthen ($*$) to give $A^2\succeq B^2$.

yurnero
  • 10,675
  • 1
    I don't know the answer to your question, but thank you for mentioning the min-max principle. I learned something new. – stressed out Jul 11 '18 at 18:35
  • 1
    $A^2\succeq B^2$ iff $Id\succeq A^{-1}BBA^{-1}$, which is equivalent to the singular values of $A^{-1}B$ being less or equal to 1. – Daniel Jul 12 '18 at 16:48
  • @Daniel That sounds promising. Please feel free to develop that into an answer. Do you know how to relate the eigenvalues of $A^{-1}B$ to those of $A$ and of $B$? – yurnero Jul 12 '18 at 17:09

0 Answers0