9
$\begingroup$

Let $X$ and $Y$ be positive semi-definite self-adjoint complex matrices of same finite order. The, is it true that $|X-Y|\leq X+Y$ where for any matrix $A$, $|A|$ is defined to be $|A|:=(A^*A)^{\frac{1}{2}}$ ?

PS. I think the answer is No. But I could not find any counterexample!

$\endgroup$
  • 1
    $\begingroup$ Consider $X,Y$ to be of trace 1 and their difference to be block diagonal $\sigma^x,\sigma^x$. Now $|X-Y|=$ identity matrix. Take the trace of both sides, the LHS gives 4 the RHS is 2. $\endgroup$ – lcv Apr 2 '19 at 9:56
  • $\begingroup$ @lcv. can you please elaborate your notations? $\endgroup$ – A beginner mathmatician Apr 2 '19 at 11:16
16
$\begingroup$

The answer is No. Here is a counter-example: $$X=\begin{pmatrix} 9 & 3 \\ 3 & 1 \end{pmatrix},\qquad Y=\begin{pmatrix} 1 & 3 \\ 3 & 9 \end{pmatrix}.$$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Partial solution. When $X$ and $Y$ commute, the answer is "Yes". Indeed, in this case, the equality in the question is equivalent to $XY+YX\geq0$, and sure this is true.

| cite | improve this answer | |
$\endgroup$
  • 4
    $\begingroup$ I went also through this inequality. However your claim is incorrect: $XY+YX\ge0$ is only equivalent to $|X+Y|^2\le(X+Y)^2$. But $A^2\le B^2$ with $A,B$ positive definite is not equivalent to $A\le B$. We only have $(A^2\le B^2)\Longrightarrow(A\le B)$. In other words, the map $t\mapsto\sqrt t$ is operator monotone over $(0,+\infty)$, but the reciprocal $x\mapsto x^2$ is not. $\endgroup$ – Denis Serre Apr 2 '19 at 16:48
  • $\begingroup$ To complete the previous comment, I confirm that $XY+YX$ is not always true for $X,Y$ positive definite when $n\ge2$. Besides the fact that it would imply that $x\mapsto x^2$ be operator monotone, a false fact, it is also related to the theory of so-called Jordan algebra. See the book by Faraut and Koranyi. $\endgroup$ – Denis Serre Apr 2 '19 at 16:51
  • $\begingroup$ @DenisSerre. The inverse is true if $A$ and $B$ commute. $\endgroup$ – Meisam Soleimani Malekan Apr 2 '19 at 16:52
  • $\begingroup$ If matrices commutte, then also the positivity of $X$ and $Y$ implies that of $XY+YX$. $\endgroup$ – Denis Serre Apr 2 '19 at 17:22
  • 2
    $\begingroup$ @Meisan. Of course they don't. $\endgroup$ – Denis Serre Apr 2 '19 at 19:31

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.