# A Matrix Inequality for positive definite matrices

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!

• 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. – lcv Apr 2 '19 at 9:56
• @lcv. can you please elaborate your notations? – A beginner mathmatician Apr 2 '19 at 11:16

The answer is No. Here is a counter-example: $$X=\begin{pmatrix} 9 & 3 \\ 3 & 1 \end{pmatrix},\qquad Y=\begin{pmatrix} 1 & 3 \\ 3 & 9 \end{pmatrix}.$$
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.
• 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. – Denis Serre Apr 2 '19 at 16:48
• 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. – Denis Serre Apr 2 '19 at 16:51
• @DenisSerre. The inverse is true if $A$ and $B$ commute. – Meisam Soleimani Malekan Apr 2 '19 at 16:52
• If matrices commutte, then also the positivity of $X$ and $Y$ implies that of $XY+YX$. – Denis Serre Apr 2 '19 at 17:22