If $A^2\succ B^2$, then necessarily $A\succ B$

I remember reading somewhere about the following properties of non-negative definite matrix. But I don't know how to prove it now.

Let $A$ and $B$ be two non-negative definite matrices. If $A^2\succ B^2$, then it necessarily follows that $A\succ B$, but $A\succ B$ doesn't necessarily leads to $A^2\succ B^2$.

How can you prove it? Thanks!

• Are you sure about the other side? Because I feel that the other direction should also be correct. – Arash Oct 1 '13 at 8:47

In general, if $$X\succ0$$ and $$Y\succeq0$$, then $$X\succ Y$$ if and only if $$\rho(X^{-1}Y)<1$$.
Now suppose $$A\succ B\succeq0$$ and $$A^2\succeq B^2$$. Then $$\rho\left(A^{-2}B^2\right)<1$$ and hence $$\rho(A^{-1}B)^2 \le\|A^{-1}B\|_2^2 =\rho\left(A^{-1}B(A^{-1}B)^\ast\right) =\rho(A^{-1}B^2A^{-1}) =\rho\left(A^{-2}B^2\right)<1.$$ Thus $$\rho(A^{-1}B)<1$$ and $$A\succ B$$.
By a limiting argument, we may also prove that $$A\succeq B\succeq0$$ if $$A^2\succeq B^2\succeq0$$. Pick any $$t>0$$. By unitarily diagonalising $$A$$, we see that $$P=A+tI$$ is positive definite and $$P^2\succ A^2$$. Thus $$P^2\succ B^2\succeq0$$ and $$P\succ B$$. Let $$t\to0$$, we get $$A\succeq B$$.
However, note that the converses to the above are not true. E.g. when $$A\succ B$$, it is not necessarily true that $$A^2\succ B^2$$, as illustrated by the following counterexample where $$\epsilon>0$$ is small: \begin{align*} A&=\pmatrix{2+\epsilon&1\\ 1&3},\\ B&=\pmatrix{1\\ &2},\\ A-B&=\pmatrix{1+\epsilon&1\\ 1&1}\succ0,\\ A^2&=\pmatrix{(2+\epsilon)^2+1&5+\epsilon\\ 5+\epsilon&10},\\ B^2&=\pmatrix{1\\ &4},\\ A^2-B^2&=\pmatrix{(2+\epsilon)^2&5+\epsilon\\ 5+\epsilon&6}\not\succ0. \end{align*}
• What is meant by the notation $x^H$? – Babai Aug 9 '16 at 5:39
• @Babai It's the conjugate transpose (aka $\color{red}{H}$ermitian transpose) of $x$, i.e. $x^\ast$. The notation $x^\ast$ is more popular in theoretical linear algebra literature, while $x^H$ is more popular in numerical linear algebra literature. – user1551 Aug 9 '16 at 9:44
• Hello, what if we have $A,B,A-B$ non-negative definite matrix (which means we may not have $A^{-1}$) , can we have $\sqrt{A}-\sqrt{B}$ also is non-negative definite matrix? Thanks. – Idele Jan 17 '17 at 15:52