Is it true that $A^2\succ B^2$ implies that $A\succ B$, but not the converse? 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?
 A: In general, if $X\succ0$ and $Y\succeq0$, then $X\succ Y$ if and only if $\rho(X^{-1}Y)<1$.
Now suppose $A,B\succeq0$ and $A^2\succ B^2$. 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, 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*}
