Can $A^2\preceq \gamma^2 B^2$ lead to that $A\preceq \gamma B$?

Question

In the question, $A$ and $B$ are positive semi-definite matrices, $\gamma\geq 0$ is a constant, and $A\preceq \gamma B$ means that $\gamma B-A$ is positive semi-definite.

We have known another fact that $A\preceq \gamma B$ cannot lead to that $A^2\preceq \gamma^2 B^2$.

Answer 1

Absorb $\gamma$ into $B$. Now you are essentially asking whether the following is true:

if $A,B$ are positive semidefinite and $A^2\preceq B^2$, then $A\preceq B$.

For any $\varepsilon>0$, define $A_\varepsilon = (A^2+\varepsilon I)^{1/2}$ and define $B_\varepsilon$ analogously. Then $A_\varepsilon$ and $B_\varepsilon$ are positive defintie and they approach $A$ and $B$ respectively when $\varepsilon\to0$. We also have $A_\varepsilon^2\preceq B_\varepsilon^2$. So, if we can prove that $A_\varepsilon\preceq B_\varepsilon$, then by taking the limit of $\varepsilon$ to zero, we get $A\preceq B$.

In other words, we may assume that $A,B$ are positive definite in the first place. Now this case has been handled in the thread If $A^2\succ B^2$, then necessarily $A\succ B$.