If $A<B$, does it follow that $A^2\leq B^2$?

Question

Suppose $A$ and $B$ are positive semidefinite matrices satisfying $A\leq B$ (meaning that $x^TAx \leq x^TBx$ for any vector $x$). Does it follow that $A^2\leq B^2$? It certainly follows if $A$ and $B$ commute, but beyond this case I am not so sure -- I suspect there is a counterexample.

Answer 1

Let's see. I assume that when you say semidefinite matrices you assume that they are symmetric (or hermitian). Here is a counterexample in this setting: Let

$$A=\left[\begin{matrix} 5 & 2\\2& 2\end{matrix}\right] \quad\text{and}\quad B=\left[\begin{matrix} 4 & 1\\1& 1\end{matrix}\right].$$

Then $A-B$ is positive semidefinite and $A^2-B^2 = \left[\begin{matrix} 12 & 9 \\ 9 & 6\end{matrix}\right]$ is indefinite. I'm sure you can make a small perturbation to have $A-B$ positive definite and $A^2-B^2$ still indefinite.

Answer 2

Try $$ A = \pmatrix{3 & 3\cr 3 & 5 \cr}, \ B = \pmatrix{1 & 2\cr 2 & 4\cr} $$