Let $A$ and $B$ be positive semidefinite matrices that satisfy $A\geq B$, where the inequality denotes that the matrix $A-B$ is positive semidefinite.
The square root operation is defined for a matrix $X$ by first changing to the eigenbasis of $X$, taking the square root of all eigenvalues (which are unique since they are all nonnegative) and converting back to the original basis.
Does it hold that $A^{1/2} \geq B^{1/2}$? Clearly, it does when $A$ and $B$ commute but is it also the case when they don't?
Best Answer
Yes, the square root is operator monotone, this is Löwner-Heinz theorem https://encyclopediaofmath.org/wiki/L%C3%B6wner-Heinz_inequality.
(Remark by the way that it is not true anymore when you take a power greater than $1$.)