2
$\begingroup$

This question already has an answer here:

If $M,N$ are two positive definite matrix st. $M<N$, is that true that $M^2<N^2$?

$\endgroup$

marked as duplicate by user1551 linear-algebra Apr 13 '16 at 23:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2
$\begingroup$

It is true if $NM=MN$; $N+M$, $M-N$ are positive $M^2-N^2=(M+N)(M-N)=(M-N)(M+N)$ apply 2 in

https://en.wikipedia.org/wiki/Positive-definite_matrix#Further_properties

$\endgroup$
2
$\begingroup$

Consider $$ A = \pmatrix{1 & 0\cr 0 & 2\cr}, \ B = \pmatrix{2+\epsilon & 1\cr 1 & 3\cr} $$ which are positive definite, with $B > A$ if $\epsilon > 0$, but $\det(B^2 - A^2) = -1 + 14 \epsilon + 5 \epsilon^2$, so for small positive $\epsilon$ (in fact $0 < \epsilon < (3\sqrt{6}-7)/5)$ we do not have $A^2 < B^2$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.