# $M^2<N^2$ if $M,N$ are two positive definite matrix [duplicate]

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

## marked as duplicate by user1551 linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 13 '16 at 23:22

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
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 $A^2 < B^2$.