# Can $U\Sigma U^T \preceq UVSV^TU^T$ lead to $( UVS^{-1}V^TU^T )^2 \preceq (U\Sigma^{-1} U^T)^2$? [duplicate]

This question already has an answer here:

Can $U\Sigma U^T \preceq UVSV^TU^T$ lead to $( UVS^{-1}V^TU^T )^2 \preceq (U\Sigma^{-1} U^T)^2$?

where $UU^T=U^TU=VV^T=V^TV=I$ and $\Sigma, S$ are square matrix only with positive elements in its diagonal and the other elements are zero, and (X)^2 is a simplified notation for matrix multiplication $XX$.

My problem is more complicated compared with Can $U\Sigma U^T \preceq UVSV^TU^T$ lead to $UVS^{-1}V^TU^T \preceq U\Sigma^{-1} U^T$?

and the difference between my problem and the previous similar one If $A^2\succ B^2$, then necessarily $A\succ B$ lie in that the two matrices in my problems have some relation while the two matrices in the above link are arbitrary.

## 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(); } ); }); }); Oct 5 '15 at 15:49

• It is a duplicate. Despite the title, Q510895 asks whether (a) "$A^2\succ B^2 \Rightarrow A\succ B$" and (b) its converse are true or not. Now, from your previous question, you should know that $U\Sigma U^T \preceq UVSV^TU^T$ is equivalent to $UVS^{-1}V^TU^T \preceq U\Sigma^{-1} U^T$. So, if you put $A=UVS^{-1}V^TU^T$ and $B=U\Sigma^{-1} U^T$, you are asking whether $A\preceq B$ implies that $A^2\preceq B^2$. This is the limiting case of part (b) of Q510895, hence effectively a duplicate. – user1551 Oct 5 '15 at 17:05
• Olivia, I see where you came from now. Unfortunately you seem to have some misunderstandings. (1) Real invertible matrices always share the same column space, which is the whole space $\mathbb R^n$. This is true in your question as well as in Q510895. (2) In general, any two real symmetric invertible matrices $A$ and $B$ can be real orthogonally diagonalised as $A=WD_1W^T$ and $B=UD_2U^T$. Let $V=U^TW$, we get $A=UVD_1V^TU^T$. So, the decomposition in your question has nothing special at all. – user1551 Oct 6 '15 at 7:50