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 Oct 5 '15 at 15:49

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  • $\begingroup$ sorry but it is not duplicate. $\endgroup$ – olivia Oct 5 '15 at 16:46
  • $\begingroup$ 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. $\endgroup$ – user1551 Oct 5 '15 at 17:05
  • $\begingroup$ Thank you. I will use your notation. In my question, at least A and B are in the same column space. However, Q510895 dont have such constraint. $\endgroup$ – olivia Oct 6 '15 at 2:45
  • $\begingroup$ 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. $\endgroup$ – user1551 Oct 6 '15 at 7:50