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Dear all,

Suppose U and V are unitary matrix, A and B are positive definite,


$UAU^{-1} < VBV^{-1}$

implies $A< B$

and vice versa?

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closed as too localized by Deane Yang, Bill Johnson, Mark Sapir, Suvrit, Mark Meckes Aug 16 '12 at 17:02

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What do you mean with $A<B$? Are you comparing term by term? Moreover, Do $U$ and $V$ diagonalize $A$ and $B$ respectively and you are comparing the diagonal terms? – Dox Aug 16 '12 at 15:47
I mean by A < B is (B-A) > 0 or (B-A) is positive definite. U and V not necessarily diagonalize A and B, just to make them block diagonal. Thanks. – hayu Aug 16 '12 at 15:59
up vote 1 down vote accepted

No: think of the case of $A$, $B$ diagonal; $U$ identity, and $V$ a permutation operator.

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thank you ^_^ – hayu Aug 16 '12 at 16:18

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