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Apr 13, 2017 at 12:19 history edited CommunityBot
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
Feb 26, 2014 at 6:36 comment added dehiker Thanks for your reply, Federico! I've modified the question to link the one on math.se. By the way, I'm wondering why there're so many minuses here :(
Feb 26, 2014 at 6:09 history edited dehiker CC BY-SA 3.0
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Feb 25, 2014 at 13:58 answer added Bazin timeline score: 0
Feb 25, 2014 at 10:54 comment added Federico Poloni Oh, and by the way, can you cross-link to the math.se question, just for reference?
Feb 25, 2014 at 10:52 comment added Federico Poloni Not sure if you call that a "relationship", but if the diagonal is real the signature of the matrix is preserved, so you can at least predict their signs. I.e., $R$ and $QRQ$ have the same number of positive, zero, and negative eigenvalues. That's probably all that you can get.
Feb 25, 2014 at 10:50 comment added Alex Degtyarev Simultaneously diagonalizable. Otherwise, there's no hope to see what $Q$ does to the eigenvectors of $R$.
Feb 25, 2014 at 10:44 comment added dehiker Q is diagonal; R is hermitian, so it's diagonalizable. Then conditions satisfied?
Feb 25, 2014 at 10:30 comment added Alex Degtyarev I mean that $R$ is also diagonal. (In the learned language, $Q$ and $R$ commute, or are simultaneously diagonalizable.)
Feb 25, 2014 at 9:55 comment added dehiker Hi, Alex! What do you mean by saying "Q and R are in a special position with respect to each other"?
Feb 25, 2014 at 9:28 comment added Alex Degtyarev No, there's no nice relation unless $Q$ and $R$ are in a special position with respect to each other.
Feb 25, 2014 at 8:48 review First posts
Feb 25, 2014 at 8:58
Feb 25, 2014 at 8:30 history asked dehiker CC BY-SA 3.0