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Jun 30, 2018 at 3:44 vote accept GA316
Jun 28, 2018 at 12:58 answer added Rodrigo de Azevedo timeline score: 5
Jun 28, 2018 at 11:03 comment added Geoff Robinson It's better to start with a positive definite diagonal matrix $B.$ Then for any orthogonal involution $Q,$ you see that $A = QBQ$ is still a positive definite matrix, and $QAQ = B,$ so that $A$ is a positive definite matrix diagonalized by the orthogonal involution $Q.$ Conversely, every positive definite matrix $A$ which may be diagonalized by an orthogonal involution $P$ has such a form, since $PAP$ is positive definite diagonal, and is inverted by the orthogonal involution $P.$
Jun 28, 2018 at 5:07 comment added GA316 @James But can you say anything more algebraically which would be help me more?
Jun 28, 2018 at 5:06 comment added GA316 @RodrigodeAzevedo the $P$ is the question has to be symmetric as it is orthogonal and involution. That much only I can understand. If you have further insight please share with me.
Jun 28, 2018 at 2:23 comment added James Geometrically, this occurs precisely when there is a reflection (through some subspace) taking the eigenspaces of A to coordinate-subspaces.
Jun 28, 2018 at 0:48 review Close votes
Jul 2, 2018 at 3:48
S Jun 28, 2018 at 0:13 history suggested Rodrigo de Azevedo CC BY-SA 4.0
Minor improvements
Jun 27, 2018 at 21:21 review Suggested edits
S Jun 28, 2018 at 0:13
Jun 27, 2018 at 6:53 history asked GA316 CC BY-SA 4.0