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S Apr 3, 2021 at 11:01 history bounty ended CommunityBot
S Apr 3, 2021 at 11:01 history notice removed CommunityBot
Mar 28, 2021 at 9:24 comment added Serguei Popov I think I didn't formulate it well: I don't need the eigenvectors to form a basis, and actually even don't need them to be distinct (the top one and a combination of bottom ones are used to construct a Lyapunov function for some stochastic process, so I don't need all of them). So, rather, it should be "let $v_1,\ldots,v_N$" be some corresponding eigenvectors; then they can be chosen in such a way that (2) holds.
Mar 28, 2021 at 0:17 answer added Jochen Glueck timeline score: 4
Mar 27, 2021 at 23:39 comment added Jochen Glueck Do you assume that $A$ is diagonalizable? (Since otherwise, not all if the eigenvectors $v_1, ..., v_N$ exist.)
Mar 26, 2021 at 10:52 history edited gmvh
Added descriptive tags
S Mar 26, 2021 at 9:10 history bounty started Serguei Popov
S Mar 26, 2021 at 9:10 history notice added Serguei Popov Draw attention
Mar 24, 2021 at 4:17 history became hot network question
Mar 23, 2021 at 21:06 answer added Noam D. Elkies timeline score: 10
Mar 23, 2021 at 20:16 history asked Serguei Popov CC BY-SA 4.0