Skip to main content
8 events
when toggle format what by license comment
Dec 7, 2019 at 17:01 history closed abx
ARG
user44191
Max Horn
Alex M.
Not suitable for this site
Dec 5, 2019 at 16:51 comment added Henrique de Oliveira Thanks @MTyson! This is actually closer to what I had in mind.
Dec 2, 2019 at 1:41 history became hot network question
Dec 1, 2019 at 20:55 review Close votes
Dec 7, 2019 at 17:01
Dec 1, 2019 at 19:32 comment added MTyson In the nondiagonalizable case, it's not even true that all eigenvectors lie in the span of products of eigenvectors. Take $A$ and $B$ to be the $2$-by-$2$ Jordan block with eigenvalue $1$. Then $e_1\otimes e_1$ and $e_1\otimes e_2-e_2\otimes e_1$ are both eigenvectors of $A\otimes B$. The problem of computing the extra eigenvectors can always be reduced to the case where $A$ and $B$ are Jordan blocks by Jordanizing both.
Dec 1, 2019 at 17:54 vote accept Henrique de Oliveira
Dec 1, 2019 at 17:49 answer added Federico Poloni timeline score: 9
Dec 1, 2019 at 17:38 history asked Henrique de Oliveira CC BY-SA 4.0