Timeline for Finding the eigenvectors of a submatrix
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 13, 2022 at 16:57 | comment | added | Toni Mhax | This is still unclear, is it $B=\begin{pmatrix}A&AU\\AU&AU\end{pmatrix}$ for a $\pm 1$ diagonal matrix $U$... | |
| Dec 10, 2022 at 17:50 | comment | added | ABB | Your notation is also so helpful. | |
| Dec 10, 2022 at 17:48 | comment | added | ABB | Thanks for your comment. I have done. | |
| Dec 10, 2022 at 17:46 | history | edited | ABB | CC BY-SA 4.0 |
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| Dec 10, 2022 at 13:29 | comment | added | Toni Mhax | I don't think you defined the fourth block as $B=\begin{pmatrix}A&AU\\AU&*\end{pmatrix}$? | |
| Dec 9, 2022 at 15:25 | comment | added | ABB | $B$ is supposed to be symmetric, so to find $B$ we need only to know left side entries that are known for us. | |
| Dec 9, 2022 at 14:33 | comment | added | Antoine Labelle | What is $b_{n+k,n+l}$? Also it seems to me that if $A$ isn't symmetric then B won't be neither since the top left $n \times n$ block of $B$ is exactly $A$. | |
| Dec 9, 2022 at 9:38 | history | edited | ABB |
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| Dec 9, 2022 at 9:35 | comment | added | ABB | Yes, extending eigenvectors of the submatrix $A$ to $B$ would be also considered as a question. But, the converse process is my concern for now. | |
| Dec 9, 2022 at 9:10 | comment | added | Beni Bogosel | I would assume the opposite. If you know the eigenvectors of $A$ then you could try to build eigenvectors for $B$, changing the sign of certain components. With a bit of luck, you could use one eigenvector for $A$ to generate two different eigenvectors for $B$. Anyways, did you try $n=2$ and test some simple cases by hand?? | |
| Dec 9, 2022 at 8:47 | history | asked | ABB | CC BY-SA 4.0 |