Timeline for How to determine there exists a unique invariant subspace for a set of matrices
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jun 25, 2013 at 3:02 | review | First posts | |||
| Jun 26, 2013 at 8:42 | |||||
| Jun 19, 2013 at 12:51 | history | edited | jeremy | CC BY-SA 3.0 |
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| Jun 19, 2013 at 12:49 | comment | added | jeremy | Hm okay yea, I mean generic in the sense that I put no restrictions on the conditions of the eigenvalues. i.e. $A_1,..A_k\in\mathcal{M}^n(\mathbb{C})$ and can have degenerate eigenvalues. | |
| Jun 19, 2013 at 12:45 | comment | added | jeremy | Oh, I mean that the only matrices which commutes with every matrix $A_1,...,A_k$ and their Hermitian conjugates $A^*_1,...,A^*_k$ are those which are scalar multiples of the identity. Sometimes this is denoted $\{A_1,...,A_k,A^*_1,...,A^*_k\}'=\mbox{span}(\mathbb{I})$. | |
| Jun 19, 2013 at 12:43 | comment | added | Peter Michor | Matrices with pairwise distinct eigenvalues are generic. | |
| Jun 19, 2013 at 12:40 | history | edited | jeremy | CC BY-SA 3.0 |
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| Jun 19, 2013 at 11:14 | comment | added | user1688 | What is the "joint commutant"? | |
| Jun 19, 2013 at 9:00 | history | asked | jeremy | CC BY-SA 3.0 |