Timeline for Why is there not a test for diagonalizability of a matrix
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Jul 7 at 6:50 | comment | added | Jos Bergervoet | And purely for the looks, $\binom{1 \ \epsilon}{1 \ 1}$ and $\binom{1 \ \epsilon}{0 \ 1}$ would be two cases that look more similar. | |
| Jul 6 at 22:27 | comment | added | Hagen von Eitzen | For me, the first part is easier to see by noting that $\begin{pmatrix}1&1\\0&1+\epsilon\end{pmatrix} $ is diagonalizable iff $\epsilon\ne 0$. | |
| Jul 6 at 6:19 | audit | First answers | |||
| Jul 6 at 6:37 | |||||
| Jul 6 at 1:24 | comment | added | David Gao | @user3716267 A correction, as pointed out by Jim Belk: the collection of diagonalizable matrices does have interior. Its interior is exactly the collection of matrices with all eigenvalues distinct, which is already dense in all matrices. | |
| Jul 6 at 1:20 | comment | added | David Gao | @JimBelk You’re right. My claim was wrong. The closure of the collection of non-diagonalizable matrices is only the collection of matrices with multiplicities in eigenvalues, the complement of $\text{Dist}$, in your notations, not the full space of $n \times n$ matrices. | |
| Jul 5 at 22:29 | comment | added | Jim Belk | @DavidGao The set $\mathrm{Dist}$ of complex $n\times n$ matrices with $n$ distinct eigenvalues is open and dense, and every such matrix is diagonalizable. Thus the interior of the set of diagonalizable matrices is nonempty, and indeed dense. The set of non-diagonalizable matrices is contained in the complement of $\mathrm{Dist}$, which is the set of matrices whose characteristic polynomial has discriminant zero. This is a subvariety of the space of complex $n\times n$ matrices with (complex) codimension one. | |
| Jul 5 at 14:27 | comment | added | user3716267 | That paints a much clearer picture! | |
| Jul 5 at 13:38 | comment | added | David Gao | @user3716267 Not sure what exactly do you mean. I’m pretty sure you can adapt these examples to show that the set of diagonalizable matrices is dense but has no interior (at least in the complex case), so in that sense it is extremely badly behaved, topologically speaking. | |
| Jul 5 at 13:27 | comment | added | user3716267 | Is there any systematic way of understanding what these discontinuities mean, and whether we can recover any useful local analysis under stricter assumptions? | |
| Jul 5 at 11:54 | comment | added | David Gao | @PaulFrost That’s not what I meant. I didn’t mean for both an $f$ s.t. $A$ is diagonalizable iff $f(A) = 0$ and a $g$ s.t. $A$ is diagonalizable iff $g(A) \neq 0$. I meant neither such an $f$ nor such a $g$ can exist. (I discussed both because the OP’s two examples, invertible matrices and unitarily diagonalizable matrices, are different. The first set is open while the second is closed.) | |
| Jul 5 at 11:51 | comment | added | Paul Frost | If $f$ exists, then the set of diagonalizable matrices must be clopen. | |
| Jul 5 at 10:44 | vote | accept | Mahammad Yusifov | ||
| Jul 5 at 10:21 | history | answered | David Gao | CC BY-SA 4.0 |