Timeline for Why is there not a test for diagonalizability of a matrix
Current License: CC BY-SA 4.0
32 events
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| 22 hours ago | audit | First questions | |||
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| Jul 12 at 20:23 | audit | First questions | |||
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| Jul 8 at 6:34 | audit | First questions | |||
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| Jul 7 at 20:40 | audit | First questions | |||
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| Jul 7 at 20:30 | audit | First questions | |||
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| Jul 7 at 4:46 | answer | added | ronno | timeline score: 8 | |
| Jul 6 at 19:57 | review | Close votes | |||
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| Jul 6 at 18:51 | answer | added | Qiaochu Yuan | timeline score: 26 | |
| Jul 5 at 17:28 | history | became hot network question | |||
| Jul 5 at 10:44 | vote | accept | Mahammad Yusifov | ||
| Jul 5 at 10:44 | history | edited | Mahammad Yusifov | CC BY-SA 4.0 |
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| Jul 5 at 10:21 | answer | added | David Gao | timeline score: 49 | |
| Jul 5 at 10:12 | comment | added | David Gao | @MahammadYusifov I’ll write an answer to resolve this, then. | |
| Jul 5 at 10:08 | comment | added | Mahammad Yusifov | @DavidGao found it really hard to formalize exactly what I want from the test, but I think your answer considers a very important type of such functions. Could you please post it as an answer and perhaps link some resources mentioning or motivating the result? | |
| Jul 5 at 10:07 | comment | added | David Gao | This means there is no polynomial function (even continuous function, since the set is not even closed/open in the Euclidean topology) in $A_{ij}$ s.t. diagonalizability is equivalent to the function being $0$ or being nonzero, like for invertibility or unitary diagonalizability. | |
| Jul 5 at 10:06 | history | edited | Mahammad Yusifov | CC BY-SA 4.0 |
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| Jul 5 at 10:05 | comment | added | David Gao | @MahammadYusifov If you meant whether there is an algebraic function that tests if a given matrix is diagonalizable, that seems closer to ask if the set of diagonalizable matrix is Zariski closed or Zariski open. This is mostly definitely false. This set is not even closed/open in the Euclidean topology. | |
| Jul 5 at 9:58 | comment | added | Mahammad Yusifov | @DavidGao I understand there are algorithms, there are algorithms for finding roots of polynomials as well, but this doesn't mean there is a generic formula for solving polynomials. I am not looking for algorithmic ways as well in the sense that Gauss algorithm can be used for testing invertibility, but in this case determinant is the algebraic function I would be talking about. I will make my statement more formal with an edit | |
| Jul 5 at 9:57 | comment | added | peek-a-boo | If the minimal polynomial is $\prod (\lambda-\lambda_i)^{m_i}$ then the size of the largest Jordan block corresponding to $\lambda_i$ is $m_i$. Hence an equivalent condition is that $A$ is diagonalizable if and only if the minimal polynomial splits and all roots have multiplicity $1$. So if you consider the minimal polynomial (which is uniquely determined by the matrix) then you get a complete answer as well. Now if you’re asking whether there is a simple “explicit” formula for these $m_i$ then the answer is 99.999999% going to be no. | |
| Jul 5 at 9:56 | comment | added | David Gao | (This is over the complex numbers, or over any algebraically closed field, for that matter. Otherwise, over reals, you need to check the characteristic polynomial can be factored into linear terms in the first place. If it cannot, then the matrix is not diagonalizable. If it can, then the same procedure as the complex case applies.) | |
| Jul 5 at 9:55 | history | edited | Mahammad Yusifov | CC BY-SA 4.0 |
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| Jul 5 at 9:54 | comment | added | David Gao | To compute the Jordan normal form, you can compute the characteristic polynomial, then factor it. Then for each eigenvalue one may compute the associated eigenvectors. There is a Jordan block larger than $1 \times 1$ iff the multiplicity of an eigenvalue in the characteristic polynomial is strictly larger than the number of linearly independent eigenvectors associated to that eigenvalue. All of these can be computed via an algorithm. (Not sure how computationally effective, but for small matrices this can be done by hand.) | |
| Jul 5 at 9:52 | comment | added | Mahammad Yusifov | @ David Gao, edited | |
| Jul 5 at 9:51 | history | edited | Mahammad Yusifov | CC BY-SA 4.0 |
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| Jul 5 at 9:50 | comment | added | David Gao | To add to MSEU’s comment, Jordan normal form is a correct thing to do. It can be computed via an algorithm, and a matrix is diagonalizable iff the Jordan normal form is a diagonal matrix. | |
| Jul 5 at 9:50 | comment | added | Mahammad Yusifov | I know, but given a matrix, how do you know the dimensions of Jordan block as a function of $A?$ | |
| Jul 5 at 9:48 | comment | added | peek-a-boo | $A$ is diagonalizable if and only if the characteristic polynomial splits and all Jordan blocks are $1\times 1$. So that’s your true/false questionnaire. | |
| Jul 5 at 9:48 | comment | added | David Gao | Technically, over complex numbers, unitarily diagonalizable is equivalent to $AA^\ast = A^\ast A$, not $A = A^\ast$. Over reals orthogonally diagonalizable is equivalent to $A = A^t$, though. | |
| Jul 5 at 9:43 | history | undeleted | Mahammad Yusifov | ||
| Jul 5 at 9:43 | history | deleted | Mahammad Yusifov | via Vote | |
| Jul 5 at 9:38 | comment | added | Mahammad Yusifov | This doesn't talk about the map which depend on $A_{ij}$ to determine diagoalizability. | |
| Jul 5 at 9:28 | history | asked | Mahammad Yusifov | CC BY-SA 4.0 |