Timeline for square matrix depending on complex value: spectral radius continous? [closed]
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 10, 2023 at 13:22 | comment | added | Iosif Pinelis | Do you have a further response to the answer below? | |
| Sep 6, 2023 at 20:58 | history | closed |
Federico Poloni Carlo Beenakker Max Horn Joseph Van Name user44191 |
Needs details or clarity | |
| Sep 6, 2023 at 16:38 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor Math Jaxing
|
| Sep 6, 2023 at 13:35 | comment | added | Iosif Pinelis | @Fynn13 : What is confusing here for you? As I said, the counterexamples above are not about the spectral radius; they are about an eigenvalue with the largest modulus. On the other hand, I believe my answer is correct. | |
| Sep 6, 2023 at 13:24 | comment | added | Fynn13 | Okey, I'm a little confused. So in my case (the spectral radius), the counter examples above are correct or your text below? | |
| Sep 6, 2023 at 13:19 | comment | added | Iosif Pinelis | @Fynn13 : In general, the spectral radius is not the same as the largest eigenvalue. In fact, if (say) all eigenvalues are complex (but not real) numbers, then a largest eigenvalue does not exist. Also, the spectral radius $\lambda_z$ is the largest modulus of the eigenvalues, not an eigenvalue with the largest modulus. In particular, an eigenvalue with the largest modulus can be a complex number that is not real, whereas the spectral radius is always a (nonnegative) real number. I have edited your post accordingly. | |
| Sep 6, 2023 at 13:18 | review | Close votes | |||
| Sep 6, 2023 at 20:58 | |||||
| Sep 6, 2023 at 13:09 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
deleted 11 characters in body; edited title
|
| Sep 6, 2023 at 13:06 | comment | added | Fynn13 | I don’t understand the difference. I am interested in the spectral radius of $A(z)$, that is the maximum of the absolute values of its eigenvalues. | |
| Sep 6, 2023 at 13:01 | comment | added | Iosif Pinelis | @CarloBeenakker : Your counterexamples are based on your misunderstanding of the question. The "largest eigenvalue" $\lambda_z$ is defined (somewhat awkwardly) in the OP as the largest modulus of the eigenvalues, not as an eigenvalue with the largest modulus. | |
| Sep 6, 2023 at 12:49 | answer | added | Iosif Pinelis | timeline score: 1 | |
| Sep 6, 2023 at 12:34 | comment | added | Carlo Beenakker | the eigenvalues themselves depend continuously on the matrix elements, so if all eigenvalues are positive taking the absolute value makes no difference and the largest eigenvalue will depend continuously on the matrix elements. | |
| Sep 6, 2023 at 12:31 | comment | added | Fynn13 | Thank you. Your examples are diagonal matrices: What if we would take just positive matrices? Lets say $A(z)\geq 0$. | |
| Sep 6, 2023 at 12:22 | comment | added | Carlo Beenakker | a counter example with real matrix elements: $2\times 2$ diagonal matrix with $-x$ and $x^2$ on the diagonal, the eigenvalue with the largest absolute value jumps from $-1$ to $1$ when $x$ crosses 1. | |
| Sep 6, 2023 at 12:01 | comment | added | Fynn13 | Thank you for the example, i will check that. Is the assertion true if we would take real values instead of complex? | |
| Sep 6, 2023 at 11:55 | comment | added | Carlo Beenakker | certainly not; for a simple counter example, take the $2\times 2 $ diagonal matrix with $iz$ and $z^2$ on the diagonal; as $z$ moves along the real axis and crosses 1 the "largest" eigenvalue jumps from $i$ to 1. | |
| Sep 6, 2023 at 11:46 | comment | added | Fynn13 | Sorry, I updated the answers to your questions! | |
| Sep 6, 2023 at 11:44 | history | edited | Fynn13 | CC BY-SA 4.0 |
added 60 characters in body
|
| Sep 6, 2023 at 11:43 | comment | added | Mikhail Katz | How do you define the largest eigenvalue $\lambda\in\mathbb C$? | |
| Sep 6, 2023 at 11:40 | history | asked | Fynn13 | CC BY-SA 4.0 |