Timeline for Continuity of eigenvector of zero eigenvalue
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Oct 19 at 20:43 | vote | accept | muddy | ||
| Mar 21 at 18:34 | vote | accept | muddy | ||
| Oct 19 at 20:41 | |||||
| Mar 17 at 17:54 | answer | added | muddy | timeline score: 0 | |
| Feb 28 at 0:13 | comment | added | muddy | I am able to convince myself that there exists a continuous $0$-eigenvector $v(t)$ that is continuous over some non-singleton interval contained in $I$. This holds only with the requirement that $A(t)$ is continuous on $I$. Whether there exists of a continuous $0$-eigenvector $v(t)$ that is continuous a.e. on the whole interval $I$ is still unknown to me. | |
| Feb 28 at 0:08 | history | edited | muddy | CC BY-SA 4.0 |
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| Feb 24 at 23:49 | comment | added | muddy | I have edited the question to ask for $0$-eigenvector $v(t)$ to be continuous a.e. on $I$, which is all I need. Not necessarily continuous everywhere on $I$. | |
| Feb 24 at 23:03 | comment | added | Achim Krause | In your example, the vector is not an eigenvector for $t=0$, as eigenvectors are required to be nonzero. If you allow the zero vector the answer is obviously yes. | |
| Feb 24 at 22:58 | comment | added | muddy | Yes, I did look through the book, but I think the book considers more general situations. I ask here to check whether researchers more familiar with the topic can provide a solution, which may be straightforward, and not too involved. What I ask in the question is for special situation of zero eigenvalue which may have more straightforward solutions. | |
| Feb 24 at 22:32 | comment | added | Pietro Majer | Have a look to Perturbation Theory for Linear Operators, by Tosio Kato. I think it is free online. | |
| Feb 24 at 21:27 | history | edited | muddy | CC BY-SA 4.0 |
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| Feb 24 at 21:17 | history | edited | muddy | CC BY-SA 4.0 |
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| Feb 24 at 21:16 | comment | added | muddy | From the example that I gave above, at $t = 0$, it is a zero vector, which cannot be an eigenvector. So based on the example, the question I would like to ask is whether there exists an $0$-eigenvector $v(t)$ that is continuous on $I$, except a set of zero measure. | |
| Feb 24 at 21:06 | comment | added | muddy | Suppose $A(t) = \left( \begin{array}{cc} t & 0 \\ 0 & 0 \end{array}\right)$ for $t \leq 0$ and $A(t) = \left( \begin{array}{cc} 0 & 0 \\ 0 & t \end{array}\right)$ for $t > 0$. This $A(t)$ is differentiable, and the $0$-eigenvector $(0, t)^T$ for $t \leq 0$ and $(t, 0)^T$ for $t > 0$ is continuous. | |
| Feb 24 at 21:04 | comment | added | muddy | In the case of your example, there is a choice of which $0$-eigenvector to choose. You can choose the $0$-eigenvector in such a way that it is continuous over $I$. This is the key in this problem, which is the existence of an $0$-eigenvector $v(t)$ of $A(t)$ which is continuous over $I$. | |
| Feb 24 at 20:56 | comment | added | Aleksei Kulikov | Yes, now take a differentiable curve passing throught the matrix which is all zeroes, so that before it all but $(1, 1)$ entry is zero, and after it all but $(2, 2)$ entry is zero. Then you can't have a continuous eigenvalue at that point. | |
| Feb 24 at 20:47 | comment | added | muddy | For any diagonal $2 \times 2$ matrix which has a zero eigenvalue. Suppose that its $(2,2)$ entry is zero. The $0$-eigenvector $(0,1)$ is continuous for $t$ in $I$. So I cannot say that it is not true. | |
| Feb 24 at 20:42 | comment | added | Aleksei Kulikov | This is not true, already for diagnonal $2\times 2$ matrices. But the question is not appropriate for MO. | |
| Feb 24 at 20:26 | history | asked | muddy | CC BY-SA 4.0 |