Talk:Generalized eigenvector

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Intuition[edit]

Can someone give an intuitive explanation of these vectors? ie. with a normal eigenvector, when the matrix is multiplied by the vector, it is simply scaled, not rotated. Is this also the case for a generalized eigenvector? What does it mean when the dimension of the null space of (A-\lambda I) is more than 1 (ie. now there are more than one eigenvectors associated with the same eigenvalue? Does it just mean that two different directions are scaled by the same amount? Or does it have stronger mean, like something about the vectors spanned by the two eigenvectors??) daviddoria (talk) 21:26, 16 September 2008 (UTC)[reply]

This has nothing to do with the dimension of

ker(A - λ I)

. Every element of that set (other than 0), is a regular eigenvector of A. The vectors spanned by two eigenvectors for the same eigenvalue are also regular eigenvectors for that eigenvalue. A generalized eigenvector v such that

(A - λ I) 2 v = 0

almost acts like a normal eigenvector, except it picks up a bit of a normal eigenvector in the action:

Av = λ v + w

, where w is an eigenvector

Aw = λ w

. A similar fact when k = 3, except in that case, w is a k = 2 generalized eigenvector, and so on.Eigenguy (talk) 01:28, 12 July 2012 (UTC)[reply]

I expected the section "for defective matrices" to provide some intuition, but I cannot comprehend it. Improvement of the presentation will be very welcome. The above comment by Eigenguy seems insightful, couldn't it be integrated into the article? (Maybe even extended?) AmirOnWiki (talk) 14:16, 21 October 2014 (UTC)[reply]

@AmirOnWiki: Defective matrix should really be the place for more explanation of a defective matrix. I believe the intuition comes from the simple examples. Frietjes (talk) 17:32, 21 October 2014 (UTC)[reply]

"Motivation of the Procedure" section[edit]

...is an opaque, dense tract. The small snippet I could stand to read at the beginning of the proof looks correct, but it's not well organized with respect to the rest of the article, and does not use a large, readable math font like the rest of the article. Worse, the recursive procedure is presented as a proof of its viability without clearly introducing the procedure itself. This section needs to be cleaned up and bridged to the easier examples in the earlier part of the article, but I'm not competent enough to do it. 128.171.78.5 (talk) 00:36, 1 October 2012 (UTC)[reply]

I completely agree with this. It is way to technical and focused on the proof, rather than the usage. When I have the time, i.e. when I'm done with my exams, I will take a look at this and see whether I'm able to explain the general procedure better. Until then, I will add a tag to get this page some attention. Erispre (talk) 13:45, 2 April 2013 (UTC)[reply]

I am working to clean it up by first fixing the excessive bolding and use of <br /> tags, but it will take some time. Especially since other editors seem to feel the need to blinding revert my edits. 198.102.153.2 (talk) 16:36, 11 November 2013 (UTC)[reply]

Cleanup[edit]

This article is a disaster, and it's been a disaster since around July 2012. It has a lot of good information, but a lot of it is also inappropriate and disorganized, and the typesetting is atrocious. It needs an expert to carefully go through it and salvage what can be salvaged. I might do it myself, but it's a bit daunting given the article's length... 50.132.4.93 (talk) 03:15, 21 May 2015 (UTC)[reply]

I agree. I think this article has far too many examples and proofs. I am tempted to clean it up myself but I would be inclined to remove most of the examples and proofs, and that might anger the editors who originally put them into the article. This definitely merits some discussion. — Anita5192 (talk) 06:49, 21 May 2015 (UTC)[reply]

Anita5192, please feel free to take an axe to it, anything would be better than the current situation. Frietjes (talk) 16:10, 24 May 2015 (UTC)[reply]

Okay, I will clean up the article as I see fit. Anyone who disagrees with my edits can discuss them here. — Anita5192 (talk) 20:30, 25 May 2015 (UTC)[reply]

Done I think the article is much more readable now and that it retains all of the important concepts. — Anita5192 (talk) 22:03, 22 June 2015 (UTC)[reply]

Normal eigenvectors are not general eigenvectors?[edit]

The definition of generalized eigenvector clearly states that normal eigenvectors should be considered generalized eigenvectors of rank one. It is also stated that, for every matrix $A$ , there exists a basis of eigenvectors. This I agree with. However, in the first example it is claimed that there is only one generalized eigenvector for a certain $2\times 2$ matrix. Only after reading the entire example it was clear to me that they didn't consider the normal eigenvector to be a general eigenvector there. I suggest that this example be altered to fit the definition. --Bib-lost (talk) 15:22, 26 May 2016 (UTC)[reply]

Good eyes! Example 1 is a passage of legacy text that was not cleaned up last year when the article was rewritten. I just rewrote it. The wording is a little cumbersome, but hopefully clearer. — Anita5192 (talk) 17:23, 26 May 2016 (UTC)[reply]

Note on example 1[edit]

Quoting: "The generalized eigenvector of rank 2 is then $\mathbf {v} _{2}={\begin{pmatrix}a\\1\end{pmatrix}}$ , where a can have any scalar value. The choice of a = 0 is usually the simplest". True, but also, a = 0 is the only value that makes $\mathbf {\begin{pmatrix}a\\1\end{pmatrix}}$ normalized. May be this is also a good reason to choose zero. Aris Makridis (talk) 08:38, 29 August 2023 (UTC)[reply]