Talk:Jordan normal form/Archive 1

Archive 1

description of the Jordan normal form is ambigous

The description of the Jordan normal form is ambigous. Maybe put in something like

${\begin{pmatrix}\lambda &1&0&\cdots &0\\0&\ddots &\ddots &&\vdots \\\vdots &\ddots &\ddots &\ddots &0\\\vdots &&\ddots &\ddots &1\\0&\cdots &\cdots &0&\lambda \end{pmatrix}}$

Mathias de Riese 10:06, 20 May 2004 (UTC)

Jordan form notation

The article needs a explanation of the jordan form notation. A newbie who is reading the article can't figure out what's the difference between a $J_{2}(4)$ and a $J_{4}(4)$. --Mecanismo 12:00, 27 November 2005 (UTC)

Editorial suggestion

The article has this portion of text:

"of the 19th and early 20th-century French mathematician Camille Jordan"

There is no need to include details on the life of Camille Jordan in the Jordan normal form article. If a user wishes to learn more about Camille Jordan, he can simply click on the wiki-link to Camille Jordan's biography, which is available in the beginnig of the article and where he will be able to access information on that subject.

So, having that in mind, I propose that, for the sakes of brevity, that piece of text is substituted with "of Camille Jordan". --Mecanismo 17:00, 27 November 2005 (UTC)

Segre characteristic/type/notation

There should be some mention of the Segre notation/type/characteristic as a quick means of seeing what the matrix looks like without writing out the whole thing. ---Mpatel (talk) 07:10, August 11, 2005 (UTC)

JohnCreighton_@hotmail.com -what does $\oplus$ denote?

I had given an incorrect link to the page on Sumset, which is denoted by that operator. But Dysprosia rightly removed it. However, Dysprosia doesn't feel like enlightening us as to what $\oplus$ means... Fresheneesz 04:33, 8 February 2006 (UTC)

No, I don't feel like enlightening you as to what it means, as it is mentioned in the article "The way the normal form is usually written is explicitly as the direct sum of block square matrices, known as Jordan blocks." I've tried to draw attention to it a little closer. Dysprosia 04:46, 8 February 2006 (UTC)

Clarifying and regularizing terminology

There appear to be three different terms that are used to describe the subject of this article. Google searches find all three being used in a mathematical sense. I also checked one mathematics dictionary, three textbooks, and one web site (not WP!). Here, listed with the most frequent first (from google), is a summary:

1. Jordan canonical form is defined by Finkbeiner, MacLane & Birkhoff, Shilov, and Mathworld.
2. Jordan normal form is given as a synonym of Jordan canonical form in Shilov in a footnote.
3. Classical canonical form is defined in James & James and given as an alternative to Jordan canonical form at Mathworld.

There were more google hits for Jordan canonical form than for Jordan normal form. By themselves, I don't consider google hit counts very authoritative, but combined with the other evidence, it seems to me that this article should be renamed to Jordan canonical form with redirects from the other terms.

I am not going to rename the article until other editors comment. However, in the article, I have clarified the terminology based on the above work. The article had been using the terms Jordan normal form and Jordan canonical form apparently interchangeably. I regularized the terminology by changing all uses of canonical to normal. I hope someone will review this edit, because it changed quite a few places and I might have broken something.

--Jtir 16:09, 21 September 2006 (UTC)

Abbreviating Jordan normal form to Jordan form

Both Shilov and Finkbeiner sometimes use the abbreviation Jordan form. The article currently uses both Jordan form and normal form as abbreviations. It seems to me that consistent use of Jordan form would be preferable because it avoids having to use either normal or canonical.

--Jtir 17:29, 21 September 2006 (UTC)

There are redirects or dabs for all meaningful two-word abbreviations. --Jtir 21:19, 21 September 2006 (UTC)

confusing grammar in lead sentence

The lead sentence reads:

In linear algebra, Jordan normal form or Jordan canonical form shows that a given square matrix M over a field K containing the eigenvalues of M can be transformed into a certain canonical form by changing the basis.

The confusing grammar can be seen by removing a few words:

In linear algebra, Jordan normal form ... shows that a ... matrix ... can be transformed ...

1. Shouldn't it read: the Jordan normal form?
2. How can a matrix show something?

The lead sentence is a theorem, not a definition. While the name of a theorem could be the title of an article, the title of this article is the name of a particular type of matrix. Is there a distinct name for the associated theorem? --Jtir 07:34, 24 September 2006 (UTC)

Here you go:

• Jordan canonical form theorem
• JSTOR: Power Bounded Strictly Cyclic Operators

"By the Jordan canonical form theorem, it suffices to prove the theorem for a single Jordan block, so consider an operator ..."

• abstract of Differential Equations: Theory and Applications with Maple

"... also has a complete collection of proofs for the major theorems ranging from the usual existence and uniqueness results to the Hartman-Grobman theorem and the Jordan canonical form theorem."

--Jtir 13:08, 24 September 2006 (UTC)

This article defines the terms Jordan matrix and Jordan block and it links to Jordan matrix, which also defines them. What is the relationship between these articles?

The article uses the terms Jordan normal form and Jordan canonical form interchangeably. That is very confusing.

--Jtir 15:19, 20 September 2006 (UTC)

zeros along the super-diagonal

The Complex matrices section says "Every entry on the super-diagonal is 1."
In general, aren't there also a few zeros along the super-diagonal? --Jtir 08:13, 24 September 2006 (UTC)

that was text added by me. you're right. i've clarified it. Mct mht 12:49, 24 September 2006 (UTC)

Thanks, that makes sense now. It might be helpful to add that there are both zeros and ones along the super-diagonal of J, with the zeros occurring at the vertex (is there a term for this?) between two adjacent Jordan blocks. In trying to sort out this subject, I am finding that describing the JNF/JCF concisely, yet accurately, is very difficult. --Jtir 13:24, 24 September 2006 (UTC)

The example matrix matrix in motivation section

It is supposed to have five repeated eigenvalues, but when I checked it with Matlab, i got a slight difference. Could it be some sort of numerical roundoff error in the matrix, or is there a mistake in the matrix? Matlab code:

A = [322 -323 -323 322; 325 -326 -325 326; -259 261 261 -260; -237 237 238 -237]; [V, D] = eig(A)

Result:

V =

  Columns 1 through 3

   0.6159             0.6159             0.6160          
   0.4400 + 0.0001i   0.4400 - 0.0001i   0.4398 + 0.0001i
  -0.3716 - 0.0001i  -0.3716 + 0.0001i  -0.3714 - 0.0001i
  -0.5376 + 0.0000i  -0.5376 - 0.0000i  -0.5378 + 0.0000i

  Column 4

   0.6160          
   0.4398 - 0.0001i
  -0.3714 + 0.0001i
  -0.5378 - 0.0000i


D =

  Columns 1 through 3

   5.0044 + 0.0044i        0                  0          
        0             5.0044 - 0.0044i        0          
        0                  0             4.9956 + 0.0044i
        0                  0                  0          

  Column 4

        0          
        0          
        0          
   4.9956 - 0.0044i

Heptor _talk 16:25, 20 January 2007 (UTC)

thanks for checking. probably a (small) numerical error. the e.g.'s (in the beginning and examples 1 and 2 later) in the current version of article are not very clean looking anyhow. feel free to replace them. Mct mht 16:53, 20 January 2007 (UTC)

I did that, but the same numbers are used in another example in the article. It should be either fixed or deleted. -- Heptor _talk 19:22, 20 January 2007 (UTC)

it'd be good if you can modify those as well. Mct mht 20:11, 20 January 2007 (UTC)

Guess what, the geometric multiplicity of eigenvalue 5 in the matrix is not 1, it is 3. Will fix it some day maybe, adding {{Disputed}} untill further. -- Heptor _talk 00:33, 2 March 2007 (UTC)

for a matrix whose JCF is J, from that section, the geometric multiplicity should be 1. i'm going to remove the tag. if for some reason one would want to put it back, tag should be restricted to the section in question. Mct mht 05:40, 2 March 2007 (UTC)

Heptor probably meant it the other way around: the matrix

$A={\begin{bmatrix}5&1&1&1\\0&5&0&0\\0&0&5&0\\0&0&0&5\end{bmatrix}}.$

which was originally in the article has geometric multiplicity 3, while the Jordan form has g.m. 1 (as Mct mht says). I did a fairly quick fix. I also removed the first example which seemed to refer to the article in a prior state and which did not seem to add to the other example.

Perhaps we should mention another matrix in the "Motivation" section. The current choice is not an easy matrix for seeing what happens. However, I don't feel inspired now. -- Jitse Niesen (talk) 13:40, 2 March 2007 (UTC)

Powers of Jordan form

There is a relatively simple form for the nth order power of Jordan matricies, could we add that here? 128.211.220.163 (talk) 19:49, 7 December 2007 (UTC)

Correction

The statement that "By construction, the union the three sets {p1, ..., pr}, {qr−s +1, ..., qr}, and {z1, ..., zt} is linearly independent." is incomplete.

On the face of it the reader might get the incorrect impression that if A, B, C are finite sets of independent vectors, then their union is a collection of independent vectors. Here is a sketch of a correction. Suppose that sum ai pi +sum bi qi +sum ci ri =0. The trick is to apply A-lambda I to this expression. The image terms which are not annihilated will be linearly independent by induction, and thus the fact that the image sum is zero will imply that certain coefficents are zero. Plug that information back into the original linear dependence - you are looking at elements in the kernel which were chosen to be independent, and the rest of the coefficients will be zero.128.97.4.96 (talk) 23:07, 25 May 2008 (UTC)

Further clarification:

0. It would have better if I had written sum ai pi +sum bj qj +sum ck rk=0

1. The suggested argument works best if there is only one eigenvalue lambda. After that you must show that the different eigenvalues determine a direct sum of generalized eigenvector spaces (see Fiedberg, Insel, Spence Linear Algebra0.

2. It helps to use "Jordan chains": diagrams of basis vectors with arrow indicating what T-lambda I does to them. See the aforementioned text.

3. The argument is a little cleaner if in the induction you show that the "last vectors" in the basis of Jordan chains, i.e., the eigenvectors in those chains of basis vectors span the kernel of the restriction of T - lambda I. Assume that v is in the indicated kernel and expand in terms of the inductively hypothesized Jordan chain eigenvectors.

4. References might be given for the more general approach normally used in algebra courses. —Preceding unsigned comment added by 128.97.4.96 (talk) 20:38, 27 May 2008 (UTC)

Error in proof corrected

The current version of proof has a small error. I have made two small changes to correct it. Tbe problem is in the proof that if R(A-\lambda I) has nontrivial intersection with N(A-\lambda I), then we need to use the "last" vector, not the lead vector to extend the chain!

I have posted a very detailed construction for my Differential Equations class, but with the same idea of proof on my website, and chanlenge the students to find error of the current web version. I'll wait to see if any talented studdents can spot the error.

Xiao-Biao Lin

sry about reverting you earlier. but the current version of proof, using the leading vectors, seems to be right. we can assume they are the leading vectors in the Jordan chain, the eigenvectors, because, by assumption they lie in the kernel of A - λ. since they are also in range of A - λ, the Jordan chain is then extended by passing to the preimage. Mct mht 13:38, 25 August 2006 (UTC)

it is true that the last vectors in the Jordan chain also lie in both kernel and range of A - λ. but if one uses the last vectors to extend the chain, a possible issue is that the following statement from the argument is no longer necessarily true:

Furthermore, qi can not be in Ran(A - λ), for that would contradict the assumption that each pi is a lead vector in a Jordan chain.

in other words, the preimage of last vectors might not take you out of Ran(A - λ). Mct mht 14:16, 25 August 2006 (UTC)

Dear Editor:

I feel the whole thing is a simple misunderstanding. I like your word "take you out of Ran(A- lambda)".

Here I can use the simple example in this same article to illustrate our difference. Suppose A is a 4x4 matrix, and we have only one Jordan block with the Jordan chain:

(A-5)p4 = p3, (A-5)p3 = p2, (A-5)p2 = p1, (A-5)p1 = 0.

Ran(A-5) has a Jordan chain consisting of 3 vectors, p3,p2 and p1 where p1 is the lead vecotr and is also the only eigenvector. The others are generalized eigenvectors. To extend the Jordan chain in Ran(A-5) to the original 4-dim space, we need to use the last vecotor p3, which is the vector that has no preimage in the space Ran(A-5). Take preimage of p3 one time, but in the larger space, you get p4, that takes you out of Ran(A-5).

If you use the eigenvector p1, which is defined to be the lead vector in the article, you will need to take preimage three times to get out of Ran(A-5).

Suggestion: The easiest way to correct the error is to define the lead vector as the "generating vector" of the Jordan chain: the one that has no preimage, usually not the eigenvector. In this example it is p4 for the 4-dim space and is p3 in Ran(A-5).

XBL

yes, after our above conversation earlier, it looked to me like the only difference was notation and terminology. :-) Mct mht 20:29, 26 August 2006 (UTC)

feel free to make changes per your suggestion above. Mct mht 20:56, 26 August 2006 (UTC)

actually, you're quite right. the difference was less minor than mere notational. i've made some corrections per your comments above. Mct mht 12:34, 27 August 2006 (UTC)

There was a problem with the section discussing the use of the "previous" Jordan matrix. Previously we had a Jordan Matrix that had eigenvalues of all 5, which is clearly different from any matrix in the previous example. 99.236.42.178 (talk) 03:48, 3 November 2008 (UTC)

Alternative canons

Perhaps it should be noted that some other authors (notably, Lang) prefer to write the Jordan canonical form with 1's on the subdiagonal, and a brief explanation given of why the two are more or less equivalent. --Druiffic (talk) 00:54, 9 December 2008 (UTC)Druiffic

Since every matrix is similar to its transpose, this isn't really a difference in any significant way. I don't know if this is worth mentioning at all. -- Taku (talk) 02:41, 9 December 2008 (UTC)

Good point. I only mention it to possibly avoid some confusion. --Druiffic (talk) 07:43, 9 December 2008 (UTC)Druiffic

For a repeated eigenvalue, there might be more than one Jordan block

Somehow the page gives the impression that each eigenvalue corresponds to only one Jordan block. Real matrices with this property is called cyclic matrices. The following matrix is already in Jordan form, it has eig-val 5 repeated five times but has two different Jordan blocks instead of one: ${\begin{bmatrix}5&1&0&0&0\\0&5&1&0&0\\0&0&5&0&0\\0&0&0&5&1\\0&0&0&0&5\end{bmatrix}}.$ — Preceding unsigned comment added by Mathractitioner (talk • contribs) 14:53, 14 January 2009 (UTC)

Possible redirect

When one searches for "Jordan basis", the search comes up with nothing and recommends "Jordan bases". Seeing as how one is the plural of the other, shouldn't "Jordan basis" automatically redirect here? 74.192.195.90 (talk) 03:30, 11 August 2009 (UTC)

SOFIxIT Algebraist 07:58, 11 August 2009 (UTC)

Done. Also, SOFIXIT -- Jalanpalmer (talk) 21:47, 21 January 2010 (UTC)