Jordan Normal Form

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Ddanndt
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Jordan Normal Form

Postby Ddanndt » Sat Apr 23, 2011 9:41 am UTC

I need some help for the Jordan Normal Form please. I understand that the Jordan Nomal form is of the form

[math]\left( {J_1\atop } { \atop } { \atop J_n} \right)[/math]

where J_i is a Jordan block . And therefore the only possible non-zero entries are the diagonal and the superdiagonal(with 1's).
But now I have to find the Jordan Normal form of the [imath]\rlap{===}idem[/imath]nilpotent matrix , A=

[math]\begin{pmatrix}{ 0 1 2} \\ {0 0 1} \\{0 0 0}

\end{pmatrix}[/math]
but I'm getting P=Id, and therefore [math]J=PA{P}^{-1}=A ??[/math]
Last edited by Ddanndt on Wed Apr 27, 2011 7:26 pm UTC, edited 1 time in total.
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jestingrabbit
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Re: Jordan Normal Form

Postby jestingrabbit » Sat Apr 23, 2011 10:21 am UTC

Don't just tell us what you got for P, tell us the process: characteristic polynomial, eigenvalues, eigenspaces or generalised eigenspaces? etc.

Also, I wouldn't call that matrix idempotent.
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Re: Jordan Normal Form

Postby DavCrav » Mon Apr 25, 2011 9:53 am UTC

jestingrabbit wrote:Also, I wouldn't call that matrix idempotent.


I guess he meant nilpotent...

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Re: Jordan Normal Form

Postby Jyrki » Tue Apr 26, 2011 12:14 pm UTC

@OP: Surely somewhere in your lecture notes/textbook there is an example showing what to do, when the characteristic polynomial has a multiple root, but there aren't enough linearly independent eigenvectors? That piece of the theory seems to have hit you full force.

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Ddanndt
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Re: Jordan Normal Form

Postby Ddanndt » Wed Apr 27, 2011 6:59 pm UTC

Yeah sorry I meant nilpotent... And for some reason i think I missed an important step in finding the vectors of the matrix :? .

First I start with [math]\begin{pmatrix} 0 \\0 \\1 \end{pmatrix}[/math] which is a basis vector of Ker(A3) and isn't in Ker(A2) . Then I find
[math]\begin{pmatrix} 0 1 2\\0 0 1 \\0 0 0 \end{pmatrix}\begin{pmatrix} 0 \\0 \\1 \end{pmatrix}=\begin{pmatrix} 2 \\1 \\0 \end{pmatrix}[/math] which is a basis vector of Ker(A2) and not in Ker(A) and

[math]\begin{pmatrix} 0 1 2\\0 0 1 \\0 0 0 \end{pmatrix}\begin{pmatrix} 2 \\1 \\0 \end{pmatrix}=\begin{pmatrix} 1 \\0 \\0 \end{pmatrix}[/math]

which is a basis vector of ker(A)

Therefore
[math]\begin{pmatrix} 1 2 0\\0 1 0 \\0 0 1 \end{pmatrix}^{-1} \begin{pmatrix} 0 1 2\\0 0 1 \\0 0 0 \end{pmatrix}\begin{pmatrix} 1 2 0\\0 1 0 \\0 0 1 \end{pmatrix}=\begin{pmatrix} 0 1 0\\0 0 1 \\0 0 0 \end{pmatrix}[/math]
Last edited by Ddanndt on Wed Apr 27, 2011 7:28 pm UTC, edited 1 time in total.
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Re: Jordan Normal Form

Postby skeptical scientist » Wed Apr 27, 2011 7:13 pm UTC

Ddanndt wrote:First I start with [math]\begin{pmatrix} 0 \\0 \\1 \end{pmatrix}[/math] which is a basis of Ker(A3).

It's not a basis. What is a basis? I'm not quite sure what you meant to write...

Other than that (and the same issue again on the next line), this looks correct.
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Ddanndt
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Re: Jordan Normal Form

Postby Ddanndt » Wed Apr 27, 2011 7:29 pm UTC

Again sorry should read "basis vector" instead of "basis" :D
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Re: Jordan Normal Form

Postby skeptical scientist » Wed Apr 27, 2011 8:14 pm UTC

Well, when you're just talking about a single vector, the term "basis vector" is kind of silly, since any nonzero vector in W can be part of a basis for W.
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