## Jordan Normal Form

For the discussion of math. Duh.

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Ddanndt
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### Jordan Normal Form

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

$\left( {J_1\atop } { \atop } { \atop J_n} \right)$

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=

$\begin{pmatrix}{ 0 1 2} \\ {0 0 1} \\{0 0 0} \end{pmatrix}$
but I'm getting P=Id, and therefore $J=PA{P}^{-1}=A ??$
Last edited by Ddanndt on Wed Apr 27, 2011 7:26 pm UTC, edited 1 time in total.
God does not care about our mathematical difficulties — He integrates empirically.
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jestingrabbit
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### Re: Jordan Normal Form

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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DavCrav
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### Re: Jordan Normal Form

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

I guess he meant nilpotent...

Jyrki
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### Re: Jordan Normal Form

@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.

Ddanndt
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Location: Paris

### Re: Jordan Normal Form

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

$\begin{pmatrix} 0 1 2\\0 0 1 \\0 0 0 \end{pmatrix}\begin{pmatrix} 2 \\1 \\0 \end{pmatrix}=\begin{pmatrix} 1 \\0 \\0 \end{pmatrix}$

which is a basis vector of ker(A)

Therefore
$\begin{pmatrix} 1 2 0\\0 1 0 \\0 0 1 \end{pmatrix}^{-1} \begin{pmatrix} 0 1 2\\0 0 1 \\0 0 0 \end{pmatrix}\begin{pmatrix} 1 2 0\\0 1 0 \\0 0 1 \end{pmatrix}=\begin{pmatrix} 0 1 0\\0 0 1 \\0 0 0 \end{pmatrix}$
Last edited by Ddanndt on Wed Apr 27, 2011 7:28 pm UTC, edited 1 time in total.
God does not care about our mathematical difficulties — He integrates empirically.
—Albert Einstein

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### Re: Jordan Normal Form

Ddanndt wrote:First I start with $\begin{pmatrix} 0 \\0 \\1 \end{pmatrix}$ 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
Posts: 60
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Location: Paris

### Re: Jordan Normal Form

God does not care about our mathematical difficulties — He integrates empirically.
—Albert Einstein

skeptical scientist
closed-minded spiritualist
Posts: 6142
Joined: Tue Nov 28, 2006 6:09 am UTC
Location: San Francisco

### Re: Jordan Normal Form

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.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson