Jordan Normal Form

FUCK! I'm getting so fucking frustrated by this. NOWHERE ON THE ENTIRE INTERNET seems to be a good, straight forward tutorial on how to find this.

Just don't give a fuck. It's useless I told you

> You essentially split a matrix into a nilpotent one
Except that you can't do it. It's uncomputable

About that pic, I've seen this video - that's not the correct translation at ALL

Yeah but my linear algebra teacherdude wants us to be able to do it.

Holy shit, OP here. I think I finally got it.

these notes seem pretty good

mavdisk.mnsu.edu/singed/Spring 2012/Linear Algebra/Effective procedure for computing Jordan Normal Form of nilpotent matrix.pdf

although the procedure is specifically for nilpotent matrices

so for an arbitrary map T with eigenvalues [math] \lambda_i [/math], you'd have to apply this procedure individually to each nilpotent operator [math]T-\lambda_i\mid_{E_i}[/math]

where the vertical bar denotes restriction to the subspace E_i and E_i is the generalized eigenspace given by [math] E_i=\cup_k Ker(T-\lambda_i)^k[/math]

>It's like they don't WANT us to understand it
It should be "It's like they DON'T want us to understand it"

>I just don't know HOW to construct the basis transformation.
just check any of the standard math textbooks the has a proof on JNF and go through the proof.
fairly sure it's useless, tho.

knowing the eigenvalues / char poly is basically enough to write down the JNF.
the exact basis transformation seems irrelevant tbqh

btw I just googled "jordan normal form" and in the 3rd link there was a working example how to get it.

OP is confirmed faggot

>knowing the eigenvalues / char poly is basically enough to write down the JNF.

this is just wrong

Thanks. So basically here are my thoughts:

Let's say you have an eigenvalue [math] \lambda [/math] and you need [math] n [/math] eigenvectors for it. Then you pick an element in [math] \ker (A - \lambda I)^n [/math] that is NOT contained in [math] \ker (A - \lambda I)^{n-1} [/math], and you just start throwing [math] (A - \lambda I) [/math] at it until you end up with 0 and then you reverse the order and you have your basis with respect to [math] \lambda [/math]. Am I wrong?

Let's say you have the square matrix [math]A [/math].

1.) Calculate the characteristic polynomial [math] p(t) = \det (t I - A) [/math]
2.) Factor the characteristic polynomial [math]p(t) = \prod_i (t - \lambda_i)^{k_i} [/math]
3.) Find a basis for [math] \ker(A - \lambda_1)[/math] then add vectors to get a basis of [math] \ker(A - \lambda_1)^2[/math] and continue until the number of vectors is equal to the algebraic mulitplicity of [math] \lambda_1 [/math].
4.) Do this for the other eigenvalues too and you have the right basis for the JNF.

Wait I have a problem with step 3.

Don't you need to have that [math] (A- \lambda I) v_n = v_{n-1} [/math] since [math] A - \lambda [/math] is nilpotent?

you're on the right track, but it's a little tricker because it might be that (A-\lambda I)^k=0 for some k

Except you can, using an algorithm based on Newton's method (it doesn't give you an eigenbasis but it does compute the nilpotent part)

I'm gonna have to let this sink in for a bit. Thanks.

it's been a long time. what invariants do you need?

I thought eigenvals + multiplicities were enough. maybe you also need the minimal polynomial?

characteristic polynomial only tells you the algebraic multiplicity of each eigenvalue (i.e., number of times it appears on diagonal), and minimal polynomial tells you the size of the largest jordan block of each eigenspace, but you also need to the know the sizes of the smaller jordan blocks

JNF is best seen as a consequence of the structure them for fg modules over a pic. See for instance Aluffi's algebra book for an exposition.

>doing diff equaitions before la

why do american do this?

QR factorization algorithm.

if you can find a fast method, then you probably work at google already.

>so at each step i, you'd need to choose a maximal set of linearly independent vectors v_j contained in Ker(T^i)-Ker(T^{i-i})

Wait, I still don't understand this. When you take multiple linearly independent vectors, you basically get multiple threads of

[eqn] (A - \lambda I)^n w, (A - \lambda I)^{n-1} w, \dots (A - \lambda I) w, (A - \lambda I)^0 w[/eqn]

When you only need one "thread" for each eigenvalue. How does that work then?