>given a non-diagonalizable matrix [math]P[/math], calculate [math]P^{k}[/math]
Can someone point me to the right direction for starting my quest of learning how to begin understanding an answer to the above question?
Just convert P into Jordan normal form.
>>7746824
That's easier said than done if you don't even know what a Jordan normal form is.
>>7746837
...
how about you look it up?
>>7746867
Come on now
>>7746867
We're on /sci/ buddy
There is no sensible reason why you would want to do this.
If standard decompositions don't work then you are dealing with a very specific type of matrix that should not be discussed here.
>>7747145
Wow fuck off
>>7747145
You know your shit but you don't know that you don't know enough of the shit there is to know.
>>7746808
If P is not diagonalizable, it still has a jordan normal form like >>7746824 said.
which means, you have D diagonal, N nilpotent and a base transform S such that
P = S^-1 (D+N) S
and therefore
P^k = S^-1(D+N)^k S
(D+N)^k can be calculated easily via the binomial theorem and the knowledge that N is nilpotent and commutes with D
>>7747284
Is it easier in the case of a non-diagonalizable matrix to find [math]P^{k}\mathbf{v}[/math] or [math]P^{k}[/math]? Or doesn't it matter?
to be honest the problem doesn't specify how efficiently you have to do it. so just multiply P by itself k times
>>7747654
Think of [math]P^{100}[/math] or [math]P^{1000}[/math]
>>7747652
well obviously, if v is an eigenvector/generalized eigenvector, it gets trivial.
otherwise you have to compute the matrix anyways
>>7747682
Shit man I basically need to make a proof involving a non-diagonalizable matrix P, and whether [math]P^{k}\mathbf{v}[/math] converges or not.
>>7747686
then you just have to prove, that the eigenvalues of P have an absolute value smaller than 1
>>7747689
I have already proven that! So now I just need to combine this with the Jordan blocks the other dude was talking about and then BOOM finally done with this assignment. You got any good books or youtube vids that give me a brief, no-bullshit explanation of this?
>>7746808
I don't even know what non diagnosiliable means but why don't you just multiply P with itself k times xD
>>7747695
well from the looks of it, the wikipedia articla seems okay.
never learned la from books though, so i cant help you with that.
I'm gonna watch the thread if you have questions
>>7747705
It's indeed a good article, I've read it now. However, I've still got one question. If we have found the Jordan form [math]J[/math] of a matrix [math]P[/math], is it easy to calculate [math]J^{k}[/math]? And how?
>>7747731
to quote myself
>>7747284
since J is the sum of a diagonal matrix and a nilpotent jordan block matrix that commute, you can use the binomial theorem (pic related) to calculate powers of it.
Powers of the diagonal part are easy to compute, since you can just take the diagonal values to the power - power of the nilpotent part will vanish after a while (lets say your biggest jordan block is 3x3, it will vanish after taking it to the power of 3)
>>7747753
That's marvellous stuff! Thank you, so if I understand it correctly, we get [math] J = D + N [/math] where D is the diagonal matrix with the eigenvalues of P on the diagonal, and N is nilpotent. So
[eqn]
J^{k} = (D + N)^{k} = \sum_{n=0}^{k} {n \choose k} D^{k} N^{n-k}
[/eqn]
Am I right?
>>7747766
exactly. be carefull with your notation of n and k though, since you're mixing them up
>>7747770
You're absolutely right. Thanks A LOT for all of this anon, it was the last piece of the puzzle for me.
>>7747766
>>7747766
Exactly! If you look closely, this should be an intuitive result too. Basically, diagonizability says that the space admits an eigenbasis (I.e. basis of eigenvalues) under the operator (your matrix), so it should be relatively easy to see that repeated application of the operator continues to scale the shifted basis (I.e. your eigenbasis) by the corresponding eigenvalues. Thus, powers of the diagonalized matrix! This has even more fun ties to generalized eigenvectors and generalized eigenbases too, which is to say that there's a corresponding (though not as nice) format to calculate these types of things even for nondiagonalizable matrices.
