motivations for eigenvalues/vectors


by eckiller
Tags: motivations
eckiller
eckiller is offline
#1
May1-05, 05:00 PM
P: 45
Hi,

I understand one of the motivations for eigenvalues/vectors is when you need
to compute A^k * x. So we like to write,

A = C*D*C^-1 and then A^k = C * D^k * C^-1, and D^k is trivial to compute.

My professor said C^-1 and C can be though of as change of coordinate
matrices. But from which basis? For example, C^-1 would take me from
*some* basis to the basis of eigenvectors. But what is this *some* basis?

Is it assumed that everything is coordinitized relative to some basis B in
R^n. And then I want to change to the basis of eigenvectors B'?
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Data
Data is offline
#2
May1-05, 05:06 PM
P: 998
Short answer: Yes.

A is the matrix of a transformation wrt some basis B. D is the matrix of the same transformation wrt the eigenbasis B'. [itex]C^{-1}[/itex] takes vectors from B to B', and C takes vectors back from B' to B.
HallsofIvy
HallsofIvy is offline
#3
May1-05, 06:58 PM
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Thanks
PF Gold
P: 38,882
A general linear transformation can be written as a matrix in a given basis. If all you are given is the matrix, then the corresponding "given basis" is <1, 0,...>, <0,1,...> , etc. The basis you are changing to is the basis consisting of the eigenvectors for the matrix A.

(A matrix can be diagonalized if and only if there exist a basis consisting entirely of eigenvectors of the matrix.)

eckiller
eckiller is offline
#4
May1-05, 09:00 PM
P: 45

motivations for eigenvalues/vectors


Thanks for clearing that up. I wish both of my linear algebra textbooks made it clear which basis we were in.
matt grime
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#5
May2-05, 08:00 AM
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P: 9,398
But it obviously goes from whatever basis A is written with respect to, and it doesn't matter what that basis is, which is why the book didn't state it.


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