
#1
May105, 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'? 



#2
May105, 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. 



#3
May105, 06:58 PM

Math
Emeritus
Sci Advisor
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.) 



#4
May105, 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.




#5
May205, 08:00 AM

Sci Advisor
HW Helper
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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