can some one explain to me why if v is in V with cordinate vector x in real ^n then v is an

eigenvector for f with eigenvalue λ iff x is an eigenvector for A_f with eigenvalue λ where

A_f is the matrix of A wrt an arbitary basis.

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- Jan 21st 2011, 04:33 PMalexandrabel90eigenvector
can some one explain to me why if v is in V with cordinate vector x in real ^n then v is an

eigenvector for f with eigenvalue λ iff x is an eigenvector for A_f with eigenvalue λ where

A_f is the matrix of A wrt an arbitary basis. - Jan 21st 2011, 04:57 PMtonio
- Jan 21st 2011, 11:05 PMFernandoRevilla
I guess you meant:

**(i)**real vector space**(ii)**endomorphism,**(iii)**basis of**(iv)**such that

**(v)**the matrix of with respect to

In that case, you have to prove which is deduced inmediately using the meaning of matrix of a linear map with respect to a determined basis.

Fernando Revilla - Jan 22nd 2011, 08:01 AMHallsofIvy
The point of this is, of course, that the concept of an "eigenvalue" is independent of the particular matrix representation you choose for a linear transformation- that "eigenvalue" is really a "Linear Algebra" concept and not just a "Matrix Algebra" concept.

The simplest way to prove this is to use the fact that the matrices representing a linear transformation in different bases are "similar". That is, that A and B represent the same linear transformation, in different bases, if and only if there exist an invertible matrix, C, such that . (C is the "change of basis" matrix, the matrix that maps the vectors in one basis into the vectors in the other basis.)

Recall that the eigenvalues of A are the solutions to "characteristic equation", . Since, for any invertible C, , we can write that as . - Jan 22nd 2011, 10:39 AMFernandoRevilla
We can prove the statement using a

__previous__result to the eigenvalues theory:

If is a vector space over , finite, is a fixed basis of and is the matrix of with respect to then,

where,

are the coordinates of and respectively, on .

Fernando Revilla