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.
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
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 .
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