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  1. #1
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    eigenvector

    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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  2. #2
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    Quote Originally Posted by alexandrabel90 View Post
    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.

    What does "v is in V with cordinate vector x in ^n" mean??

    Tonio
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    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) V real vector space (ii)   f:V\rightarrow V endomorphism, (iii) B=\{e_1,\ldots,e_n\} basis of V (iv) v\in V such that

     x=\begin{pmatrix}x_1\\ \vdots\\{x_n}\end{pmatrix} \quad (v=x_1e_1+\ldots+x_ne_n)

    (v) A_f the matrix of f with respect to B

    In that case, you have to prove f(v)=\lambda v \Leftrightarrow A_{f}x=\lambda x which is deduced inmediately using the meaning of matrix of a linear map with respect to a determined basis.


    Fernando Revilla
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  4. #4
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    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 A= CBC^{-1}. (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", \left|A- \lambda I\right|= 0. Since, for any invertible C, I= CIC^{-1}, we can write that as \left|A- \lambda I\right|= \left|CBC^{-1}- \lambda CIC^{-1}\right|= \left|C(B- \lambda I)C^{-1}\right| = \left|<br />
C\right|\left|B- \lambda I\right|\left|C^{-1}\right|= \left|B- \lambda I\right|= 0.
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    We can prove the statement using a previous result to the eigenvalues theory:


    If V is a vector space over \mathbb{K} , \dim V=n finite, B is a fixed basis of V and A_f is the matrix of f with respect to B then,

    Y=A_f X

    where,

    X=\begin{bmatrix}x_1\\ \vdots\\{x_n}\end{bmatrix}\; ,\quad Y=\begin{bmatrix}y_1\\ \vdots\\{y_n}\end{bmatrix}

    are the coordinates of x\in V and f(x)\in V respectively, on B .



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