eigenvector

**alexandrabel90** · January 21st 2011, 03:33 PM

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.

**tonio** · January 21st 2011, 03:57 PM

Originally Posted by alexandrabel90

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

**FernandoRevilla** · January 21st 2011, 10:05 PM

Originally Posted by alexandrabel90

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

**HallsofIvy** · January 22nd 2011, 07:01 AM

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 $\left|A- \lambda I\right|= \left|CBC^{-1}- \lambda CIC^{-1}\right|= \left|C(B- \lambda I)C^{-1}\right|$ $= \left|<br /> C\right|\left|B- \lambda I\right|\left|C^{-1}\right|= \left|B- \lambda I\right|= 0$ .

**FernandoRevilla** · January 22nd 2011, 09:39 AM

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 $B$ .

Fernando Revilla

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