# Free choice of variables

1. Mar 31, 2013

### synkk

Hi, I don't quite understand when finding eigenvectors there is usually a free choice of variables to pick, and you can sub in any number to find an eigenvector. Could anyone please explain how this works (and why you can sub in any number), as this usually comes up in vector problems with 3 variables also.

2. Mar 31, 2013

### Ray Vickson

For ANY value $c \neq 0$ the vector
$$\pmatrix{x\\y} = c \pmatrix{-5\\1}$$
is an eigenvector. So, If I want, I can choose
$$\pmatrix{5000\\-1000}\text{ or } \pmatrix{-1/3 \\1/15}\text{ or } \cdots .$$

3. Apr 1, 2013

### HallsofIvy

In fact, the set of all eigenvectors of liear transformation A, corresponding to a given eigenvalue, form a subspace.

If u and v are both eigenvectors corresponding to eigenvalue $\lambda$ then so is au+ bv for any scalars a and b:
$$A(au+ bv)= aA(u)+ bA(v)= a(\lambda u)+ b(\lambda v)= \lambda (au+ bv)$$.

Last edited by a moderator: Apr 1, 2013