Why is there a free choice of variables when finding eigenvectors?

[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.

 

[Ray Vickson]

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.

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
$$\pmatrix{5000\\-1000}\text{ or } \pmatrix{-1/3 \\1/15}\text{ or } \cdots .$$

 

[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:
$$A(au+ bv)= aA(u)+ bA(v)= a(\lambda u)+ b(\lambda v)= \lambda (au+ bv)$$.