Let A be a square matrix, show that any non-zero linear combination of two eigenvectors $v$ and $w$, corresponding to the same eigenvalue, is also an eigenvector.
First I'll show what I did..
1) Let $V$ and $W$ be eigenvectors of A with corresponding eigenvalue $\lambda$
2) If we use the characteristic equation for both eigenvectors, we get…
$Av = \lambda v$
$Aw = \lambda w$
3) Now if we use $v$ and $w$ as vectors for a linear combination with the eigenvalue $\lambda$ we get
$\lambda v + \lambda w = x$
4) now we just have to prove that $x$ is an eigenvector with corresponding eigen value $\lambda$
this is where im stuck.. i need someone to give me a useful hint on what to do next..
did i even take the proper steps here?
any help will be appreciated
Best Answer
First remember that $A$ is linear, so we have:
$$A(av+bw) = A(av)+A(bw) = aAv+bAw =a(\lambda v) +b(\lambda w)= \lambda (av+bw)$$
so if $v$ and $w$ are eigenvectors with eigenvalue $ \lambda $ then $av+bw$ is also eigenvector with eigenvalue $\lambda $.