# Prove that any non zero linear combination of two eigenvectors is also an eigenvector

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

• the matrix multiplication is linear, thus for any vectors $v,w$ and any scalar $c$ we have that $A(cv+w)=cAv+Aw$. Apply the linearity of $A$ to your case and, voilá! Jan 2 '18 at 20:31

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 $\lambda$.

• Thank you very much, i understand now! Jan 2 '18 at 20:50

Note that

$$A( v + w)= \lambda v + \lambda w=\lambda(v+w)$$

thus

$$A( av + bw)= \lambda av + \lambda bw=\lambda(av+bw)$$