1. ## Eigenvectors

Let x and y be eigenvectors of a matrix A such that is also an eigenvector of A. Is an eigenvector of A? Prove or disprove.

I know that and and but I'm not sure where I can go from there without being given that x and y correspond to the same eigenvalue... I started with: but I don't know how to proceed. Please help!!

OP edit after getting help: Thank you--I didn't think of the 2 cases, it becomes easy!

2. Originally Posted by casanova
Let x and y be eigenvectors of a matrix A such that is also an eigenvector of A. Is an eigenvector of A? Prove or disprove.

I know that and and but I'm not sure where I can go from there without being given that x and y correspond to the same eigenvalue... I started with: but I don't know how to proceed. Please help!!

$\underline{\mbox{First case}}$ : both eigenvectors are linearly dependent $\Longrightarrow y = \alpha x\,,\,\alpha\in\mathbb{F}=$ the definition field, but then:

i) they both have the same eigenvalue $\lambda$;

ii) $x\pm y=(1\pm \alpha)x\Longrightarrow A(x\pm y)=A((\alpha\pm 1)x)=(\alpha\pm 1)Ax=(\alpha\pm a)\lambda x$

$\underline{\mbox{Second case}}$: both eigenvectors are lin. independent, with eigenvalues $\lambda_1\,,\,\lambda_2$ but then:

$\lambda_1 x+\lambda_2 y=Ax+Ay=A(x+y)=\lambda_2(x+y)=\lambda_3x+\lambda_3 y\Longrightarrow (\lambda_1-\lambda_3)x+(\lambda_2-\lambda_3)y=0$ $\Longrightarrow \lambda_1=\lambda_3=\lambda_2$ , by lin. independence...

Now you just end the argument to show that with the given info, anyway $x-y$ is an eigenvector of A.

Tonio