Prove that if $x,y$ are eigenvectors of the same eigenvalue, then $\langle x,y \rangle \neq 0$.

Question

I'm trying to prove the following statement:

"Let $T$ be a linear operator over a finite-dimensional inner product space. Show that if $x$ and $y$ are eigenvectors of the same eigenvalue, then $\langle x,y \rangle \neq 0$".

My attempt: I was trying to prove this statement by contrapositive, assuming $\langle x,y \rangle = 0$, and showing that either $x$ or $y$ aren't eigenvectors with the same eigenvalue as the other. I planned to do this by assuming W.L.O.G. that $T(x) = \lambda x$, and then show that $T(y) \neq \lambda y$. The problem is that I don't know how to combine the eigenvector hypothesis with the inner product. Could anyone tell me how I could do this? Thank you!

Answer 1

Let $T=I\in\mathbb{R}^{2\times2}$, $x=e_1$, $y=e_2$...

Answer 2

This is certainly false. Take $T$ to be the identity operator. Then every non-zero vector is an eigenvector with eigenvalue $1$.