I'm having difficulties with this homework problem:
Suppose $T$ is a self adjoint linear operator on a vector space $V$. Let $\lambda \in \mathbb{F}$ (where $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$; naturally $V$ is a vector space over $\mathbb{F}$) and $\epsilon > 0$. Prove that if there exists $v \in V$ such that $||v|| = 1$ and $||Tv-\lambda v|| < \epsilon$, then $T$ has an eigenvalue \lambda' such that |\lambda - \lambda'|<\epsilon.
I thought I had proven it, but I assumed that $v$ is an eigenvector of $T$ with eigenvalue \lambda' (in fact then it is quite simple, since it turns out $||Tv-\lambda v|| = |\lambda - \lambda'|$). So here is my question: can we assume without loss of generality that $v$ is an eigenvector with eigenvalue \lambda'? I suspect the answer is no, so if not I'd also appreciate a push in the right direction.