Show there is an eigevalue within a certain value of $\lambda$. Here is the problem: Let $V$ be a finite dimensional inner product space with $T:V\to V$ self adjoint transformation. Suppose there exists $v\in V$ a unit vector such that
$$\|Tv-\lambda v\|<\epsilon$$
Show that there must exist an eigenvalue $\lambda'$ of $T$ such that $$\lambda -\lambda'<\epsilon$$
Here's my attempt: Let $v$ be as supposed, then
$$\|Tv-\lambda v\|^2=\langle Tv-\lambda v, Tv-\lambda v\rangle=\|Tv\|^2-2\lambda \langle v,Tv\rangle +\lambda^2 \|v\|^2=\|Tv\|^2-2\lambda \langle v,Tv\rangle +\lambda^2$$
Here we use the fact that $T$ is self adjoint so its eigenvalues are real so theres no need to worry about conjugates, as well as the fact that $v$ is unit.
Now, let $v=\sum a_i v_i$ where $v_1,...,v_i,...,v_n$ are orthonormal eignevectors ($T$ is self adjoint so we use the spectral theorem):
Continuing from above:
$$\|Tv\|^2-2\lambda \langle v,Tv\rangle +\lambda^2=\|\sum a_i \lambda_i v_i\|^2-2\lambda \sum a_i\lambda_i +\lambda^2$$
where we use orthogonormality of the eigenvectors.
Now i get stuck however. Is this on the right track? Any help is appreciated.

Since $T$ is self adjoint, we have a diagonalizing orthonormal basis $
\{v_1,\dots,v_n\}$ corresponding to eigenvalues $\lambda_1,\dots,\lambda_n$.  We write $v=\sum_{i=1}^n\alpha_i v_i$, with $\|v\|^2=\|\sum_{i=1}^n\alpha_i v_i\|^2=\sum_{i=1}^n \alpha_i^2=1$.
We know that $\|Tv-\lambda v\|^2<\varepsilon^2$. Expanding, we have: $$\|Tv-\lambda v\|^2=\|T(\sum_{i=1}^n\alpha_i v_i)-\lambda(\sum_{i=1}^n\alpha_i v_i)\|^2=\|\sum_{i=1}^n\alpha_i\lambda_i v_i-\lambda(\sum_{i=1}^n\alpha_i v_i)\|^2=\|\sum_{i=1}^n\alpha_i(\lambda_i-\lambda) v_i\|^2$$ Since $\{v_1,\dots,v_n\}$ is an orthonormal basis, we can rewrite:$$\|\sum_{i=1}^n\alpha_i(\lambda_i-\lambda) v_i\|^2=\sum_{i=1}^n\alpha_i^2|\lambda-\lambda_i|^2<\varepsilon ^2$$
Let $\lambda'$ be the eigenvalue for which $|\lambda-\lambda_i|^2$ is minimized. Than: $$|\lambda-\lambda'|=|\lambda-\lambda'|\sum_{i=1}^n\alpha_i^2\leq\sum_{i=1}^n\alpha_i^2|\lambda-\lambda_i|^2<\varepsilon ^2$$
And the conclusion follows.
