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.

Continue to use orthonormality of the orthornomal basis (i.e. Pythagoras theorem) from what you have so far to show that
$$
\Vert Tv - \lambda v \Vert^2 = \sum_i |a_i|^2 \lambda_i^2 - 2 \lambda \sum_i |a_i|^2 \lambda_i + \lambda^2 = \sum_i |a_i|^2 (\lambda_i - \lambda)^2,
$$
where we use the fact that $\sum_i |a_i|^2 = \Vert v \Vert^2 = 1$. (Also with a minor correction where the cross term should also have $|a_i|^2$ instead of $a_i$.)
Now, suppose that for all eigenvalues, we have $|\lambda_i - \lambda| \geq \epsilon$. (We have to assume that the relevant norm is being used.) Then the above formula shows that
$$
\Vert Tv - \lambda v \Vert^2 \geq \epsilon^2 \sum_i |a_i|^2 = \epsilon^2 ,
$$
a contradiction to our initial assumption. Hence we conclude that there must exist some eigenvalue $\lambda_i$ such that $|\lambda_i - \lambda| < \epsilon$.
