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.
 A: 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 $|\lambda_i - \lambda| < \epsilon$.
A: 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.
A: You can assume that $T$ is diagonal, that is $T= \operatorname{diag} ( \lambda_1,...,\lambda_n)$.
Then $\sum_k |\lambda - \lambda_k|^2 v_k^2 < \epsilon^2$. If $|\lambda - \lambda_k| \ge \epsilon$ for all $k$ then we would have
$\sum_k |\lambda - \lambda_k|^2 v_k^2 \ge \epsilon^2$, a contradiction.
Addendum:
Since $T$ is self adjoint we can choose a basis of orthonormal vectors $u_1,...,u_n$ corresponding to eigenvalues $\lambda_1,...,\lambda_n$. If we let $U = \begin{bmatrix} u_1 & \cdots & u_n \end{bmatrix}$ we see that $T U = U \operatorname{diag} ( \lambda_1,...,\lambda_n)$ and so
$U^* TU = \operatorname{diag} ( \lambda_1,...,\lambda_n)$.
Since $U$ is unitary, $\|T v - \lambda v\| = \|   \operatorname{diag} ( \lambda_1,...,\lambda_n) U^* v - \lambda U^*v\|$, or letting $v' = U^* v$ (noting that $\|v'\| = \|v\|$) we have  $\| \operatorname{diag} ( \lambda_1,...,\lambda_n) v' - \lambda v'\| <\epsilon$.
