Suppose V is finite dimensional, $\mathcal{T} \in \mathcal{L}(V)$ and $\lambda \in \mathbb{F}$. Prove that there exists $\alpha \in \mathbb{F}$ such that $|\alpha - \lambda| < \frac{1}{1000}$ and $T-\alpha I$ is invertible.
So I have seen a solution to this question on another site and I am not able to make sense of how it was even put together. This is the solution:
Let $\alpha_i\in\mathbb{F}$ such that:
$$\left|\alpha_i-\lambda\right| = \frac{1}{1000+i},\quad i=1,\cdots,\dim V+1.$$
These $\alpha_i$ exist and are different from each other since $\mathbb{F} =\mathbb{R} \ or \ \mathbb{C}$ Note that each operator on $V$ has at most dim V distinct eignevalues by 5.13. Hence there exists some $i\in\{1,2,\cdots,\dim V+1\}$ such that $\alpha_i$ is not an eigenvalue of T. Then by 5.6, $T-\alpha I$ is invertible.
Questions: How does this solution work? I understand that we have to create the $\alpha$, but how to illustrate that this works? I'm not sure how the bound was chosen and I don't get how invertiblity is shown. In essence I am lost.....
For Reference:
Theorem 5.6 Equivalent conditions to be an eigenvalue: Suppose V is finite-dimensional, $T \in \mathcal{L}(V),$ and $\lambda \in \mathbb{F}.$ Then the following are equivalent:
(a) $\lambda$ is an eigenvalue of T ;
(b) $T-\lambda I$ is not injective;
(c) $T- \lambda I$ is not surjective;
(d)$T- \lambda I$ is not invertible
Theorem 5.13: Number of eigenvalues Suppose V is finite-dimensional. Then each operator on V has at most dim V distinct eigenvalues.
As long as $\alpha$ is not one of the finitely many eigenvalues of $T$, $T-\alpha I$ is invertible.