Exercise 5.A.13 (Approximation of eigenvalues)

Question

Suppose $V$ is finite-dimensional, $T \in \mathcal{L}(V)$, 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.

Célio Passos · Accepted Answer

\begin{proof}
Choose $\alpha$ such that it is not an eigenvalue of $T$ and $\left| \alpha - \lambda \right| < \frac{1}{1000}$. This $\alpha$ exists because there are infinite values of $\alpha$ satisfying such inequality, but only finitely many eigenvalues, because $V$ is finite-dimensional. Then, suppose $v \in \operatorname{null}(T - \alpha I)$. That means $Tv = \alpha Iv = \alpha v$. But $\alpha$ is not an eigenvalue of $T$. Therefore $v$ must be the $0$ vector, proving $T - \alpha I$ is injective.
\end{proof}

Exercise 5.A.13 (Approximation of eigenvalues)

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Exercise 5.A.13 (Approximation of eigenvalues)

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