# Exercise 5.A.11 in Linear Algebra Done Right, 4th edition, Sheldon Axler.

This is exercise 11 which comes from Linear Algebra Done Right, Section 5A, 4th edition, Sheldon Axler.

Suppose $$V$$ is finite-dimensional, $$T \in \mathcal{L}(V)$$, and $$\alpha \in \mathbf{F}$$. Prove that there exists $$\delta > 0$$ such that $$T -\lambda I$$ is invertible for all $$\lambda \in \mathbf{F}$$ such that $$0 < |\alpha - \lambda| < \delta$$.

Here, $$\mathbf{F}$$ denotes either the set of real numbers or complex numbers. I'm not sure where to start with this exercise. Here's a few things I know, but don't know how to use for this problem:

• $$\lambda$$ is not an eigenvalue $$\iff T -\lambda I$$ is invertible
• $$T$$ has at most $$\text{dim} \ V$$ distinct eigenvalues.

Could someone provide a hint to get me started?

• Is it given that $A := T - \lambda I$ is invertible? If so, you can use the geometric series formula to write $(A + rI)^{-1}$ . Commented May 1 at 3:32
• @Kakashi I don't think so. The way I'm interpreting the question is that we must find a $\delta > 0$ in order for $T - \lambda I$ to be invertible. Commented May 1 at 3:34
• I see, the finite-dimensionality is key. Note $T - \lambda I$ is invertible if and only if $p(\lambda) = \det(T - \lambda I) \neq 0$. Since $p$ is a polynomial of degree $n = \dim(V)$, the result follows. Commented May 1 at 3:40
• What is F? Supposedly a field with a real-valued absolute value operation, so I guess that would be a subfield of $\Bbb C$. Why not just take $\Bbb C$? Commented May 1 at 3:43
• Plot your $\dim V$ eigenvalues in the complex plane. Now describe the eligible $\lambda$s and figure out what $\delta$ will work. Commented May 1 at 3:45

This has little to do with linear algebra. There are only finitely many eigenvalues, so if there are any eigenvalues distinct from $$\alpha$$, there is one such eigenvalue$$~\mu$$ with $$|\alpha-\mu|$$ minimal among them. Then if you take $$\delta$$ to be that minimal absolute value, there are by construction no eigenvalues$$~\lambda$$ with $$0<|\alpha-\lambda|<\delta$$. Of course if there are no eigenvalues distinct from $$\alpha$$ at all, you can take whatever positive $$\delta$$ you like,

• I think I'm following what you laid out. What I'm confused about is the exercise asks for all $\lambda \in \mathbf{F}$, not just the eigenvalues. Is this implemented in your solution? Commented May 1 at 4:15
• @PaulAsh Replace "$T-\lambda I$ is invertible" by the equivalent "$\lambda$ is not an eigenvalue of $T$", and it says "for all $\lambda$ with $0<|\alpha-\lambda|<\delta$, $\lambda$ is not an eigenvalue of $T$" which I took the liberty to rephrase as "there are no eigenvalues$~\lambda$ of $T$ that satisfy $0<|\alpha-\lambda|<\delta$". Commented May 3 at 7:36

To simplify thinks lets ignore $$\alpha$$ for the time being (or assume $$\alpha=0$$

$$T-\lambda I$$ is invertible if $$\lambda$$ is not an eigenvalue.

Suppose T has no non-zero eigenvalues.

Let $$\mu$$ be the eigenvalue of the smallest absolute value.

For all $$\lambda$$ where $$|\lambda| < |\mu|,$$

$$\lambda$$ cannot be an eigenvalue. And $$T-\lambda I$$ is non singular.

But, suppose $$T$$ is zero and has an eigenvalue equal to $$0.$$

Then, again with $$\mu$$ as the non-zero eigenvalue of the smallest absolute value

With $$0<|\lambda| < |\mu|$$ again $$\lambda$$ cannot be an eigenvalue.

Re-introducing $$\alpha$$...

For any $$\alpha$$ (which may or may not be an eigenvalue), there is some available neighborhood around $$\alpha$$ such that $$\lambda$$ is not an eigenvalue.

• "Non-zero" should be "non-$\alpha$". Commented May 1 at 3:54
• @MarcvanLeeuwen I was trying to introduce the idea considering $\alpha=0$ (by not even discussing $\alpha$) to show that there should be some available neighborhood of $\lambda$ in this simplified case, and once that has been introduced, can be generalized to any $\alpha$ Commented May 1 at 3:56
• I need some time digesting your solution. I will respond back with any questions sometime this week. Thank you. Commented May 1 at 4:18
• Thanks for your answer. Couldn't there exist $\lambda \in \mathbf{F}$ such that $|\mu| < |\lambda|$? Commented May 3 at 0:19
• @PaulAsh That isn't what you have to show. You have to show that there exists a $\delta$ such that $0< |\alpha - \lambda| < \delta \cdots$ This is more a question of real analysis than it is linear algebra. Commented May 3 at 0:35

Since $$\dim V =n$$, $$T$$ can have up to $$n$$ different eigenvalues, written in decreasing order as $$\lambda_1,\ldots, \lambda_n$$. Therefore for $$\lambda>\lambda_1$$ or $$\lambda<\lambda_n$$, $$T-\lambda I$$ is invertible. It is also the case that if $$\lambda \in (\lambda_{i+1},\lambda_{i})$$ then $$T-\lambda I$$ is invertible. Denote $$\frac{\lambda_i+\lambda_{i+1}}{2}$$ as $$\alpha$$ and $$\frac{\lambda_i-\lambda_{i+1}}{2}>0$$ as $$\delta$$. Then if $$|\alpha-\lambda|<\delta$$, $$T-\lambda I$$ is invertible.