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I've been stuck on a linear algebra problem from Axler's Linear Algebra Done Right. The problem statement is as follows:

Suppose $F=\mathbb{C}$, $V$ is finite-dimensional, $T \in \mathcal{L}(V)$, all the eigenvalues of $T$ have absolute value less than $1$, and $\epsilon > 0$. Prove that there exists a positive integer $m$ such that $\left\|T^m(v)\right\| \le \epsilon\left\|v\right\|$ for every $v \in V$. (Source: Axler 6.B. Problem 16)

I have done some work on the problem by proving the statement for diagonalizable $T$. This holds because then there is an eigenbasis $e_1,...,e_n$ so there exists $a_j$ such that $v=a_1e_1+...+a_ne_n$ so $\left\|T^m(v)\right\|=|a_1\lambda_1^{m}e_1+...+a_n\lambda_n^{m}e_n|$ and since there exists all $\lambda_j$ are less than $1$, by WOP, $\exists{m_j}$ such that each $\lambda_j^{m_j} < \epsilon$ so taking $\max(m_1,...,m_j)=m$ gives a value of $m$ such that $\left\|T^m(v)\right\|$ is less than $\epsilon\left\|v\right\|$.

I'm not quite sure how I should generalize to all linear operators $T \in \mathcal{L}(V)$ from here though. Any help is appreciated.

Hasek
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Brian J. Lee
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  • The argument is right. You should do somewhat similar over invariant subspaces and obtain the same answer. Btw, when you said absolute value you meant module, right? Or is it implicit that all eigenvalues are real? ā€“ astro Aug 09 '20 at 17:19
  • I meant the norm for $|a_1\lambda_1^me_1+...+a_nĪ»_m^ne_n|$ that was a typo. I don't quite get what you mean and how that would extend to linear operators. The eigenvalues do not necessarily have to be real but yes $|a_1|$ refers to the magnitude. ā€“ Brian J. Lee Aug 09 '20 at 18:05
  • When the operator is not diagonalizable you can use an eigenspace of higher dimension than 1 and generalize whatever happens in there. I mean, even when is not straightforward from projection onto an eigenline, you have that inside an eigenspace vector within get uniformly contracted even when they could be rotating or something, so in the end of the day you can adapt the diagonaizable setting to the most general one. Moreover, if your equality up there holds, apply triangular inequality and get your result without any harder work. ā€“ astro Aug 09 '20 at 18:19

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Let $A$ be a matrix of the linear operator $T$ in some basis.

Recall that diagonalizable matrices are dense in the set of all complex matrices, i.e. for every $X\in\mathbb{C}^{n\times n}$ there exists a sequence of diagonalizable matrices matrices $Y_n$ such that for every $n\in\mathbb{N}$ $||X-Y_n||<\frac{1}{n}$.

Let $Y_n$ be a sequence of diagonalizable matrices approximating $A^m$. We have $$||A^m(v)||=||(A^m-Y_n)(v)+Y_n(v)||\leq||A^m-Y_n||||v||+\varepsilon||v||<\left(\frac{1}{n}||v||+\varepsilon||v||\right)\to\varepsilon||v||\mbox{ as }n\to\infty$$ that proves the result for the case of arbitrary $T\in\mathcal{L}(V)$.

Hasek
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