Linear Algebra Done Right Exercise 6.B.16

Question

I am trying to solve Exercise 16 from Section 6.B of the third edition of Linear Algebra Done Right by Axler.

Suppose $ \mathbf{F} = \mathbf{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 $ \lVert T^m v \rVert \leq \epsilon \lVert v \rVert $ for every $ v \in V $.

($V$ is an inner product space.)

I think this is a quick corollary of the implication $\rho(T) < 1 \implies \lim_{m \to \infty} \lVert T^m \rVert = 0 $. However, at this point in the textbook we have discussed neither spectral radius nor operator norm; for this reason I think Mr. Axler had another approach in mind.

The title of Section 6.B is 'Orthonormal Bases', so I tried to look for a solution which only uses basic properties of orthonormal bases. Since $V$ is a finite-dimensional complex vector space, Schur decomposition (Theorem 6.38 in the text) ensures there is an orthonormal basis of $V$ with respect to which $T$ has an upper-triangular matrix. I managed to prove the result in the special case where this matrix is diagonal and also in the special case where $\dim V = 2$, but I haven't been able to prove the general case.

Can anyone see an 'orthonormal basis approach'?

I did find this post. Unfortunately, I think there is a mistake in that answer. At least, I can't see how the equality $$ \sum_{k=1}^n \langle T^j(v), e_k \rangle e_k = \sum_{k=1}^n a_k \lambda_k^j e_k $$ follows from $\langle T^j(e_k), e_k \rangle = \lambda_k^j$.

Answer 1

Here is one possible approach that does not directly involve the notions of operator norm and equivalence of norms. First of all, in $\mathbb C^n$ equipped with the Euclidean norm, we have the following observations:

1. The product of a diagonal matrix and a strictly upper triangular matrix is strictly upper triangular, regardless of the order of multiplication.
2. The product of $n$ strictly upper triangular $n\times n$ matrices is always zero.
3. For any $n\times n$ complex matrix $A$, there exists a constant $C$ (that depends on $A$ and $n$) such that $\|Ax\|\le C\|x\|$ for all $x\in\mathbb C^n$. We don’t need any topological argument to justify this. In fact, if we denote by $|A|,|x|$ the entrywise absolute values of $A$ and $x$ and denote by $e$ the all-one vector, we have $$ \|Ax\|\le\big\||A||x|\big\|\le\big\||A|\big(\|x\|e\big)\big\| \le\left(\max_i\sum_j|a_{ij}|\right)\sqrt{n}\|x\|. $$
4. If $D\in\mathbb C^{n\times n}$ is a diagonal matrix and $x\in\mathbb C^n$, then $\|Dx\|\le\max_i|d_{ii}\|x\|$.

Now return to your question. By picking a suitable orthonormal basis, we may identify $T$ with an upper triangular matrix whose diagonal elements have moduli $<1$, $x$ with a vector in $\mathbb C^n$, and the norm $\|\cdot\|$ on $V$ with the Euclidean norm on $\mathbb C^n$.

Let $D$ and $F$ be respectively the diagonal and off-diagonal parts of the matrix. Since $T$ is upper triangular, $F$ is strictly upper triangular. Also, by assumption, we have $\rho:=\max_i|d_{ii}|<1$. Let $C>0$ be a constant such that $\|Fx\|\le C\|x\|$ for all $x\in\mathbb C^n$. It follows from the observations above that when $m\ge n$, $$ \|T^mx\|=\|(D+F)^mx\|\le\sum_{k=0}^{n-1}\binom{m}{k}C^k\rho^{m-k}\|x\|.\tag{1} $$ To illustrate, suppose $n=2$ and $m=3$. Then $DFF=FDF=FFD=F^3=0$ by observations 1 and 2. Therefore $$ \begin{aligned} \|T^3x\| &=\|(D+F)^3x\|\\ &=\|(D^3+DDF+DFD+FDD+DFF+FDF+FFD+F^3)x\|\\ &=\|(D^3+DDF+DFD+FDD)x\|\\ &\le\|D(D(Dx))\|+\|D(D(Fx))\|+\|D(F(Dx))\|+\|F(D(Dx))\|\\ &\le\rho(\rho(\rho\|x\|))+\rho(\rho(C\|x\|))+\rho( C(\rho\|x\|))+C(\rho(\rho\|x\|))\\ &=\sum_{k=0}^{1}\binom{3}{k}C^k\rho^{3-k}\|x\|.. \end{aligned} $$ Now, for any fixed integer $k\ge0$, since $\lim_{m\to\infty}\binom{m}{k}\rho^m=0$, $(1)$ gives $\lim_{m\to\infty}\|T^mx\|=0$. Hence the result follows.

Answer 2

Denote by:

• $n$ the dimension of $V$,
• $T=D+N$ ($D$ diagonalizable, $N^n=0,$ $ND=DN$) the Jordan-Chevalley decomposition of $T$,
• $\lambda:=$$\|D\|$$<1$ the largest absolute value of the eigenvalues of $T$,
• $\mu:=\max(1,\|N\|,\|N^2\|,\dots,\|N^{n-1}\|)$.

For every $m\ge n,$ $$T^m=\sum_{k=0}^{n-1}\binom mkD^{m-k}N^k$$ hence $$\|T^m\|\le\mu\sum_{k=0}^{n-1}\binom mk\lambda^{m-k}\le\mu n m^{n-1}\lambda^{m-(n-1)}.$$ When $m\to\infty,$ this upper bound tends to $0.$

Answer 3

This is not really an orthonormal basis approach, but you can certainly use the Schur decomposition to prove the result.

Suppose $T=QUQ^*$ is the Schur decomposition, with $U$ upper triangular. Since $T^n = QU^nQ^*$, it suffices to show that $U^n \to 0$. I am using the induced Euclidean norm.

Let $D(r) = \operatorname{diag}(r_1,...,r_d)$ (with $r_k >0$) and note that the $ij$ entry of $D(r)UD(r)^{-1}$ is ${r_i \over r_j} U_{ij}$. The diagonal elements are unchanged, we can choose $r_d = 1$, and the remaining $r_k$ such that the off diagonal elements are arbitrarily small.

Note that $\| \operatorname{diag}(U_{11},...,U_{dd}) \| = \max_k |U_{kk}| < 1$.

In particular, since $|U_{ii}| < 1$ and the matrix norm is continuous, there is some $r$ such that $\| D(r)UD(r)^{-1} \| < 1$, and since the induced norm is submultiplicative, we see that $\| D(r)U^nD(r)^{-1} \| \le \| D(r)UD(r)^{-1} \|^n$ and hence $U^n \to 0$.