TITLE: Question on upper triangular matrix with complex eigenvalues with modulus less than 1
QUESTION [6 upvotes]: This is problem 16, Section 6.B from Linear Algebra Done Right, 3rd Edition.
Suppose the field is $\mathbb{C}$, $V$ is finite-dimensional, $T \in  \mathcal{L}(V)$, all the eigenvalues
of $T$ have absolute value less than 1, and $\epsilon > 0$.
 Show that there exists a positive integer $m$ such that $||T^m v|| < \epsilon ||v|| \; \forall \; v \in V$.
At this point in the book Jordan form has not been proved, so all I can use is Schur's Theorem that guarantees upper triangular matrix with eigenvalues on the diagonal.
Could anyone give me a hint on how to proceed ? I tried computing powers of $T$, but while I get $\lambda^m$ on diagonal and same upper triangular structure, I am unable to manage the off-diagonal entries and control their growth.
Note that in this case if matrix is split as diagonal and strictly upper diagonal matrix as $T = D + U$, $D$ and $U$ do not commute because entries on diagoals of $D$ are unequal.

REPLY [9 votes]: Here is what I had in mind when I added that exercise to the third edition of Linear Algebra Done Right:
By Schur's Theorem (which is in the same section as this exercise), there is an orthonormal basis of $V$ such that $T$ has an upper-triangular matrix $A$ with respect to that basis. Orthonormality is important so that we can compute norms. We will work with $A$ instead of $T$ and $z \in \mathbf{C}^n$ instead of $v$.
The entries on the diagonal of $A$ are the eigenvalues of $T$. Thus all the diagonal entries of $A$ have absolute value less than $1$.
If we replace each entry in $A$ with its absolute value and each coordinate of $z$ with its absolute value, then $\|A^mz\|$ gets larger or remains the same for each positive integer $m$, and $\|z\|$ remains the same. Thus we can assume without loss of generality that each entry of $A$ is nonnegative.
If we now replace each entry on the diagonal of $A$ with the largest element on the diagonal of $A$, then $\|A^mz\|$ gets larger or remains the same for each positive integer $m$. Thus we can assume without loss of generality that there exists a nonnegative number $\lambda < 1$ such that every entry of the diagonal $A$ equals $\lambda$.
We can write
$$
A = \lambda I + N,
$$
where $N$ is an upper-triangular matrix whose diagonal entries all equal $0$. Thus $N^n = 0$.
Let $m$ be a positive integer with $m > n$. Then
\begin{align}
A^m &= (\lambda I + N)^m \\[6pt]&= \lambda^m I + m \lambda^{m-1} N + \tfrac{m(m-1)}{2} \lambda^{m-2} N^2 \\[6pt]
&\quad + \tfrac{m(m-1)(m-2)}{2 \cdot 3} \lambda^{m-3} N^3 +  \dots + \tfrac{m(m-1) \dotsm (m - n + 1)}{(n-1)!} \lambda^{m-n+1} N^{n-1},
\end{align}
where the other terms in the binomial expansion do not appear because $N^k = 0$ for $k \ge n$. In the expression above, the coefficients of the $n$ matrices $I$, $N$, $N^2$, $N^3$, $\dots$, $N^{n-1}$ are all bounded by
$$
\frac{m^n \lambda^m}{\lambda^{n-1}}.
$$
Think of $n$ and $\lambda$ as fixed. Then the limit of the expression above as $m \to \infty$ equals $0$ (because $0 \le \lambda < 1$).
Because the sum above has only a fixed number $n$ of terms, we thus see that by taking $m$ sufficiently large, we can make all the entries of the matrix $A^m$ have absolute value as small as we wish. In particular, there exists a positive integer $m$ such every entry of $A^m$ has absolute value less than $\epsilon/n$.
The definition of matrix multiplication and the Cauchy-Schwarz Inequality imply that for each $j$, the entry in row $j$, column $1$ of $Az$ has absolute value at most $\epsilon \|z\|/\sqrt{n}$. Thus $\|Az\| \le \epsilon \|z\|$, as desired.
This exercise is, in my opinion, one of the harder exercises in the third edition of Linear Algebra Done Right if one uses only the tools available at that stage of the book. The proof becomes much clearer and cleaner if one uses the tools associated with the spectral radius in a Banach algebra, which shows the power of those tools.