If $\|A\| < 1$, does that imply $A$ is nilpotent?

Question

Suppose $\|A\| < 1$ where $\| \cdot \|$ is the operator norm on matrices, intuitively, $\lim\limits_{k \to \infty} A^k$ goes to zero $\Rightarrow$ $A$ is nilpotent

But is this indeed the case? Under what condition is $A$ both nilpotent and $\|A\|<1$

Answer 1

Note that by definition, $A$ is nilpotent means that there exists a finite $k$ such that $A^k = 0$. The property that you're referring to, it seems, is that $A$ should satisfy $\lim_{k \to \infty} \|A^k\| = 0$.

Indeed, any matrix $A$ such that $\|A\|<1$ will satisfy $\lim_{k \to \infty}\|A^k\| = 0$, since $\|A^k\| \leq \|A\|^k$. However, a matrix $A$ will be nilpotent (if and) only if its only eigenvalue is zero.

An equivalent condition is that $A$ is nilpotent if and only if for every $\epsilon > 0$, there exists a $k$ such that $\|A^k\|^{1/k} < \epsilon$. Or, if you prefer: $A$ is nilpotent if and only if for every $\epsilon > 0$, there exists a submultiplicative matrix norm $\|\cdot\|$ such that $\|A\| < \epsilon$.

Keep in mind however that if $A$ is $n \times n$, then $A$ is nilpotent if and only if $A^n = 0$. Often, this is a much easier condition to check. Note also that $A$ can be nilpotent with $\|A\| > 1$. In particular, if $A$ is nilpotent, then so is $kA$ for any $k>0$.