# Bound on the norm of a matrix power

Suppose we have the square matrix $A$ and we know that its spectral radius $\rho(A)$ is less than $1$, therefore matrix $A$ is stable. How can we prove that $\exists \gamma \in(0,1)$ and $\exists M >0$ such that $$\|A^k\|\leq M\gamma^k, \:\:\:\: \forall k\geq0$$ What I tried so far is $\|A^k\|=\|A\dots A\|\leq\|A\|\dots \|A\| =\|A\|^k$ so taking $\gamma=\|A\|$ I should be close to the above inequality, but I am not sure it is correct.

• What did you try? Where do you get stuck? Did you try to apply the definition? Do you know about Jordan normal form? – TZakrevskiy Dec 11 '17 at 14:59
• @TZakrevskiy I have updated my question. No, I don't know about Jordan normal form and I don't know how I could use it – cholo14 Dec 11 '17 at 15:12
• The JNF is not necessary to answer your question. What you did is correct - for the 2-norm. A norm for which $\|AB\|≤\|A\|\|B\|$ holds is called submultiplicative. Note that in your case you want to prove it for an arbitrary norm. – P. Siehr Dec 11 '17 at 15:14
Let $\lVert \cdot \rVert$ be a matrix norm on $\mathbb{C}^{n \times n}$. I assume that we know Gelfand's formula: $$\rho(A) = \lim\limits_{k\to\infty}\lVert A^k\rVert^{1/k}.$$ We want to prove that for any $\gamma > \rho(A)$ there exists $M = M(\gamma) \ge 1$ such that $$\lVert A^k \rVert \le M \gamma^k, \quad k \in \mathbb{N}.$$ It follows from Gelfand's formula that there exists $k_0$ such that for any $k = k_0 +1, k_0 + 2, \ldots,$ there holds $$\lVert A^k \rVert < \gamma^k.$$ It suffices now to take $$M := \max\{\lVert A^k \rVert/\gamma^k : k = 1, \ldots, k_0 \}.$$
When $\rho(A) < 1$ we can take $\gamma \in (\rho(A), 1)$.