# Show that $A^{n}\to0$ if and only if $\|A\|^{n}\to0$

Let $\boldsymbol{A}$ be a square matrix. Show that $\lim_{n\to\infty}\boldsymbol{A}^{n}=0$ if and only if $\lim_{n\to\infty}\|\boldsymbol{A}\|^{n}=0$ for the spectral radius or for some operator norm.

• Does the problem actually ask you to show this? It's not true. (To give a counterexample we need to know what definition of $||A||$ you're using...) – David C. Ullrich Jun 28 '15 at 21:32
• Only one side is always true, since $\|A^n\| \leq \|A\|^n$, then if $\|A\|^n \to 0$ then $\|A^n\| \to 0$. – Alonso Delfín Jun 28 '15 at 22:11
• If you added "for some operator norm $\|\cdot\|$", then the result would be true since for any $\epsilon>0$ there is an operator norm $\|\cdot\|$ such that $\|A\|\leq\rho(A)+\epsilon$ and $A^n\to 0$ as $n\to\infty$ iff $\rho(A)<1$. – Algebraic Pavel Jun 28 '15 at 23:27

Let $A=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ with the $l_1$ norm, then $\|A\| = 1$. Since $A^2 = 0$, we see that $\lim_n \|A^n\| = 0$, but we have $\lim_n \|A\|^n = 1$.

• +1, great example. The $l_2$ norm is also $1$, the spectral radius is $0$. – Orest Bucicovschi Jun 28 '15 at 22:34

The spectral radius is not a norm; one reason is because there are nonzero matrices all of whose eigenvalues are zero. (Such matrices can't be diagonalizable, but nondiagonalizable matrices exist!) copper.hat gave one example.

Your property does hold if you replace $\| \cdot \|$ with the spectral radius.

If $A^n \to 0$, then there exists a norm, which can be chosen to be induced by some vector norm, such that $\| A \|^n \to 0$. This is ultimately because $\rho(A)$ is the infimum of $\| A \|$ over all possible operator norms. One direction (that the spectral radius is a lower bound) is easy to see by homogeneity and the operator norm property. The other direction (that the spectral radius is the greatest lower bound) is harder to prove.

• +1, you nailed the correct statement . Good to learn about the spectral radius! It seems that you can restrict to operator norms coming from some $l_2$ norm, and it's enough to reason on Jordan cell, which is not that hard. – Orest Bucicovschi Jun 29 '15 at 2:23

For $A$ matrix denote by $\rho(A)$ the spectral radius of $A$, $= \max \{ |\lambda_i|\}$, with $\lambda_i$ are the eigenvalues of $A$.

We have $A^n \to 0$ if and only $\rho(A)^n \to 0$ if and only if $\rho(A)<1$. For proof one can use the Jordan canonical form.

Now if $||\cdot ||$ is any algebra norm on $M_n(\mathbb{R})$ ( for example coming from a norm on $\mathbb{R}^n$), then $||A||< 1$ implies $A^n \to 0$, because $||A^n|| \le ||A||^n$. The converse is not true, as the example of @copper.hat: shows. The norm of the operator $A\colon e_1 \mapsto \alpha e_2 \mapsto 0$ can be made as large as wanted, while $A^2 = 0$.

Notes:

The spectral radius is not a norm for the algebra $M_n(\mathbb{R})$ if $n \ge 2$; one can have $A$, $B$ nilpotent ( $\rho(A) = \rho(B) = 0$) and $\rho(AB) =$ large; take $A$ the one above, $B= A^{t}$.

If $A^n \to 0$ then the convergence is exponential. Moreover, the series $\sum_{n\ge 0} A^n$ is also covergent, with sum $(I-A)^{-1}$.

If use the norm for matrix as

$$\|A\|=\sqrt{\sum\limits_{i,j=1}^{n}|a_{ij}|^2}$$

We can prove

$$\|A\|^n \to 0\implies \|A^n\| \to 0$$ Converse is not true as a counter example is given by copper.hat.

First we prove the following:

Lemma: $\hspace{2 mm}\|AB\|\leqslant\|A\|\|B\|$

Prove: \begin{align} \|AB\|^2&=\sum\limits_{i,j=1}^{n}\left|\sum\limits_{k=1}^na_{ik}b_{kj}\right|^2 \\ &\leqslant\sum\limits_{i,j=1}^{n}\left(\sum\limits_{k=1}^n|a_{ik}|^2\sum\limits_{k=1}^n|b_{kj}|^2\right)\tag{Cauchy-Schwarz} \\ &=\sum\limits_{i,j=1}^{n}\left(\sum\limits_{k,l=1}^n|a_{ik}|^2|b_{lj}|^2\right) \\ &=\sum\limits_{i,k=1}^{n}|a_{ik}|^2\sum\limits_{l,j=1}^n|b_{lj}|^2 \\ &=\|A\|^2\|B\|^2 \\ \end{align} Now by lemma $$\|A^n\|\leqslant\|A\|^n \hspace{5 mm} \text{and so} \hspace{5 mm} \|A^n\| \to 0\hspace{5 mm} \text{as} \hspace{5 mm} \|A\|^n\to 0$$