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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.

cholo14
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2 Answers2

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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 $$ M := \max(\{1\} \cup \{\lVert A^k \rVert/\gamma^k : k = 1, \ldots, k_0 \}). $$

When $\rho(A) < 1$ we can take $\gamma \in (\rho(A), 1)$.

user539887
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  • I think there's a slight technical issue in this argument: it is entirely possible that $M < 1$, since we are only assuming that $|A^k| \leq \gamma^k$ for $k \geq k_0$. However, it could also be that $|A^k| \leq \tfrac{\gamma^k}{2}$ when $k < k_0$ (at least we did not assume that $k_0$ was minimal so it could happen). And as written you will have $M \leq 1/2$ even though perhaps for some terms $k \geq k_0$ $|A^k| \in (0.5 \gamma^k, \gamma^k]$. Hence, I think you technically want $\max{M, 1}$, not just $M$ – Drew Brady Sep 17 '20 at 05:14
  • Alternatively, you can assume that $k_0$ is the smallest integer such that $|A^k| \leq \gamma^k$ for all $k \geq k_0$ in which case, there must be some $k < k_0$ for which $|A^k| > \gamma^k$, and therefore $M > 1$ automatically. (Of course, this is without loss, so I am perhaps just being a bit pedantic here.) – Drew Brady Sep 17 '20 at 05:16
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    @DrewBrady: I added $1$ in the definition of $M$. Indeed, when $A = 0$ then $M$ as defined in the earlier version of my answer would be zero. Thank you! – user539887 Sep 17 '20 at 08:09
  • looks good to me! – Drew Brady Sep 17 '20 at 19:18
  • Is it possible to take $\gamma=\rho(A)$? In this case, maybe a new proof is needed? – Shi James May 13 '22 at 10:38
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    In general, no. Take $A = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}$. Then $\rho(A) = 1$ but $\lVert A^k \rVert \to \infty$ as $k \to \infty$ (whatever norm you use). Perhaps for $A$ with trivial Jordan blocks that should hold? – user539887 May 14 '22 at 11:37
  • @LorenzoPompili Indeed, in the latter case the norm of $A^k$ is, for large $k$, realized as $\lVert A^k u \rVert$, where $u$ is any eigenvector corresponding to the eigenvalue with maximum absolute value. Filling all the details, however, appears to be quite involved. Perhaps it would be good to formulate it as a new answer. Will you try? – user539887 May 12 '23 at 12:58
  • @user539887 I did that. I wanted to say that in your answer, $M$ does not only depend on $\gamma$, but also on $A$ (see my answer). Maybe you could change that – Lorenzo Pompili May 12 '23 at 19:18
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I have been asked to explain things better.

The formula in the question holds for all $\gamma>\rho(A)$ and some $M>0$ which depends on $\gamma$ (… and on $A$, as shown in the end), as it has been shown in a different answer.

If the matrix is non-diagonalizable, in general the formula does not hold for $\gamma=\rho(A)$. One example, shown in the comments, is the matrix $$ A=\left[\begin{array}2 1& 1\\ 0 & 1\end{array}\right], $$ for which $\rho(A)=1$, but $\|A^k\|^{1/k}\to\infty$.

If the matrix $A$ is Hermitian, we can take $\gamma=\rho(A)$ since fot Hermitian matrices $\rho(A)=\|A\|_{\mathcal L(\ell^2)}$, where the latter is the norm induced by the Hermitian product on $\mathbb C^n$ (see this post). In this case, $M=1$. If you change norm, you need to change $M$ of course, but it is nice that $M$ can be taken to be $1$ with some norm induced by a norm on $\mathbb C^n$ (with an arbitrary norm $M$ can be anything, you simply rescale the definition of the norm by a multiplicative factor).

If the matrix $A$ is diagonalizable… it is still true that you can take $\gamma=\rho(A)$, but unfortunately this time the number $M$ has to depend on the matrix itself (so I was kinda wrong in the comments). In fact, if $A=VDW$, where $D$ is diagonal, $V,W$ are invertible, then $\|\cdot\|_*:=\|V\cdot W\|$ is still a norm (not necessarily a norm induced by some norm on $\mathbb C^n$). We also have $\|A^k\|=\|D^k\|_*$ and $\rho(A)=\rho(D)$. It follows $$ \|D^k\|_*\leq M\rho(D)^k, $$ simply because the inequality is true with the norm $\|\cdot\|_{\mathcal L(\ell^2)}$ and since all norms are equivalent. So, the claim follows. In this case though, $M$ cannot be independent of $A$ in general, for any norm you choose. Take the matrix

$$ A=\left[\begin{array}2 0& r^{-1}\\ r & 0\end{array}\right]. $$

You can make any norm of $A$ arbitrarily large, but the spectral radius is still $1$. This makes me conclude that the constant $M$ also depends on $A$ in general, even when $\gamma>0$.

Lorenzo Pompili
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  • I see that my comment of yesterday is not necessarily correct, namely, I have doubts regarding the realization, for large $k$, of $\lVert A^k \rVert$ as the norm of $\lVert A^k u \rVert$ for any $u$ belonging to the eigenspace corresponding to all the eigenvalues with the largest absolute value. Indeed, if that eigenspace has dimension one then it is O.K., but the situation is more complicated in higher dimensions. – user539887 May 14 '23 at 12:08
  • Consider the matrix $$ A = \begin{pmatrix} 1 & -\frac{2}{\varepsilon} \ 0 & - 1 \end{pmatrix}. $$ Its eigenvalues are $1$ and $-1$, and $A^k$ is either $A$ or $I$ depending on whether $k$ is odd or even. The action of $A^k$ on $(0, 1)$ equals, for odd $k$, $(-\frac{2}{\varepsilon}, - 1)$. So, for such $k$, $\lVert A^k \rVert > 1$, where $\lVert \cdot \rVert$ is the matrix norm induced by $\ell_2$, or $\ell_1$, or else $\ell_{\infty}$-norm on $\mathbb{R}^2$. Therefore, the proof should be subtler than I thought yesterday. – user539887 May 14 '23 at 12:24