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In the Spectral radius wikipedia article in section Matrices there is a lemma, what states that:

Lemma. Let $A \in \mathbb{C}^{n \times n}$ be a complex-valued matrix, $\rho(A)$ its spectral radius and $\| \cdot \|$ a consistent matrix norm, then, for each $k \in \mathbb{N}$: $\rho(A) \leq \|A^k\|^{1/k}$.

Now take a look at the definiton of a consistent matrix norms.

A matrix norm $\| \cdot \|_{ab}$ on $K^{m \times n}$ is called consistent with a vector norm $\| \cdot \|_a$ on $K^n$ and a vector norm $\| \cdot \|_b$ on $K^m$ if: $$\|Ax\|_b \leq \|A\|_{ab}\|x\|_a$$ for all $A \in K^{m \times n}$, $x \in K^n$.

Okay. Back to the Spectral radius article to the proof of the lemma.

Proof. Let $(v, \lambda)$ be an eigenvector-eigenvalue pair for a matrix $A$. By the sub-multiplicative property of the matrix norm, we get: $$| \lambda |^k \|v\| = \| \lambda^k v \| = \| A^k v \| \leq \|A^k\| \cdot \|v\|$$ and since $v \neq 0$ for each $\lambda$ we have $| \lambda |^k \leq \|A^k\|$ and therefore $\rho(A) \leq \|A^k\|^{1/k}$.

So! Why do we need the consistency of matrix norm? We use this property here $\| A^k v \| \leq \|A^k\| \cdot \|v\|$. But am I wrong with the following stronger statement?

Lemma. Let $A \in \mathbb{C}^{n \times n}$ be a complex-valued matrix, $\rho(A)$ its spectral radius and $\| \cdot \|$ an arbitrary matrix norm, then, for each $k \in \mathbb{N}$: $\rho(A) \leq \|A^k\|^{1/k}$.

Proof. Let $(v, \lambda)$ be an eigenvector-eigenvalue pair for a matrix $A$. Multiply from the right the eigenequation with $v^T$, and write it in the form for $A^k$ in the $\| \cdot \|$ matrix norm. So we get $\|A^k(v v^T)\|=\| \lambda^k(v v^T) \|$. Now $$\|A^k\| \cdot \|v v^T \| \geq \|A^k (v v^T) \| = \| \lambda^k (v v^T) \| = | \lambda^k | \cdot \|v v^T\|.$$ And since $\| v v^T \| \neq 0$ for each $\lambda$ we have $| \lambda |^k \leq \|A^k\|$ and therefore $ \rho(A) \leq \|A^k\|^{1/k}$ for all $k \in \mathbb{N}$.

And here we just used the sub-multiplicativity and the abolute homogenity of matrix norms.

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  • $\begingroup$ Do you mean to say, for each $k \in \mathbb{N} : \|A^k\|^{1/k} \leq \rho(A)$? $\endgroup$ – snar Aug 27 '14 at 21:39
  • $\begingroup$ @snarski no, the reverse of it, but sorry about the mistake! I corrected! $\endgroup$ – user153012 Aug 27 '14 at 21:43
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    $\begingroup$ That is true for Banach algebras too. $\endgroup$ – AD. Aug 27 '14 at 21:45
  • $\begingroup$ @user153012 there's nothing wrong with your more general proof. Not sure why the wiki author decided to stop with that particular lemma. Note: it is also common to consider the matrix $v \mathbf{1}^T$, where $\mathbf 1$ is the column vector of $1$s $\endgroup$ – Omnomnomnom Aug 27 '14 at 21:47
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    $\begingroup$ I don't see how your statement is stronger. Every submultiplicative matrix norm defined on $M_n(\mathbb{C})$ is consistent with some vector norm. Your version of the lemma --- which has assumed a potentially stronger condition (that the matrix norm is submultiplicative) --- is a potentially weaker result, or at best an equally strong one when compared with Wikipedia's version. $\endgroup$ – user1551 Sep 14 '14 at 7:30

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