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I am given a lemma and it states:

Lemma: Let $A$ be a real $n\times n$ matrix. Then given any $e > 0$, there is a $norm$ such that $norm(A) \leq p(A) + e$, where $p(A)$ is the spectral radius of $A$.

They then go on to state that based on this lemma, if $p(A) < 1$, then $norm(A) < 1$ for the correct choice of the norm.

What I'm looking for help on is understanding how they came to that conclusion based only on that lemma. For example, given $e =0.001$, then the lemma states there is a norm such that $norm(A)\leq p(A) + .001$.

Since it is a "or equal" then this means there is the possibility that $norm(A) = p(A) + 0.001$, and in such a case then $norm(A) > p(A)$. So it seems that based on the lemma, then $p(A) < 1$ does not necessarily always imply that $norm(A) < 1$.

I even tried to do a proof of this but still came up with the same result:

$$\begin{align} p(A)<1&\implies p(A)+e<1+e\\ &\implies norm(A)\leq p(A)+e<1+e\\ &\implies norm(A)-e\leq p(A)<1. \end{align} $$

This pretty much says the same thing, if $norm(A) - e \leq p(A)$, then $p(A) < 1$, but if $p(A)$ close to $1$ by a difference less than $e$, then $norm(A) > 1$ in the "equal to" case where $norm(A) - e = p(A)$.

Now of course you could pick a smaller e, but the way the lemma reads it says that for a given $e$, you can find a norm s.t. $norm(A) \leq p(A) + e$, so the "or equal" part will still get you using a counter example I gave similar to the $e = .001$ above

Now I have seen a different proof on wikipedia that shows if $p(A) < 1$ then $norm(A) < 1$, so I am not disputing that fact, but that proof gets into looking at entries of the Jordan Normal Form. So my point is, yes $p(A) < 1$ then $norm(A) < 1$, BUT I don't understand how you can conclude that based only on the Lemma I gave at the beginning. It's these big jumps in rational that confuse me.

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    The flaw in your reasoning is you're trying to think of $e$ as a fixed value independent of $A$, which clearly doesn't work. For any given matrix $A$, you have to choose $e$ the right way, as Jonas' answer explains.2010-10-26
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    I see. I thought it was the opposite based on the wording of the lemma, that "given any e", then "there is a norm such that" i.e. there is a norm you can pick that satisfies the inequality. Is the lemma worded poorly or am I reading it wrong?2010-10-26
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    It is worded correctly. I think the problem is that you're thinking too much about what it says for some particular choices of $e$. As you said, when $e=.001$, the lemma will give you a norm(A) $\leq$ p(A) +.001. But when you said it might be equal, that is not a problem, because you can just take a smaller $e$ and apply the lemma to get another norm that does what you want.2010-10-26

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Apply the lemma with $e=\frac{1}{2}(1-p(A))$.

Also, the fact that it is not strict inequality is insignificant. Given $e$, you could apply the lemma with $e/2$ to get strict inequality with $e$.