# How does norm(A) <= p(A) + e imply norm(A) < 1 if p(A) < 1?

I am given a lemma and it states:

Lemma: Let A be a real n x n matrix. Then given any e > 0, there is a norm such that norm(A) <= 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 = .001, then the lemma states there is a norm such that norm(A) <= p(A) + .001

Since it is a "or equal" then this means there is the possibility that norm(A) = p(A) + .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:

p(A) < 1 ->
p(A) + e < 1 + e ->
norm(A) <= p(A) +e < 1 + e
norm(A) - e <= p(A) < 1


This pretty much says the same thing, if norm(A) - e <= 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. A <= 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. – Rahul Oct 26 '10 at 3:57
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? – AaronLS Oct 26 '10 at 5:01
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. – Jonas Meyer Oct 26 '10 at 5:09

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 $e$.