Math Help - Square matrix and induced norm

1. Square matrix and induced norm

Let T be any square matrix and let $\left\| \cdot \right\|$ denote any induced norm. Prove that

$lim_{n \rightarrow _{\infty}} \left\| T^{n} \right\| ^{1/n}$ exists and equals $inf _{n=1,2,\cdots } \left\| T^{n} \right\| ^{1/n}$

I am not sure how I go about proving that the limit exists.
For the infimum, I think it has something to do with the fact that

$\left\| T^{n} \right\| = sup_{x \neq 0} \frac{\left\| T^{n} x\right\|}{\left\| x \right\|}$
But I don't know how this information helps with the solution of the problem.

2. This isn't an answer; I'm just retyping your equations which are broken for me (hotlinking I think)...

Originally Posted by math8
Let T be any square matrix and let $||\cdot||$ denote any induced norm. Prove that $\lim_{n\to\infty}||T^n||^{1/n}$ exists and equals $\inf_{n=1,2,\cdots} ||T^n||^{1/n}$.

I am not sure how I go about proving that the limit exists.
For the infimum, I think it has something to do with the fact that

$||T^n|| = \sup_{x\neq0} {||T^nx||\over||x||}$

But I don't know how this information helps with the solution of the problem.

This isn't an answer; I'm just retyping your equations which are broken for me (hotlinking I think)...
Originally Posted by math8
Let T be any square matrix and let $\|\cdot\|$ denote any induced norm. Prove that $\lim_{n\to\infty}\|T^n\|^{1/n}$ exists and equals $\inf_{n=1,2,\ldots} \|T^n\|^{1/n}.$

4. Thanks to Maddas and to Opalg.

I can understand the proof of the existence of $lim_{n \rightarrow \infty }\left\| T^{n} \right\| ^{1/n}$.

$lim_{n \rightarrow \infty }\left\| T^{n} \right\| ^{1/n}$ exists and is equal to $\rho (T)$, the spectral radius of T.

But how do we show that $\rho (T) = inf_{n=1,2, \cdots } \left\| T^{n} \right\| ^{1/n}$ ?

5. Originally Posted by math8
Thanks to Maddas and to Opalg.

I can understand the proof of the existence of $lim_{n \rightarrow \infty }\left\| T^{n} \right\| ^{1/n}$.

$lim_{n \rightarrow \infty }\left\| T^{n} \right\| ^{1/n}$ exists and is equal to $\rho (T)$, the spectral radius of T.

But how do we show that $\rho (T) = inf_{n=1,2, \cdots } \left\| T^{n} \right\| ^{1/n}$ ?
This follows from the fact the norm has to be submultiplicative, so that $\|S^2\|\leqslant\|S\|^2$ for any matrix S.

Suppose that, for some n, $\|T^n\|^{1/n}=\lambda<\rho(T)$. Then $\|T^n\|=\lambda^n$, $\|T^{2n}\| = \|(T^n)^2\| \leqslant \|T^n\|^2= \lambda^{2n}$, and by induction $\|T^{2^kn}\|\leqslant\lambda^{2^kn}$ for all k.

Therefore $\|T^{2^kn}\|^{1/(2^kn)}\leqslant\lambda$. let $k\to\infty$ to see that $\rho(t)\leqslant\lambda$ — contradiction.

6. Square matrix and induced norm

I might be wrong but according to that proof, does this just mean that $\rho (T)$ must be a lower bound for $\{ \left\| T^{n} \right\| ^{1/n} : n,1,2, \cdots \}$ ?

How do we show that $\rho (T)$ is actually the greatest lower bound? I mean is there an n such that $\rho (T) = \left\| T^{n} \right\| ^{1/n}$ ?

7. Originally Posted by math8
I might be wrong but according to that proof, does this just mean that $\rho (T)$ must be a lower bound for $\{ \left\| T^{n} \right\| ^{1/n} : n,1,2, \cdots \}$ ?

How do we show that $\rho (T)$ is actually the greatest lower bound? I mean is there an n such that $\rho (T) = \left\| T^{n} \right\| ^{1/n}$ ?
Yes, it shows that $\rho (T)$ must be a lower bound for $\{ \left\| T^{n} \right\| ^{1/n} : n,1,2, \cdots \}$. But I thought you had already shown that $\rho (T) = \lim_{n\to\infty} \left\| T^{n} \right\| ^{1/n}$. Taken together, those results show that $\rho (T)$ is the greatest lower bound.

8. Oh, I didn't realize that. I actually thought that for that result, one actually has to show that there is an n such that $\rho (T) = \left\| T^{n} \right\| ^{1/n}$. (without the limit)

But I guess you're right.

Thanks a lot!