Square matrix and induced norm

• Apr 10th 2010, 10:44 AM
math8
Square matrix and induced norm
Let T be any square matrix and let $\displaystyle \left\| \cdot \right\|$ denote any induced norm. Prove that

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

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

$\displaystyle \left\| T^{n} \right\| = sup_{x \neq 0} \frac{\left\| T^{n} x\right\|}{\left\| x \right\|}$
But I don't know how this information helps with the solution of the problem.
• Apr 10th 2010, 12:54 PM
This isn't an answer; I'm just retyping your equations which are broken for me (hotlinking I think)...

Quote:

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

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

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

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

• Apr 11th 2010, 12:22 AM
Opalg
Quote:

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

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

• Apr 12th 2010, 01:26 PM
math8
Thanks to Maddas and to Opalg.

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

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

But how do we show that $\displaystyle \rho (T) = inf_{n=1,2, \cdots } \left\| T^{n} \right\| ^{1/n}$ ?
• Apr 13th 2010, 12:17 AM
Opalg
Quote:

Originally Posted by math8
Thanks to Maddas and to Opalg.

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

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

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

This follows from the fact the norm has to be submultiplicative, so that $\displaystyle \|S^2\|\leqslant\|S\|^2$ for any matrix S.

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

Therefore $\displaystyle \|T^{2^kn}\|^{1/(2^kn)}\leqslant\lambda$. let $\displaystyle k\to\infty$ to see that $\displaystyle \rho(t)\leqslant\lambda$ — contradiction.
• Apr 14th 2010, 06:35 AM
math8
Square matrix and induced norm
I might be wrong but according to that proof, does this just mean that $\displaystyle \rho (T)$ must be a lower bound for $\displaystyle \{ \left\| T^{n} \right\| ^{1/n} : n,1,2, \cdots \}$ ?

How do we show that $\displaystyle \rho (T)$ is actually the greatest lower bound? I mean is there an n such that $\displaystyle \rho (T) = \left\| T^{n} \right\| ^{1/n}$ ?
• Apr 14th 2010, 08:43 AM
Opalg
Quote:

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

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

Yes, it shows that $\displaystyle \rho (T)$ must be a lower bound for $\displaystyle \{ \left\| T^{n} \right\| ^{1/n} : n,1,2, \cdots \}$. But I thought you had already shown that $\displaystyle \rho (T) = \lim_{n\to\infty} \left\| T^{n} \right\| ^{1/n}$. Taken together, those results show that $\displaystyle \rho (T)$ is the greatest lower bound.
• Apr 14th 2010, 01:56 PM
math8
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 $\displaystyle \rho (T) = \left\| T^{n} \right\| ^{1/n}$. (without the limit)

But I guess you're right.

Thanks a lot!