Co-norm of an invertible linear transformation on R^n

ianchenmu
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Homework Statement


|\;| is a norm on \mathbb{R}^n.
Define the co-norm of the linear transformation T : \mathbb{R}^n\rightarrow\mathbb{R}^n to be
m(T)=inf\left \{ |T(x)| \;\;\;\; s.t.\;|x|=1 \right \}
Prove that if T is invertible with inverse S then m(T)=\frac{1}{||S||}.


Homework Equations


n/a


The Attempt at a Solution


I think probably we need to do something with the norm, but I still can't get it... So thank you.
 
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Equalities of the form inf X = A are often proved by showing that the two inequalities inf X ≤ A and inf X ≥ A both hold. Together they imply equality of course. One of these proofs will typically use that inf X a lower bound of X (consider an arbitrary member of X), and the other will typically use that inf is the greatest lower bound of the set.

How is ##\|S\|## defined? Can you prove anything about the relationship between ##\|S\|## and ##\|T\|##?

Edit: I have so far only proved the inequality ##m(T)\leq 1/\|S\|##. The idea that I think looks the most promising for a proof of the equality is to take a closer look at the set ##\{|Tx|:|x|=1\}##. What is its infimum? What is its supremum?
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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