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$|\;|$ 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 $m(T)=\frac{1}{||S||}$.

(I think probably we need to do something with the norm, but I still can't get it... So thank you.)

Ian
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1 Answers1

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On the one hand, $$ m(T) = \inf_{\|x\|=1} \|T(x)\| = \inf_{\|S(T(x))\|=1} \|T(x)\| = \inf_{\|S(y)\| = 1} \|y\|, $$ where for $y$ with $\|S(y)\| = 1$, $1 = \|S(y)\| \leq \|S\|\|y\|$, and hence $\|y\| \geq \|S\|^{-1}$.

On the other hand, $$ \|S\| = \sup_{\|x\|=1} \|S(x)\| = \sup_{\|T(S(x))=1\|} \|S(x)\| = \sup_{\|T(y)\|=1}\|y\|, $$ where for $y$ with $\|T(y)\|=1$, $1 = \|T(y)\| \geq m(T)\|y\|$, and hence $\|y\| \leq m(T)^{-1}$.

Given this, what can you conclude about the relations between $m(T)$ and $\|S\|^{-1}$?

Branimir Ćaćić
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  • But I only get $||S||m(T)\leqslant 1$, what about the other direction? – Ian Mar 17 '13 at 03:58
  • In the first paragraph, since you're taking an infimum, you get that $m(T) \geq |S|^{-1}$. In the second paragraph, since you're taking a supremum, you get that $|S| \leq m(T)^{-1}$, and hence that $m(T) \leq |S|^{-1}$. – Branimir Ćaćić Mar 17 '13 at 04:14
  • And why $||S||=\underset{||x||=1}{sup||S(x)||}$? – Ian Mar 17 '13 at 04:19
  • That's precisely the definition of the operator norm... – Branimir Ćaćić Mar 17 '13 at 04:23
  • But why it's the operator norm, not the Euclidean norm? Is that because operator norm is just the Euclidean norm for matrix? – Ian Mar 17 '13 at 04:56
  • In this particular context, the co-norm is being defined "dually" to the operator norm. There are certainly other norms you can define on square matrices (i.e., operators on finite-dimensional inner product spaces), such as the Frobenius norm, but the operator norm is extremely natural, and turns out to be what generalises most readily to the context of infinite-dimensional Banach and Hilbert spaces.

    Of course, on a finite-dimensional vector space, such as that of $n \times n$ real matrices, any two norms are equivalent; see http://en.wikipedia.org/wiki/Matrix_norm#Equivalence_of_norms

     – Branimir Ćaćić Mar 17 '13 at 05:38
  • In any event, the operator norm is the norm on linear transformations that makes this problem actually work. – Branimir Ćaćić Mar 17 '13 at 05:41
  • I'm not seeing why ||T(y)||>=m(T)||y|| in the second part. –  Mar 17 '13 at 13:38
  • Because $m(T) = \inf_{|y|=1}|T(y)| = \inf_{y \neq 0} \frac{|T(y)|}{|y|}$, so that for general $y \neq 0$, $m(T) \leq \frac{|T(y)|}{|y|}$, and hence $|T(y)| \geq m(T)|y|$. – Branimir Ćaćić Mar 17 '13 at 19:12