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

ianchenmu

Member
Feb 3, 2013
74
$|\;|$ 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.)
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779
first, the title should be "co-norm of an invertible linear transformation on R^n" and here is the question:


$|\;|$ 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.)
Let's take a look at a generic matrix $A$ that has $\lambda_M$ as the eigenvalue with the largest magnitude.
Then $A^T A$ is symmetric with eigenvalues that are the squares of the eigenvalues of $A$.
According to the spectral theorem a symmetric real matrix can be diagonalized into $B D B^T$, where $D$ is a diagonal matrix, and $B$ is an orthogonal matrix.
It follows that:

$$\begin{aligned}
\lVert A \rVert^2 &= \max_{\lVert x \rVert=1} \lVert Ax \rVert^2 & (1) \\
&= \max_{\lVert x \rVert=1} x^T A^T A x & (2) \\
&= \max_{\lVert x \rVert=1} x^T B D B^{-1} x & (3) \\
&= \max_{\lVert y \rVert=1} y^T D y & (4) \\
&= \max_{\lVert y \rVert=1} \lambda_M^2 \lVert y \rVert^2 & (5) \\
&= \lambda_M^2 & (6) \\
\end{aligned}$$

Now apply this reasoning to both $m(T)$ and $\lVert S \rVert$...