State matrix norm (respect to Matrix) with matrix norm (respect to infinity)

Question

$$ \begin{array}{l}{\text { We have following matrix }} \\ {\qquad T=\left(\begin{array}{cc}{-3} & {1} \\ {1} & {2}\end{array}\right)} \\ {\text { a) Show that, with }\|x\|_{T}=\|T x\|_{\infty} \text { a vector norm is defined for } \mathbb{R}^{2} \text { }} \\ {\text { (Hint: Properties of a norm.) }}\end{array} $$ $$ \begin{array}{l}{\text { b) Sketch the unit circle } B_{T}=\left\{x \in \mathbb{R}^{2}\|x\|_{T} \leq 1\right\}} \\ {\text { c) State the }\|\cdot\| T \text { assigned matrix norm }} \\ {\quad\|A\|_{T}:=\max _{x \neq 0} \frac{\|A x\|_{T}}{\|x\|_{T}}} \\ {\text { with the matrix norm}\|\cdot\|_{\infty} \text { }}\end{array} $$

Trying to do c) my hint is to reform it until there is only a maxx norm but how do i do that?

Answer 1

Hint. Try to rewrite $\max_{x\ne0}\frac{\|Ax\|_T}{\|x\|_T}$ in the form of $\max_{y\ne0}\frac{\|By\|_\infty}{\|y\|_\infty}$ where $B$ is some matrix that depends on $T$ and $A$, and $y$ is some vector depending on $T$ and $x$. It follows that $\|A\|_T=\|B\|_\infty$.

Full solution:

! For every nonzero vector $x$, let $y=Tx$. Then $$ \frac{\|Ax\|_T}{\|x\|_T} =\frac{\|TAx\|_\infty}{\|Tx\|_\infty} =\frac{\|TAT^{-1}y\|_\infty}{\|y\|_\infty}.\tag{1} $$ Since $T$ is invertible, there is a one-to-one correspondence between $x$ and $y$. Therefore, the maximum of the LHS of $(1)$ over all nonzero vectors $x$ is equal to the maximum of the RHS over all nonzero vectors $y$. It follows that $$ \|A\|_T=\max_{x\ne0}\frac{\|Ax\|_T}{\|x\|_T} =\max_{y\ne0}\frac{\|TAT^{-1}y\|_\infty}{\|y\|_\infty} =\|TAT^{-1}\|_\infty. $$