$$
\begin{array}{l}{\text { We have following matrix }} \\ {\qquad T=\left(\begin{array}{cc}{-3} & {1} \\ {1} & {2}\end{array}\right)} \\ {\text { a) Show that, with }\|x\|_{T}=\|T x\|_{\infty} \text { a vector norm is defined for } \mathbb{R}^{2}} \\ {\text { (Hint: Properties of a norm.) }} \\ {\qquad \text { b) Sketch the unit circle } B_{T}=\left\{x \in \mathbb{R}^{2}\|x\|_{T} \leq 1\right\}}\end{array}
$$
So i have $$ \mid \mid x \mid \mid_{T} = max\{\mid -3a+b \mid, \mid a+2b \mid \} $$ so -3a +b = 1, a+2b =1 ? How do i have to continue to sketch the unit circle?
Best Answer
You have $|-3a+b|=1$ and $|-3a+b|\ge|a+2b|$ or $|a+2b|=1$ and $|a+2b|\ge|-3a+b|$.
For the first case, it could be $-3a+b=1$ or $-3a+b=-1$.
In the former case we have $1\ge|a+2b|=|a+2(1+3a)|=|7a+2|,$
so $-\dfrac37\le a\le-\dfrac17$ and $b=1+3a$, which is a line segment in the $ab$-plane.
I will leave you to figure out the case $-3a+b=-1$
and the two other cases $a+2b=1$ and $a+2b=-1$.
Altogether, you should get a parallelogram in the $ab$-plane for the "unit circle."