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

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?

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  • $\begingroup$ you have actually $|-3a+b|=1$ and $|-3a+b|\ge|a+2b|$ or $|a+2b|=1$ and $|a+2b|\ge |-3a+b|$ $\endgroup$ – J. W. Tanner Nov 4 '19 at 17:49
  • $\begingroup$ Well thanks for that. How do i get to the values with that? $\endgroup$ – Rack Cloud Nov 4 '19 at 18:01
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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."

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  • $\begingroup$ Jesus thanks a lot mate! $\endgroup$ – Rack Cloud Nov 4 '19 at 19:10
  • $\begingroup$ wait isn't the case with -3a+b=-1 just the $$a \leq \frac{-3}{7}$$ ? $\endgroup$ – Rack Cloud Nov 4 '19 at 19:18
  • $\begingroup$ Ok guess i got it for -3a+b=-1 i have $$ \frac{1}{7} \leq a \leq \frac{3}{7} b= -1+3a$$ For the next 2 i have to start with b = ? $\endgroup$ – Rack Cloud Nov 4 '19 at 19:36
  • $\begingroup$ now i got $$ -\frac{1}{7} \leq a \leq \frac{3}{7} $$ with b = $$\frac{1}{2} - \frac{1}{2} a $$ And $$ -\frac{3}{7} \leq a \leq \frac{1}{7} $$ with b = $$-\frac{1}{2} - \frac{1}{2} a $$ $\endgroup$ – Rack Cloud Nov 4 '19 at 19:59

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