# Sketch unit circle for matrix norm

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

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?

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

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."

• Jesus thanks a lot mate! – Rack Cloud Nov 4 '19 at 19:10
• wait isn't the case with -3a+b=-1 just the $$a \leq \frac{-3}{7}$$ ? – Rack Cloud Nov 4 '19 at 19:18
• 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 = ? – Rack Cloud Nov 4 '19 at 19:36
• 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 = $$-\frac{1}{2} - \frac{1}{2} a$$ – Rack Cloud Nov 4 '19 at 19:59