# Sketch a set with respect to infinity norm

$$\text { For } x=\left(x_{1}, x_{2}\right) \in \mathbb{R}^{2} \text { and } 1

$$\begin{array}{l}{\text { c) Sketch for }} \\ {\qquad A=\left(\begin{array}{ll}{1} & {4} \\ {0} & {2}\end{array}\right)} \\ {\text { the set }\left\{A x:\|x\|_{\infty}=1\right\} \text { . }} \\ {\text { A linear mapping like for example } x \rightarrow A x \text { maps lines }} \\ {\text { on lines and the intersection of 2 lines maps on the intersection point }} \\ {\text { of their image line }}\end{array}$$

stuck with that Trying to figure out what i have to do to solve that. My thoughts are that x1 has to be 1 or x2 has to be 1 so i could take both of em in a R^2 and multiply that?

• First, can you sketch the set $\{x:\|x\|_\infty=1\}$? – user856 Oct 29 '19 at 3:34
• already done that – Rack Cloud Oct 29 '19 at 3:40
• Square on all 1 and -1 points on a x,y axis – Rack Cloud Oct 29 '19 at 3:42

Ok. From the comments it seems you have a picture of $$\{x: ||x||_\infty = 1\}$$. Now look at your linear transformation; it takes the regular unit vector along the $$x$$ direction, i.e. $$\binom 1 0$$ to again $$\binom 1 0$$, and the regular unit vector along the $$y$$ direction, i.e. $$\binom 01$$ to the vector $$\binom 42$$.
Draw a picture of this transformation; just think about tilting the plane in such a way the the $$x$$-axis stays along the $$x$$-axis but the $$y$$-axis now aligns with $$\binom 42$$. Then look back at your picture $$\{x: ||x||_\infty = 1\}$$ and draw in the lines. I don't want to give you the full answer (with picture and all), but comment if you have further questions.