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$$ \text { For } x=\left(x_{1}, x_{2}\right) \in \mathbb{R}^{2} \text { and } 1<p<\infty \text { we define following norms: } \\ \|x\|_{1} :=\left|x_{1}\right|+\left|x_{2}\right| \\\|x\|_{p} :=\left(\left|x_{1}\right|^{p}+\left|x_{2}\right|^{p}\right)^{\frac{1}{p}} \text { } \\\|x\|_{\infty} :=\max \left\{\left|x_{1}\right|,\left|x_{2}\right|\right\} $$

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

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?

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

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