$$ \begin{array}{l}{\text { We have following matrix }} \\ {\qquad T=\left(\begin{array}{cc}{-3} & {1} \\ {1} & {2}\end{array}\right)} \\ {\text { a) Show that, with }\|x\|_{T}=\|T x\|_{\infty} \text { a vector norm is defined for } \mathbb{R}^{2} \text { }} \\ {\text { (Hint: Properties of a norm.) }}\end{array} $$ $$ \begin{array}{l}{\text { b) Sketch the unit circle } B_{T}=\left\{x \in \mathbb{R}^{2}\|x\|_{T} \leq 1\right\}} \\ {\text { c) State the }\|\cdot\| T \text { assigned matrix norm }} \\ {\quad\|A\|_{T}:=\max _{x \neq 0} \frac{\|A x\|_{T}}{\|x\|_{T}}} \\ {\text { with the matrix norm}\|\cdot\|_{\infty} \text { }}\end{array} $$
Am trying to do a. So if the vector norm is 0 the inside has to be 0. Is the vector norm the square root and the inside ^2 ? I literally have no clue how to proof the properties.. Thanks for any help.
They are defining a new norm in $\mathbb{R}^2$ by the expression $$\|x\|_T=\|Tx\|_{\infty}.$$ Take into account that if $x=(a,b)$ then $$\|(a,b)\|_{\infty}=\max\{|a|,|b|\}$$ in your particular case $$\|x\|_T=\|Tx\|_{\infty}=\max\{|-3a+b|,|a+2b|\}.$$