$T=\pmatrix{a &-b \\ b & a} \in \mathbb{R}^{2 \times 2} \cong L_b (\mathbb{R}^2,\mathbb{R}^2)$
Where $L_b (\mathbb{R}^2,\mathbb{R}^2)$ denotes the linear and bounded functions from $\mathbb{R}^2 $ into $\mathbb{R}^2$.
I want to determine the operator-norm of $T$: $\|T\|$ when $\mathbb{R}^2$ is associated with $\|.\|_2$.
Operatornorm: $$\sup_{{\|v\|}_V=1}\|f(v)\|_W$$
$T \pmatrix{x \\ y}=\pmatrix{ax-by \\ bx+ay}$
$\Bigg\|\pmatrix{ax-by \\ bx+ay}\Bigg\|_2=\sqrt{(ax-by)^2+(bx+ay)^2}=\sqrt{(x^2+y^2)(a^2+b^2)} \le \sqrt{a^2+b^2}$. (Because $\|x^2+y^2\|_2$ is $1$ per definition)
For the other direction: $x:=1, y:=0$.
Then $T \pmatrix{x \\ y}=\sqrt{a^2+b^2}$ so $T \pmatrix{x \\ y} \ge \sqrt{a^2+b^2}$.
Therefore alltogether $\|T\|=\sqrt{a^2+b^2}$.
Is that correct?
Furthermore I want to determine the operator norm of $T=\pmatrix{3 & 5} \in \mathbb{R}^{1x2} \cong L_b(\mathbb{R}^2,\mathbb{R}^1)$ when $\mathbb{R}^2$ is associated with $\|.\|_\infty$ and $\mathbb{R}$ with $|.|$
$T \pmatrix{x \\ y}=\pmatrix{3x+5y}$
$|3x+5y| \le 3|x|+5|y| \le 8 \max\{|x|,|y|\}$ therfore $\|T\| \le 8$.
For the other direction choose $x=y=1$, then $\|T\| \ge 8$ and alltogether $\|T\|=8$.
Is that correct?
Thanks for your help!
Both of your computations are correct.
In the first case there is a slightly simpler argument:
$\Bigg\|\pmatrix{ax-by \\ bx+ay}\Bigg\|_2=\sqrt{(ax-by)^2+(bx+ay)^2}=\sqrt{(x^2+y^2)(a^2+b^2)}=\sqrt{a^2+b^2}$
for all $\pmatrix{x \\ y}$ with $||\pmatrix{x \\ y}||_2=1$.
Hence $\|T\|=\sqrt{a^2+b^2}$.