How can I find a norm of a linear transformation $T(x,y) = (ax+cy, bx+dy)$? Let a linear transformation $T : \mathbb{C}^2 \to \mathbb{C}^2$ s.t
$T(x,y) = (ax+cy, bx+dy)$
where $a,b,c,d \in \mathbb{C}$.
Now, find the norm of T equipped with ($\mathbb{C}^2$ , $l^1(\{1,2\})$ norm),  ($\mathbb{C}^2$ , $l^{\infty}(\{1,2\})$ norm) each.
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I tried it several ways, using $z=a + bi$ of $z=r_1e^{\theta_1}$ and etc.
However, since a,b,c,d,x,y are all complex, finding maximizing condition with
$|ax+by|, |cx+dy|$ is too hard.
(radius and argument of x and y can change. so it has too many changing variables. also, computing $|ax+by| + |cx+dy|$ or max$(|ax+by|, |cx+dy|)$ requires $$\big|\sqrt{(ax+by)^2}\big|$$
Is there any method to compute the norm?
 A: It seems the following.
Surprisingly, the answers turned out very similar. 
$\ell^1$-norm.
$$\|(x,y)\|=|x|+|y|.$$
$$\|T(x,y)\|= \|(ax+cy,bx+dy)\|=| ax+cy |+|bx+dy|\le$$ $$|ax|+|cy|+|bx|+|dy|=|a||x|+|c||y|+|b||x|+|d||y|=
(|a|+|b|)|x|+(|c|+|d|)|y|\le$$ $$ \max\{|a|+|b|,|c|+|d|\}(|x|+|y|)=\max\{|a|+|b|,|c|+|d|\}\|(x,y)\|.$$
From the other side, if $|a|+|b|\le |c|+|d|$ and $x=0$ then $$\|T(x,y)\|= \|(cy,dy)\|=|cy|+|dy|=|c||y|+|d||y|=(|c|+|d|)|y|=(|c|+|d|)\|(x,y)\|.$$
A case $|a|+|b|>|c|+|d|$ is considered similarly. Thus 
$$\|T\|=\max\{|a|+|b|,|c|+|d|\}.$$
$\ell^\infty$-norm.
$$\|(x,y)\|=\max\{|x|,|y|\}.$$
$$\|T(x,y)\|= \|(ax+cy,bx+dy)\|=\max\{| ax+cy |,|bx+dy|\}\le$$ $$\max\{| ax|+|cy |,|bx|+|dy|\}=
\max\{| a||x|+|c||y |,|b||x|+|d||y|\}\le$$ $$\max\{| a|+|c|,|b|+|d|\}\max\{|x|,|y|\}.$$
From the other side, if $|a|+|c|\le |b|+|d|$ and $x=y$ then $$\|T(x,y)\|= \|(a+c)y,(b+d)y)\|=
\max\{|(a+c)y|,|(b+d)y)|\}=$$ $$\max\{|a+c||y|,|b+d||y|\}=|b+d||y|=|b+d|\|(x,y)\|.$$
A case $|a|+|c|>|b|+|d|$ is considered similarly. Thus 
$$\|T\|=\max\{|a|+|c|,|b|+|d|\}.$$
