calculate the norm of the following operator on $\mathbb{C}^2$ It is well known that if  $A\in \mathcal{B}(F)$, then
$$\|A\|:=\displaystyle\sup_{\|x\|=1}\|Ax\|.$$
and
$$\omega(T):=\displaystyle\sup_{\|x\|=1}|\langle Ax, x\rangle|,$$
where $F$ is a  complex Hilbert space and  $\mathcal{B}(F)$ is the algebra of all bounded linear operators on $F$.

Consider the following operator on $\mathbb{C}^2$
$$T=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}.$$
  I want to prove that
   \begin{align*}
\omega(T)
&=\sup\left\{||x|^2+y\overline{x}+|y|^2|\,;\;(x,y)\in \mathbb{C}^2,\;\text{and}\;|x|^2+|y|^2=1 \right\}=\tfrac{3}{2},
\end{align*}
  and
  \begin{align*}
\|T\|
&=\sup\left\{\sqrt{|x+y|^2+|y|^2}\,;\;(x_1,x_2)\in \mathbb{C}^2,\;\text{and}\;|x|^2+|y|^2=1 \right\}=\frac{\sqrt{5}+1}{2}.
\end{align*}

Attempt: 
Clearly, for all $(x,y)\in \mathbb{C}^2$ such that $|x|^2+|y|^2=1$, we have
$$||x|^2+y\overline{x}+|y|^2|\leq \frac12(|x|^2+|y|^2)+|x|^2+|y|^2=\frac32.$$
Hence,
$$\omega(T)\leq\frac32.$$
To calculate $\|T\|$, I try to use the following result:
$$\|T\|=\max\{\sqrt{\lambda};\;\lambda\in \sigma(T^*T)\},$$
where $\sigma(A)$ denotes the spectrum of an operator $A$.
Clearly
$$T^*T=\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}.$$
Also
$$\sigma(T^*T)=\{\frac{\sqrt{5}+3}{2},\frac{3-\sqrt{5}}{2}\}.$$
Hence,
$$\|T\|^2=\frac{\sqrt{5}+3}{2}.$$
 A: For the second, let's see why for all $z,w\in\mathbb{C}$ with $|z|^2+|w|^2=1$ it is $\sqrt{|z+w|^2+|w|^2}\leq(\sqrt{5}+1)/2$. Let $a=|z|, b=|y|\in\mathbb{R}$. It is obviously $\sqrt{|z+w|^2+|w|^2}\leq\sqrt{1+2ab+b^2}$ while $a,b$ satisfy $a^2+b^2=1$. Now we can use Lagrange multipliers to solve a classical problem. I won't go into details, but it will work (at least a graph calculator says so).
Now let's see that there exist $z,w\in\mathbb{C}$ such that $|z|^2+|w|^2=1$ and $\sqrt{|z+w|^2+|w|^2}=(\sqrt{5}+1)/2$. Rewrite this equivalently as $4(|z+w|^2+|w|^2)=6+2\sqrt{5}$. This is equivalent to $2(|w|^2+2\text{Re}(z\overline{w}))=1+\sqrt{5}$. Now let $w$ be a real number, say $c\in(0,1)$. Then we have that $2c^2+2c\text{Re}(z)=1+\sqrt{5}$, hence $\text{Re}(z)=\frac{1+\sqrt{5}-2c^2}{2c}$. Let for convenience $c$ be the solution to the equation $2t^2=1+\sqrt{5}$ (it does indeed have a solution in $(0,1)$). Then $\text{Re}(z)=0$. So now since $z$ is imaginary, it is $z=iy$ and $|z|=|y|$. solve $c^2+y^2=1$ for $y$ and note that it has a solution in $(0,1)$ too. the pair $(iy,c)\in\mathbb{C}^2$ makes this supremum a maximum.
