<s>[INST] Find the supremum of $|z|^2+\text{Re}(\overline{z}w)$ subject to $|z|^2+|w|^2=1$. I am trying to find the operator norm of $$A=\begin{pmatrix}3 & 1 \\
1 & 1\end{pmatrix}\in M_n(\mathbb{C}).$$This is what I have so far: If $|z|^2+|w|^2=1$, then $$||A\begin{pmatrix}z\\w\end{pmatrix}||_2=\sqrt2\sqrt{1+4(|z|^2+\text{Re}(\overline{z}w))}$$ Since $|z|^2+\text{Re}(\overline{z}w)\in\mathbb{R}$ and $\sqrt{2}\sqrt{1+4x}$ is an increasing function of $x$, we need to maximise $|z|^2+\text{Re}(\overline{z}w)$. In other words,  $$||A||=\sqrt2\sqrt{1+4m},$$ where $m=\text{sup}\lbrace|z|^2+\text{Re}(\overline{z}w): |z|^2+|w|^2=1\rbrace$.
But how do I go about finding $m$? I'm a bit rusty on this stuff, do I need to use Lagrange multipliers?
 [/INST] You can use Lagrange multipliers. Let$$f(x,y,z,t)=|x+yi|^2+\operatorname{Re}\bigl((x-yi)(z+ti)\bigr)=x^2+y^2+xz+yt$$and let$$g(x,y,z,t)=|x+yi|^2+|z+iy|^2=x^2+y^2+z^2+t^2.$$So, you should solve the system$$\left\{\begin{array}{l}\nabla f(x,y,z,t)=\lambda\nabla g(x,y,z,t)\\g(x,y,z,t)=1\end{array}\right.$$There are several families of solutions:

*

*points of the form $\left(x,-\frac{1}{2} \sqrt{-4 x^2-\sqrt{2}+2},-\left(1+\sqrt{2}\right) x,\frac{1}{2} \left(1+\sqrt{2}\right)\sqrt{-4 x^2-\sqrt{2}+2}\right)$, at which $f$ takes the value $\frac12-\frac1{\sqrt2}$;

*points of the form $\left(x,\frac{1}{2} \sqrt{-4x^2-\sqrt{2}+2},-\left(1+\sqrt{2}\right) x,-\frac{1}{2} \left(1+\sqrt{2}\right) \sqrt{-4 x^2-\sqrt{2}+2}\right)$, at which $f$ also takes the value $\frac12-\frac1{\sqrt2}$;

*points of the form $\left(x,-\frac{1}{2}\sqrt{-4 x^2+\sqrt{2}+2},\left(\sqrt{2}-1\right) x,-\frac{1}{2} \left(\sqrt{2}-1\right) \sqrt{-4 x^2+\sqrt{2}+2}\right)$, at which $f$ takes the value $\frac12+\frac1{\sqrt2}$;

*points of the form $\left(x,\frac{1}{2}\sqrt{-4 x^2+\sqrt{2}+2},\left(\sqrt{2}-1\right) x,\frac{1}{2} \left(\sqrt{2}-1\right) \sqrt{-4 x^2+\sqrt{2}+2}\right)$, at which $f$ also takes the value $\frac12+\frac1{\sqrt2}$.

So, the maximum is $\frac12+\frac1{\sqrt2}$.
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