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?
 A: Without multipliers: $A$ is a hermitian matrix (that is satisfies $A^*=A$) so it is diagonalizable in an orthonormal basis. Here the eigenvalues of $A$ are $2+\pm\sqrt{2}$, and there exists a unitary matrix $U$ such that $A=U^* D U$, where $D=diag(2+\sqrt{2},2-\sqrt{2})$.
Since $U$ is unitary , it preserves the euclidean norm. In particular, the norm of $A$ and the norm of $D$ are the same. But for a real diagonal matrix, it is easy to see that the operator norm  is given by the eigenvalue of maximum absolute value. So here the norm of $A$ is $2+\sqrt{2}$ (which agrees with José Carlos Santos result, after multiplying by $\sqrt{2}$).
This method generalizes in arbitray dimensions, provided your matrix is Hermitian.
A: Let $z=x+iy$ and $w=u+iv$ so that $|z|^2+\Re(\overline zw)=x^2+y^2+xu+yv$ is subject to $x^2+y^2+u^2+v^2=1$. Then we have $\mathcal L=x^2+y^2+xu+yv-\lambda(x^2+y^2+u^2+v^2-1)$ so that \begin{align}\mathcal L_u&=x-2\lambda u=0\implies x=2\lambda u\\\mathcal L_v&=y-2\lambda v=0\implies y=2\lambda v\\\mathcal L_x&=2x+u-2\lambda x=0\implies(2-2\lambda)2\lambda u+u=0\\\mathcal L_y&=2y+v-2\lambda y=0\implies(2-2\lambda)2\lambda v+v=0.\end{align} Either $u=v=0$ or $(2-2\lambda)2\lambda+1=0$. If the former then $w=0$ so $|z|^2+\Re(\overline zw)=|z|^2=1$.
If $(2-2\lambda)2\lambda+1=0$ then $\lambda=(1\pm\sqrt2)/2$ so that $|z|^2+\Re(\overline zw)=4\lambda^2(u^2+v^2)+2\lambda(u^2+v^2)$ subject to $4\lambda^2(u^2+v^2)+u^2+v^2=1$. This means that $u^2+v^2=1/(1+4\lambda^2)$ so $|z|^2+\Re(\overline zw)=(2\lambda+4\lambda^2)/(1+4\lambda^2)=\lambda$. Hence $\max\{|z|^2+\Re(\overline zw)\}=(1+\sqrt2)/2$.
A: 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 $\frac12+\frac1{\sqrt2}$.
