$\omega(A_{\beta})\geq \frac{1}{2}$ for all $\beta$ and $\|A_{\beta_0}\|<1$ for some $\beta_0$

Question

Let the matrix $A_\beta=\begin{pmatrix} \beta & 1+\beta \\ 0 & 2\beta \end{pmatrix}$ with $\beta\in \mathbb{C}$.

I want to calculate the numerical radius and the operator norm of $A_\beta$. If the calculus of $\|A_{\beta}\|$ and $\omega(A_{\beta})$ is difficult. Is it possible to prove that $\omega(A_{\beta})\geq \frac{1}{2}$ for all $\beta$ and to find some $\beta_0$ such that $\|A_{\beta_0}\|<1$.

The numerical radius of $A_\beta$ is given by \begin{align*} \omega(A_\beta) &=\sup\left\{\left|\beta|x|^2+(1+\beta)y\overline{x}+2\beta|y|^2\right|\,;\;(x,y)\in \mathbb{C}^2, |x|^2+|y|^2=1 \right\}. \end{align*} Clearly, one may observe that for all $(x,y)\in \mathbb{C}^2$ with $|x|^2+|y|^2=1$, we have $$\left|\beta|x|^2+(1+\beta)y\overline{x}+2\beta|y|^2\right|\leq \tfrac{|\beta|}{2}+2|\beta|.$$ Hence, $\omega(A_\beta)\leq\tfrac{|\beta|}{2}+2|\beta|.$

The operator norm of $A_\beta$ is given by \begin{align*} \|A_\beta\|^2 &=\sup\left\{\left|\beta x+(1+\beta)y\right|^2+4\left|\beta\right|^2\left|y\right|^2\,;\;(x,y)\in \mathbb{C}^2, |x|^2+|y|^2=1 \right\}. \end{align*}

Answer 1

For the numerical radius.

Let $\beta=a+ib$; then $w(A)=\sqrt{-m/4}$ where $m$ is the $\min$ of the real roots of $pol(l)$, the following polynomial of degree $4$ in the unknown $l$.

if you are interested, I can give you the details.

Example: when $a=4,b=-7$, we obtain $\omega(A)\approx 17.9883$.

EDIT 1. Answer to the OP. Of course, your conjecture is false. Yet, we can obtain an explicit form for your numerical radius; indeed, $pol$ can be factorized as follows

$pol=lf_1(l)f_2(l)$ where

The root of $f_1$ is always $\geq 0$ and, therefore, is not convenient. Thus $m$ is the smallest of the $2$ roots of $f_2$, that is

EDIT 2. To the OP. As I wrote to you, it's now your job. Use formal calculation software like Maple; if it is too expensive, then you can use Sage for free; it is not very pleasant to use but very efficient.

Here are the results.

$\omega(A)\geq 1/2$ with equality only for $\beta=0$.

${||A||_2}^2=$

Its minimum is

and is only reached in $\beta=$

Answer 2

The question is in two parts, the straight operator norm of $A$ is the largest eigenvalue of $|A|$ modulus of $A$ which has the representation you gave as supremum. The second is trickier for the numerical radius whether it is known that for $2\times 2$ complex matrices the numerical range $W$ (the set in the plane) is a known ellipse given the matrix. The numerical radius is the largest number in modulus $\in W$.

(A long comment )

As it's said when the numerical range is determined we have the equation of the ellipse curve, a second degree in terms of $x$ and $y$, now finding the numerical radius is maximizing a second degree polynomial in $x$ by Pythagoras. Other answers will fruitful also.