Let $$X=\begin{pmatrix} r & x\\ y & r\end{pmatrix} \in M_n(\mathbb{C}).$$ Here $r> 0$, $x, y \in \mathbb{C}$. Calculate the numerical radius $w(X)$ of $X$ where $$w(X) := \sup\limits_{\Vert\alpha\Vert=1}\vert\langle X\alpha,\alpha\rangle\vert$$
Comments: I can see that $w(X)\leq r+\frac{\vert x\vert+\vert y\vert}{2}$ but could not prove the equality. Moreover it is clear that $W(X-rI)$ is an ellipse with length of major axis $\vert x\vert+\vert y\vert$ and the length of the minor axis $\vert y\vert$ i.e $w(X-rI)=\frac{\vert x\vert+\vert y\vert}{2}$ but after that I am clueless to calculate numerical radius of $X$ from that.
Any comment is highly appreciated.
Let $A=\begin{pmatrix}r&a+ib\\c+id&r\end{pmatrix}$ where $r>0$ and real $a,b,c,d$ are given and $U=\{u=[p+iq,s+it]^T;p^2+q^2+s^2+t^2=1,p,q,s,t\in\mathbb{R}\}$. Then $f:u\in U\mapsto |u^*Au|^2$ is a homogeneous polynomial of degree $4$ and we seek its maximum under one constraint of degree $2$.
METHOD 1. We can use the "NLPSolve" Maple's software which solves optimization problems by iterative methods. We can choose as initial point an eigenvector of modulus $1$ associated to the eigenvalue of $A$ of maximum modulus: $u_0=[p_0+iq_0,s_0+it_0]^T$. Note that we may assume that $t_0=0$ and (eventually) we may add the constraint $t=0$.
Often, we obtain a correct solution; yet, the problem may become instable and the obtained limit point may be not convenient. For example
$A_0=\begin{pmatrix}5&-3+19i\\-4-15i&5\end{pmatrix}$.
Of course, the main question is not to find $\tau(A_0)$, the $\max$ associated with $A_0$, but to find a choice of the initial point which works in all cases; this does not prevent from calculating an approximation of $\tau(A_0)$...
EDIT. METHOD 2. With the Lagrange's method; if $f$ reaches its $\max(f)=M$ in $u_0=(p_0,q_0,s_0,t_0)$, then there is $l$ s.t. in $u_0$
$(*)$ $\dfrac{\partial f}{\partial p}+l p=\dfrac{\partial f}{\partial q}+l q=\dfrac{\partial f}{\partial s}+l s=\dfrac{\partial f}{\partial t}+l t=0$ and $t=0,p^2+q^2+s^2=1$.
Using $p\dfrac{\partial f}{\partial p}+q\dfrac{\partial f}{\partial q}+\cdots=4f(u_0)$, we deduce that $4M+l=0$.
The sequel is more technical. Using the Grobner basis theory, we formally solve (with Maple for example) the system $(*)$. $l$ is some root of a polynomial. For our above example $A_0$, we obtain
Finally $w(A)^2=M=-l_0/4$ where $l_0$ is the smallest real root of the above polynomial, that is, $M=2002.4808/4=500.6202$.
Note that the upper bound for $w(A)^2$ given by the OP is $500.8545$.