I am reading Chapter 22 of the book "A Hilbert space problem book", written by Paul R. Halmos.The topic is about "Numerical range".
According to the book, the numerical range of an operator $A: \mathbb{H} \rightarrow \mathbb{H}$, in which $\mathbb{H}$ is a Hilbert space, is defined by $$W(A) = \left \{ z \in \mathbb{C} \mid z = \left<Af, f \right> \text{where} \|f\| =1\right\}.$$
In example, the author gives matrix A \begin{pmatrix} 0 & 0\\ 1 & 1 \end{pmatrix} And he claims that the numerical range of $A$ is the closed elliptical disc with foci 0, 1, two axes are 1 and $\sqrt{2}$.
I dont know how he can give that answer.
What I tried is the following.
Let $f=(f_1,f_2)$ be some vector in $\mathbb{C}^2$.
Calculate casually, we obtain $W(A) = \left \{z \in \mathbb{C} \mid z = f_1 f_2 + f_2^2 \text{ where } f_1^2 + f_2^2 =1\right \}$.
Could you please give me the equation of desired elliptical disc? I tried many times but not effective.
Thank you in advace
Here's a sketch of the calculation. As I suggested before, write $f_1=wf_2$ for $w\in\Bbb C$. The condition that $|f_1|^2+|f_2|^2=1$ translates to $|f_2|^2(1+|w|^2)=1$. Now, we're considering $$z=f_1\bar f_2 + |f_2|^2 = |f_2|^2(1+w) = \frac{1+w}{1+|w|^2}.$$
Set $w=te^{i\theta}$. Then you can check that $z=z(t,\theta)=\dfrac1{1+t^2}\big(1+t\cos\theta,t\sin\theta\big)$. As a map from $\Bbb R_+\times [0,2\pi]$ to $\Bbb R^2$, a little calculation shows $z$ drops rank when $2t\cos\theta = 1-t^2$. With a bit of algebra, you can check that, substituting that relation, we get $z=(x,y)$ where $$x=\frac12\cdot\frac{3-t^2}{1+t^2} \quad\text{and}\quad y^2 = \left(\frac t{1+t^2}\right)^2\left(1-\frac14\left(\frac1t-t\right)^2\right).$$
I don't quite see how to do the algebra pro-actively, but you can in fact substitute and check that this parametric curve gives you precisely $$2(x-\tfrac12)^2 + 4 y^2 = 1,$$ which is the ellipse with axes $1$ and $\sqrt 2$ and centered at $x=1/2$, and this has foci at precisely $z=0$ and $z=1$.