I understand that the numerical range of a $2 \times 2$ matrix is a convex set. However, when I tried with the identity operator, the numerical range is a circle. According to enter link description here, the circle is not convex. Hence, I really confuse about that.
What I did is the following.
Suppose that $f = \begin{pmatrix} f_1\\ f_2 \end{pmatrix}$ is a unit vector in $\mathbb{R}^2$.
Then $If= \begin{pmatrix} f_1\\ f_2 \end{pmatrix}$. So $\langle If,f\rangle = |f_1|^2 + |f_2|^2 = 1$.
Then the set of $z = \langle If,f\rangle$ is the unit circle. However, the unit circle is not convex.
Could you please show me what my mistake is?
Thank you in advance.
The numeric range of $I$ is not the unit circle, it is the singleton $\{1\}$, which is a convex set.