# Is the numerical range of Identity operator convex?

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?

The numeric range of $$I$$ is not the unit circle, it is the singleton $$\{1\}$$, which is a convex set.