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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.

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    $\begingroup$ Do not delete your question after it has been answered. It is considered very rude. $\endgroup$ – Joel Reyes Noche Nov 6 '18 at 13:30
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The numeric range of $I$ is not the unit circle, it is the singleton $\{1\}$, which is a convex set.

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