Numerical range of normal matrices Let $F_1, F_2$ be two normal matrices, we consider
$$W(F_1,F_2)=\{(\langle F_1 y\; ,\;y\rangle,\langle F_2 y ,\;y\rangle):y \in F,\;\;\|y\|=1\}.$$

If $F_1F_2=F_2F_1$. Is $W(F_1,F_2)$ convex?

Thank you!
 A: It is quite simple: if all $(F_1,\cdots, F_d)$ commute with each other and normal, they all have the same orthonormal eigenbases $(v_1,\cdots,v_n)$ and every state vector $x$ can be decomposed:
$$x=\displaystyle\sum_{i=1}^n x_i v_i$$
Average value of $F_j$ over this state is simply
$$\displaystyle\sum_{i=1}^n |x_i|^2 \lambda(F_j)_i,$$
where $\lambda(F_j)_i$ is the eigenvalue of $F_j$ corresponding to the eigenvector $v_i$.
Now, if there are two points $A,B \in W(F_1,\cdots, F_k)$, there are state vectors $x,y$ such that image of $x$ is $A$ and image of $y$ is $B$.
We would like to prove that for any point $C$ on the line segment from $A$ to $B$ there exists a state vector $z$ corresponding to $C$. But
$C=\alpha A+(1-\alpha)B,\;(0<\alpha<1)$ and all the corresponding averages (coordinates of $C$) have form
$$\displaystyle\sum_{i=1}^n(\alpha |x_i|^2 + (1-\alpha)|y_i|^2) \lambda(F_j)_i,$$
so clearly the state vector $z$ defined as
$$z=\displaystyle\sum_{i=1}^n\left(\sqrt{\alpha |x_i|^2 + (1-\alpha)|y_i|^2}\right) v_i,$$
corresponds to the point $C$.
