Numerical range of a normal matrix is the convex hull of its eigenvalues

Question

Show that the numerical range of a normal matrix is the convex hull of its eigenvalues. That is, if $A\in M_n$ is a normal matrix with eigenvalues $x_1, x_2, \ldots, x_n$ then $$W(A) = \left\{t_1x_1+t_2x_2+...+t_nx_n:\;t_i\geq 0,\;\sum_{i=1}^nt_i=1\right\}.$$

Answer 1

The numerical range is invariant w.r.t. unitary transformations, that is, if $Q$ is unitary then $$ \begin{split} W(Q^*AQ)&=\{x^*(Q^*AQ)x:x^*x=1\}=\{(Qx)^*A(Qx):x^*x=1\}\\ &=\{y^*Ay:y^*y=1\}=W(A). \end{split} $$ If $A$ is normal then $A=Q^*\Lambda Q$ for some unitary $Q$ and a diagonal $\Lambda:=\mathrm{diag}(\lambda_i)_{i=1}^n$ where $\Lambda$ has the eigenvalues of $A$ on its diagonal. So $$ W(A)=W(\Lambda)=\{x^*\Lambda x:x^*x=1\}. $$ Let $x:=[x_1,\ldots,x_n]^T$. We have $$\tag{1} x^*\Lambda x=\sum_{i=1}^n\lambda_i\bar{x}_ix_i=\sum_{i=1}^n\lambda_i|x_i|^2=\sum_{i=1}^n\lambda_i t_i, $$ where $t_i:=|x_i|^2$. Note that $t_i\geq 0$ and since $x^*x=1$, $\sum_{i=1}^nt_i=1$, which shows that $x^*\Lambda x$ in (1) is a convex combinations of $\lambda_i$, $i=1,\ldots,n$. Consequently, $W(A)$ consists of convex combinations of the eigenvalues of $A$.