Let $A\in\mathbb{C}^{m\times m}$ be an arbitrary matrix, then it follows from the Hausdorff-Toeplitz theorem that the numerical range of $A$, $W(A)$, contains the convex hull of the spectrum of $A.$ However, I was going trying to come up with an explicit proof in the case where the eigenvalues are distinct and came to the following.
Say $\text{Spec}(A) = \{\lambda_1, \lambda_2,...\lambda_m\}$ and consider a convex combination $s = \sum_{i=1}^ma_i\lambda_i$ with $\sum_{i=1}^ma_i = 1,\quad a_i\geq 0.$ Then the theorem says that there exists $y = (b_1,b_2,..,b_m)^T\in\mathbb{C}^{m}$ such that $s = \sum_{i=1}^ma_i\lambda_i = x^TAx = \sum_{i,j=1}^m\lambda_i(b_ib_jx^T_ix_j),$ where $x = \sum_{i=1}^mb_ix_i$ and $x_i$ is an eigenvector for $\lambda_i.$ So this amounts to solving the equation: $$\text{diag}(a_1, a_2, ...,a_m) = CB\quad(\dagger)$$ where $C$ is the hermitian matrix $C = (x_i^Tx_j)$ and $B = yy^T$ is the symmetric, rank -1 matrix that is the outer product of $y$ with itself and $||x||_2 = 1 = \sum\limits_{i,j=1}^mb_ib_jx_i^Tx_j.$
Now, I know that the equation $(\dagger)$ is solvable because of the Hausdorff-Toeplitz theorem. But this precisely means the following:
If $a_i$ are just arbitrary non-negative numbers that add up to $1,$ and $C$ is a Hermitian matrix that's obtained by using an arbitrary basis of $\mathbb{C}^{m}$ by setting $(C)_{ij} = x^T_ix_j$, then $B= C^{-1}\text{diag}(a_1,a_2,...a_m)$ is rank $1$ matrix that can be written as $B = yy^T$, an outer product of a vector.
So, my question is if there is a way to show the above statement directly or have I made the mistake somewhere and came to a wrong conclusion?