Let $F(A)=\{x^*Ax\mid x\in \mathbb{C}^n,\|x\|=1\}$ where $A\in M_n(\mathbb{C}).$ Describe $F(A)$ when $A$ is Hermitian. Let $F(A)=\{x^*Ax\mid x\in \mathbb{C}^n,\|x\|=1\}$ where $A\in M_n(\mathbb{C}).$ Describe $F(A)$ when $A$ is Hermitian.
I checked on Wikipedia and it is said that the $F(A)=[a,b]$ where $a$ is the smallest eigenvalue and $b$ is the largest eigenvalue. However, no proof is given to justify this statement. The author of my textbook simply states that it is a closed interval. 
I know that $F(A)$ contains its eigenvalues, I also know that it contains its diagonal entries, but I am not sure why the aforementioned property holds. Thinking in terms of topology, we want to show that $F(A)$ is bounded and closed but this seems to be hard in this case.
 A: You need compactness and connectedness: the sphere $S=\{ x\in \mathbb C^n : \|x\| = 1\}$ is connected and compact, while
$$f:S\to \mathbb R, \  \ x\mapsto x^* Ax$$ 
is continuous, so the image is connected and compact. 
A: You can also show that if $A \in M_n$ is normal, then $F(A)$ equals the convex hull its eigenvalues.
This uses the Spectral Theorem (i.e., any normal matrix is unitarily diagonalizable; $A = UDU^*$):
\begin{align}
F(A)
    &= \{{x^*Ax : x \in \mathbb C^n, ||x||_2=1}\} \\
    &= \{{x^*UDU^*x : x \in \mathbb C^n, ||x||_2=1 }\} \\
    &= \{y^*Dy : y\in\mathbb C^n, ||y||_2=1\} \tag{$*$} \\
    &= \left\{{\begin{bmatrix} \overline{y}_1 & \cdots & \overline{y}_n \end{bmatrix} \begin{bmatrix} \lambda_1 & & 0 \\ & \ddots & \\ 0 & & \lambda_n \end{bmatrix} \begin{bmatrix} y_1 \\ \vdots \\ y_n \end{bmatrix} : y\in\mathbb C^n, ||y||_2=1}\right\} \\
    &= \left\{{\sum_{i=1}^n |y_i|^2 \lambda_i : \sum_{i=1}^{n} |y_i|^2 = 1}\right\} \\
    &= \mathcal H(\sigma(A)) \,,
\end{align}
where $(*)$ follows because $U^*$ is unitary and therefore an isometry.
If $A$ is also Hermitian, then its eigenvalues $\lambda_1 \leq \cdots \leq \lambda_n$ are real, so $\mathcal H(\sigma(A)) = [\lambda_1, \lambda_n]$.
