$F(A)$ is a subset of a straight line iff there is a Hermitian matrix $H$ and complex numbers $\alpha, \beta$ such that $A = \alpha H +\beta I$. Let $A$ be an $n \times n$ complex matrix. Show that $F(A)$ is a subset of a straight line if and only if there is a Hermitian matrix $H$ and complex numbers $\alpha, \beta$ such that $A = \alpha H +\beta I$.
Here $F(A)$ is the field of values of $A$ or numerical range of $A$.
For a Hermitian matrix $H$ we know that the numerical range is closed interval on a real line say $[a,b]$ where $a$ is the smallest eigenvalue and $b$ is the largest eigenvalue.
Also $F(A-\beta I) = F(A) - \beta.$
So wlog we can assume that the straight line passes through origin.
 A: Let
$$
\ell(z_1,z_2):=\{z\in\mathbb{C}:z=z_1+(z_2-z_1)t, \; t\in[0,1]\}
$$
be a line segment connecting two complex numbers $z_1$ and $z_2$ in the complex plane.
We need two facts:

Fact 1 There are real $\xi_1$ and $\xi_2$ such that $F(H)=\ell(\xi_1,\xi_2)$ if and only if $H$ is Hermitian.

This is directly related to the fact that $x^*Hx\in\mathbb{R}$ for all $x$ if and only if $H=H^*$. As you know $\xi_1$ and $\xi_2$ are the extremal eigenvalues of $H$.

Fact 2 If $\alpha,\beta\in\mathbb{C}$, then $F(\alpha A+\beta I)=\alpha F(A)+\beta$.

This is simple to show using the definition of $F$.
Now if $A=\alpha H+\beta I$ for complex $\alpha$ and $\beta$ and Hermitian $H$, we have
$$
F(A)=F(\alpha H+\beta I)=\alpha F(H)+\beta=\alpha\ell(\lambda_\min,\lambda_\max)+\beta
=\ell(\alpha\lambda_\min+\beta,\alpha\lambda_\max+\beta).
$$
On the other hand, if $F(A)=\ell(z_1,z_2)$, $z_1\neq z_2$, then
$$
F(A)=\ell(z_1,z_2)=\alpha\ell(0,1)+\beta, \quad \alpha:=z_2-z_1,\quad\beta:=z_1.
$$
So
$$
\ell(0,1)=\frac{1}{\alpha}[F(A)-\beta]=\frac{1}{\alpha}F(A-\beta I).
$$
Fact 1 gives that
$$
H:=\frac{1}{\alpha}(A-\beta I)
$$
is Hermitian. So $A=\alpha H + \beta I$.
