How would one prove (or disprove) that there is no such graph $G$ with $\lambda$ as an eigenvalue?
I tried setting up a system of equations to see if it's possible for $-\frac{1}{2}$ to be an eigenvalue for some $G$.
Trying to solve $Av=\lambda v$ where $A$ is the adjacency matrix
$0 +x_{12}v_2 + \ldots + x_{1n}v_n = -\frac{v_1}{2}$
$\vdots$
$x_{n1}v_1 + x_{n2}v^2 + \ldots + 0 = -\frac{v_n}{2}$
But I'm not sure what to do from here (or if there's a better way to do this). Any advice would be appreciated!