Let $S = \{y_1a_1+y_2a_2~|~-1 \leq y_1,y_2 \leq 1\}$ where $a_1 , a_2 \in \mathbb{R}^2$. Show $S$ is a polyhedron. Assume $a_1,a_2$ are linearly independent.
Now I believe this question was asked before, but Please check if my solution is ok. By set theory chasing, $$S = \{x \in \mathbb{R}^2~|~x = y_1a_1+y_2a_2, -1 \leq \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} \leq 1 \}$$
Write $A = (a_1 ~~~ a_2)$, then $A$ is invertible. hence we can write $S$ as $$S = \{x \in \mathbb{R}^2~|~x = A\begin{bmatrix} y_1 \\ y_2 \end{bmatrix}, -1 \leq \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} \leq 1 \}$$ $$S = \{x \in \mathbb{R}^2~|~A^{-1}x = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix}, -1 \leq \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} \leq 1 \}$$ $$S = \{x \in \mathbb{R}^2~|~-1 \leq A^{-1}x \leq 1\}$$
This is an intersection of 2 polyhedron, hence it is a polyhedron.