Is $Ape_1+Aqe_2$ where A (3x3) matrix, considered as a linear combination of $e_1,e_2$ 

$$\alpha=-8$$

Eigenvectors: 
$$e_1 = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} \text{ and }
e_2 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$$

What  I did :
(i)
$x ∈ V \implies x$ of the form $pe_1 + qe_2$
$$Ax = A(pe_1 + qe_2) = p(Ae_1) + q(Ae_2)=Ap (e_1) + Aq (e_2)$$
This raised a question

Over here:
$$Ap (e_1) + Ap (e_2)$$
$Ap$ and $Aq$ aren't numbers , they are matrices, so it is valid to say that $Ax$ contains a linear combination of $e_1,e_2$ ? 
Are matrices treated as numbers over here?

This Thought has disturbed My Proof, please help.
 A: Let's talk about this question:

Let $A$ denote some $n \times n$ matrix.  Let $e_1,e_2$ denote two eigenvectors of $A$.
  Let $V$ denote the linear space spanned by $e_1$ and $e_2$.
Prove that, for any vector $x$ belonging to $V$, the vector $Ax$ also belongs to V.

I would prove the above as follows: note that $e_1,e_2$ are eigenvectors.  By the definition of an eigenvector, this means that there exist two scalars ("numbers", not matrices) $\lambda_1,\lambda_2$ such that
$$
A e_1 = \lambda_1 e_1, \quad A e_2 = \lambda_2 e_2
$$
We may write any element $x \in V$ in the form $x = a_1 e_1 + a_2 e_2$ (where $a_1,a_2$ are numbers, not vectors).  We note that for any such element $x$, we have
$$
Ax = A(a_1 e_1 + a_2 e_2) = a_1 A e_1 + a_2 A e_2 = a_1 \lambda_1 e_1 + a_2 \lambda_2 e_2 = (a_1 \lambda _1)e_1 + (a_2 \lambda _2) e_2
$$
Since $a_1\lambda_1$ and $a_2 \lambda_2$ are just numbers, we can conclude from the above equation that $Ax$ is an element of $V$.

Indeed, as you say, $Ape_1$ is not a scalar multiple of $e_1$.  So, what you had done was not sufficient to answer the question.
