Showing the "boundaries" of linear mapping $\mathbb{R}^{2} \rightarrow \mathbb{R}^{n}$ From S.L Linear Algebra:

Let $A$, $B$ be two vectors in $\mathbb{R}^2$ forming a basis of
  $\mathbb{R}^2$. Let $F: \mathbb{R}^2 \rightarrow \mathbb{R}^n$  be a
  linear map. Show that either $F(A)$, $F(B)$ are linearly independent,
  or the image of $F$ has dimension $1$, or the image of $F$ is $\{O\}$.


I've taken following theorems from the book to try and construct the answer (proofs of theorems are omitted):

Theorem 3.1. Let $F: V \rightarrow W$ be a linear map whose kernel is $\{O\}$ . If $v ,
 ... ,v_n$ are linearly independent elements of $V$, then $F(v_1), ...
 ,F(v_n)$ are linearly independent elements of $W$.

Theorem 3.2. Let $V$ be a vector space. Let $L: V \rightarrow W$ be a linear map of $V$ into another space $W$. Let $n$ be the
  dimension of $V$, $q$ the dimension of the kernel of $L$, and $s$ the
  dimension of the image of $L$. Then $n = q + s$. In other words,
$$dim \, V=dim\, Ker \, L + dim\, Im \, L$$


This is how far I got:

My Progress: First, we consider that, if $s=dim \, Im \, F$ and $q=dim \, Ker \, F$, then $dim \, V=s+q$
  (see Theorem 3.2). Then, For $F(A)$, $F(B)$ to be linearly independent, kernel
  must be trivial (see Theorem 3.1), and hence a linear map must be injective.
  If linear map is not injective, mapping is only surjective and hence we can conclude that $n \in \{0, 1\} < 2$ in a linear mapping $\mathbb{R}^{2} \rightarrow \mathbb{R}^{n}$. 
Considering that $n=s+q$ ($dim \, V=dim \, Im \, F + dim \, Ker \, F$) and that $n \in \{0, 1\} < 2$, either both kernel and image are
  $0$-dimensional, or only one of them has a dimension of $1$ (and the other one is zero-dimensional).
Also, using Theorem 3.2, we can derive following equations for dimensions of vector space, image under $F$ and kernel under $F$ respectively:
   $$\textrm{(1)} \, dim \, V= dim \, Ker \, F + dim \, Im \, F$$
$$\textrm{(2)} \, dim \, Im \, F = dim \, V - dim \, Ker \, F$$
$$\textrm{(3)} \, dim \, Ker \, F = dim \, V - dim \, Im \, F$$
  where $dim \, V \in \{0, 1\}$. (dimension of vector space is either $1$ or $0$)

In conclusion:
If $dim \, V = 0$, both kernel and image are nildimensional (0 dimensional) scalars. 
If dimension of vector space is not $0$ and if $dim \, V = 1$, then $(dim \, Im \,F = 1 \land dim \, Ker \,F = 0) \,  \lor \, (dim \, Ker \,F = 1 \land dim \, Im \,F = 0$). (either dimension of the image of $F$ 
   is $1$ and dimension of kernel is $0$, or vice versa).
But this is still insufficient information to prove the assertion given above.

How can I use the information above to show that the image must either have dimension of $1$ or else the kernel must be trivial ($/{0/}$)?
It seems like, in this particular case, if kernel has dimension $1$ then it must be trivial. Is there any correlation?
Thank you!
 A: I think the issue is that you write $n=1$ or $n=0$. If I understood your notation correctly, then $n$ is the dimension of a given space - hence not a variable. However, the integers $s,q$ are not fixed in your setting as you should prove something exactly about these. I would also recommend to think about examples of spaces that has dimension zero. If need be, I can of course expand, but imho this might be a good first push for you to revisit your attempt.
A: First to clarify what exactly your question is asking. Your question asks us to consider a linear map from $\mathbb{R}^2$ to $\mathbb{R}^n$, for some integer $n$. I believe the question does not ask you to consider different values of $n$, but instead concerns how a linear map from $\mathbb{R}^2$ to some vector space might behave. 
Rephrasing, let $W$ be any vector space, and let $F : \mathbb{R}^2 \to W$ be a linear map. We would then like to show that either $F(A), F(B)$ are linearly independent, the image of $F$ is of dimension $1$, or $F$ is the zero map.
You are right to want to make extensive use of Theorem 3.2. It is a rather useful result, so I shall do the same. Since $F$ is a linear map from $\mathbb{R}^2$ to $\mathbb{R}^n$, $F$ is going to play the role of $L$, $\mathbb{R}^2$ will be $V$, and $\mathbb{R}^n$ will be $W$ in the theorem. The theorem then says that 
$$\dim V = \dim \ker L + \dim \text{im } L.$$
In our case, this means that 
\begin{align}
 \dim \mathbb{R}^2 &= \dim \ker F + \dim \text{im } F\\
\implies 2 &= \dim \ker F + \dim \text{im } F \text{ (The dimension of $\mathbb{R}^2 = 2$)}.
\end{align}
So no matter how $F$ acts on $A$ and $B$, it is always true that the sum of the dimension of its kernel and the dimension of its image will be $2$. We know that the dimension of a vector space must be an integer and cannot be negative, so this restricts our list of possible choices for $\dim \ker F$. Then $\dim \ker F = 0, 1, $ or $2$ (if $2$ is the sum of two positive numbers, what other choices do we have?).
Now we just see what happens to our linear map in each case. If $\dim \ker F = 0$, then what does $F$ look like? As you mentioned before, if $\ker F = \{0\}$, then $F(A), F(B)$ are linearly independent by Theorem 3.1 (what is the dimension of the image in this case?). 
If you answered the question above, then you will see that if $\dim \ker F = 1$, it follows that $\dim \text{im } F = 1$ as well (use one of your formulas from (1) to (3)). 
Lastly, we have the case where $\dim \ker F = 2$. Presumably, we know that $\ker F$ is a subspace of $\mathbb{R}^2$ and since $\dim \ker F = 2$, it follows that $\ker F = \mathbb{R}^2$. What does this say about $\text{im } F$? (Take a vector in $\mathbb{R}^2$ and see what $F$ does to it, also use one of your formulas from (1) to (3) like before).
