Linear Algebra Problem 
$A$,$B$ are two vectors forming a basis in $\mathbb{R}^2$ and
  $F:\mathbb{R}^2\rightarrow \mathbb{R}^n$ a linear map. Show that
  either $F(A)$ and $F(B)$ are linearly independent, or the image of $F$
  has dimension $1$, or the image of $F$ is $\{0\}$

I have done two of the three cases, I cannot find a way for the second case. 
Let $v\in \mathbb{R}^2$ be $aA + b B$ with $a,b$ reals. Then the image of any $v$ is $$F(aA+bB) = aF(A) + bF(b) .$$
Assume $F(A)$ and $F(B)$ are not zero vectors. Then $aF(A) + bF(b)=0$ iff $a=b=0$ as $aA + b B=0$ iff $a=b=0$ and that $F(0)=0$.
If $F(A) = F(b)=0$ then any $v = aA + b B \implies F(v) =0 $  and the image of $F$ would be $\{0\}.$
However, I am unable to prove the third case, i.e its image would have dimension $1$ if not above cases. 
This is a problem from Lang's Linear Algebra, which I am trying to learn over the weekend so don't have any other help for trivial doubts.
 A: Your reasoning is incorrect.
You are correct that if $a$ and $b$ are zero, then $aF(A)+bF(B)=0$. However, it is not true that if $aF(A)+bF(B)=0$ then you must have $a=b=0$. The error lies in that even though you are right that $aF(A)+bF(B)=F(aA+bB)$, you have no warrant for assuming that just because the image is equal to $0$ then it follows that the vector $aA+bB$ must be zero. This is not true in general.
Instead, notice that the image of $F$ is spanned by $F(A)$ and $F(B)$. Then consider the following two possibilities:


*

*Either $F(A)$ and $F(B)$ are linearly independent;

*or else $F(A)$ and $F(B)$ are linearly dependent.


If $F(A)$ and $F(B)$ are linearly independent, there is nothing to do.
If $F(A)$ and $F(B)$ are linearly dependent, and both are zero, then there is nothing to do: the image is $0$.
And if they are linearly dependent and at least one is nonzero, then either $F(B)$ is a scalar multiple of $F(A)\neq 0$, in which case you should show that the image of $F$ consists of all multiples of $F(A)$ and nothing more (so it is one dimensional); or else $F(A)=0$ and $F(B)\neq 0$, in which case you should show that the image of $F$ consists of all multiples of $F(B)$, and so is one dimensional.
A: If $F(A)$ and $F(B)$ are not linearly independent, there are scalars $c_1, c_2$ with $c_1 F(A)+c_2 F(B) = 0$ and $c_1 \ne 0$ or $c_2 \ne 0$, and by switching $A$ and $B$ we may assume without loss of generality that $c_1 \ne 0$.
You can then write $F(A)$ in terms of $F(B)$ (do you see how?) and using that, write an arbitrary linear combination of $F(A)$ and $F(B)$ in terms of just $F(B)$. But if $A$ and $B$ are a basis for $\mathbb{R}^2$, then any vector in $\mathbb{R}^2$ can be written as a linear combination of $A$ and $B$, and the image of that under $F$ will be a linear combination of $F(A)$ and $F(B)$, which by the above is just some scalar multiplied by $F(B)$. The two cases to consider are $F(B)=0$ and $F(B) \ne 0$, the latter of which has dimension 1 image and the former of which has image {0}.
There are quite a few details to be filled in, so hopefully I haven't spoiled it for you.
A: SINCE A,B FORM A BASIS OF R^2, THEIR IMAGES SPAN THE RANGE OF F.
THAT IS 
RANGE OF F=SPAN{F(A), F(B)} . CLEARLY DIM RANGE OF F IS EITHER 0,1 OR 2.
