What is the Corollary of the Nucleus and Image Theorem?

[Portuga]

TL;DR

Consider $u, v \in \mathbb{R}^2$ such that $\left\{ u,v \right\}$ is a $\mathbb{R}^2$ basis. If $F: \mathbb{R}^2 \rightarrow \mathbb{R}^n$ a linear transformation, show that one of the following alternatives verifies:

(a) $\left\{ F(u), F(v) \right\}$ is linearly independent. (b) $\dim \text{Im} (F) = 1$ (c) $\text{Im} (F) = {o}$

I tried hard to understand what this author proposed, but I feel like I failed miserably. My attempt of solution is here:
Item (a) is verified in the case where $n = 2$, since $F$ being a linear transformation, by the Corollary of the Nucleus and Image Theorem, $F$ takes a basis of $\mathbb{R}^2$ to a $\mathbb{R}^2$ basis. Thus, $\left \{ F (u), F (v) \right\}$ is a basis of $\mathbb{R}^2$.
Item (b) occurs in the case where $n = 1$, because $\dim \mathbb{R}^2 = 2, \dim \mathbb{R} = 1$, and by the Nucleus and Image theorem,
$$
\begin{align*}

& \dim\mathbb{R}^{2}=\dim\ker\left(F\right)+\dim\mathbb{R}\\

\Rightarrow & 2=\dim\ker\left(F\right)+1\\

\Rightarrow & \dim\ker\left(F\right)=2-1=1.

\end{align*}
$$
Item (c) is verified in case $n = 0$.
I am pretty sure that I am very far away of the author's intention with this exercise, so please, any help would be appreciated.

 

[pasmith]

The problem is not to find an n for which each case is possible, but to show that each case is possible for every n \geq 2 (With n = 1 the first is impossible, since there is at most one linearly independent vector in \mathbb{R}^1.)

There are clearly two possibilities for n \geq 2: Either F(u) and F(v) are linearly independent or they are not. If they are, we have case (1).

So suppose they are not linearly independent. Then there exist non-zero scalars A and B such that 0 = AF(u) + BF(v). What can you say about Au + Bv in this case?

The posbbility that F(u) = F(v) = 0 is not excluded.

 

[Portuga]

Thank you very much! Now it's clear!