Linear transformation problem

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Portuga
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TL;DR Summary
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.
 
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  • #2
pasmith
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The problem is not to find an [itex]n[/itex] for which each case is possible, but to show that each case is possible for every [itex]n \geq 2[/itex] (With [itex]n = 1[/itex] the first is impossible, since there is at most one linearly independent vector in [itex]\mathbb{R}^1[/itex].)

There are clearly two possibilities for [itex]n \geq 2[/itex]: Either [itex]F(u)[/itex] and [itex]F(v)[/itex] 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 [itex]A[/itex] and [itex]B[/itex] such that [tex]0 = AF(u) + BF(v)[/tex]. What can you say about [itex]Au + Bv[/itex] in this case?

The posbbility that [itex]F(u) = F(v) = 0[/itex] is not excluded.
 
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  • #3
Portuga
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Thank you very much! Now it's clear!
 

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