# Linear transformation problem

• A
Portuga
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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Homework Helper
2022 Award
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.

• • jedishrfu and Portuga
Portuga
Thank you very much! Now it's clear!

• jedishrfu