A specific solution to Exercise 1 in Serge Lang Linear Algebra, Chapter 3 Section 3.

The answer given by the solution manual is somehow vague, therefore I wrote down my answer there for future reviewing.

Exercise 1. 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 $\{O\}$.

Solution:

Because $\dim V = \dim \ker F + \dim \text{im} F$, $\dim V = 2$.

1. If $F(A)$ and $F(B)$ are linearly independent, then $F(A)$ and $F(B)$ forming a basis of $\text{im} F$. The $\text{im}F = 2$ and $\ker F = 0$.

2. If $F(A)$ and $F(B)$ are linearly dependent, i.e. $F(A) = k_1 F(B)$. then $t_1 F(A) + t_2 F(B) = t_1 k_1 F(B) + F(B) = (t_2 + t_1 k_1) F(B)$.

2-1. If $t_1F(A) + t_2F(B) = O$, then $(t_2 + t_1 k_1) F(B) = O$.  Because $t_1 F(A) + t_2 F(B) = F(t_1 A + t_2 B)$ and $A,B$ are a basis of $\mathbb{R}^2$, therefore for all $ v \in \mathbb{R}^2$, $F(v) = O$, that is, $\text{im}F = 0$, i.e. $\ker F = 2$.

2-2. If $t_1F(A) + t_2F(B) \neq O$, then $(t_2 + t_1 k_1)F(B) \neq O$. Because $t_1 F(A) + t_2 F(B) = F(t_1 A + t_2 B) = (t_2 + t_1 k_1)F(B) \neq O$, therefore $F(B)$ is a basis of $\text{im}F$. That is, $\dim \text{im}F = 1$ and $\dim \ker F = 1$.