Condition for mapping linearly independent vectors to linearly independent vectors Let $f \colon U \to V$ be a linear map and $\{u_1, u_2, \ldots, u_n\}$ be the set of linearly independent vectors in $U$.
Then the set $\{f(u_1), f(u_2),\ldots,f(u_n)\}$ is linearly independent iff
a. $f$ is one-one & onto
b. $f$ is one-one
c. $f$ is onto
d. $U = V$
I came to conclusion that $f$ has to be one-one.
This is what I did. 
We have $$a_1 u_1 + a_2 u_2+ \ldots + a_n u_n = 0.$$
For $\{f(u_1), f(u_2),\ldots,f(u_n)\}$ to be independent, we have to have $$a_1 f(u_1) + a_2 f(u_2) + \cdots + a_n f(u_n) = 0.$$
Simplifying we get $$f(a_1 u_1 + a_2 u_2 + \cdots + a_n u_n) = 0$$
i.e. $f(0)=0$. 
This implies that $\ker f = \{0\}$ i.e. $f$ has to be one-one.
Is my logic correct? Also what's the correct answer and why?
 A: In the light of @Gerry's comment, take the following linear transformation: $$T:\mathbb R^3\to\mathbb R^2$$ $$T(a_1,a_2,a_3)=(a_1-a_2,2a_3)$$ $N(T)=\{(a,a,0)\mid a\in\mathbb R\}$ and $R(T)=\mathbb R^2$. So a,b,d is wrong. Search for another for c.
A: $T:V\rightarrow W$ be a linear map  such that $KerT=\{0\}$. Then  $T$ maps any basis of $V$ to a basis of $W$. $V,W$ be finite dimensional vector spaces over the field $F$, $\dim W\le \dim V$
A: OP has clarified that one wants a proof that if $f$ is linear and one-one then the image of a linearly independent set is linearly independent. 
So, assume $u_1,\dots,u_n$ linearly independent, assume $f$ linear and one-one, and assume $$a_1f(u_1)+\cdots+a_nf(u_n)=0\tag1$$ By linearity of $f$, we have $$f(a_1u_1+\cdots+a_nu_n)=0\tag2$$ Every linear map $f$ satisfies $$f(0)=0\tag3$$ Since $f$ is assumed one-one, (2) and (3) imply $$a_1u_1+\cdots+a_nu_n=0\tag4$$ Since $u_1,\dots,u_n$ are assumed linearly independent, (4) implies $$a_1=\cdots=a_n=0\tag5$$ So we have proved that under the hypotheses, (1) implies (5). But that's precisely the definition of linear independence of $f(u_1),\dots,f(u_n)$. 
