Confused about small detail in rank-nullity theorem

[psie]

TL;DR

I'm reading about the rank-nullity theorem in Linear Algebra by Friedberg et al. At the end of the proof, they claim that the vectors that span the range of the linear transformation and are linearly independent are distinct. With their definition of linear dependence, I can't seem to resolve this.

Consider the rank-nullity theorem. We want to prove that for a linear transformation $\mathsf T:\mathsf V\to\mathsf W$, $$\operatorname{nullity}(\mathsf T)+\operatorname{rank}(\mathsf T)=\operatorname{dim}(\mathsf V).$$We have a basis $\{v_1,\ldots,v_k\}$ of the null space $\mathsf N(\mathsf T)$ and extend it to basis for $\mathsf V$, namely $\beta=\{v_1,\ldots,v_k,v_{k+1},\ldots,v_n\}$. The meat of the proof is to show $S=\{\mathsf T(v_{k+1}),\ldots,\mathsf T(v_{n})\}$ is a basis for the range $\mathsf R(\mathsf T)$. Here's the last bit of the proof, after the authors have already shown that $S$ spans $\mathsf R(\mathsf T)$.

Now we prove that $S$ is linearly independent. Suppose that $$\sum_{i=k+1}^nb_i\mathsf T(v_i)=0\quad\text{for }b_{k+1},\ldots,b_n\in F.$$Using the fact that $\mathsf T$ is linear, we have $$\mathsf T\left(\sum_{i=k+1}^nb_iv_i\right)=0.$$So $\sum_{i=k+1}^nb_iv_i\in\mathsf N(\mathsf T)$. Hence there exist $c_1,\ldots,c_k\in F$ such that $$\sum_{i=k+1}^nb_iv_i=\sum_{i=1}^kc_iv_i\quad\text{or}\quad \sum_{i=1}^k(-c_i)v_i+\sum_{i=k+1}^nb_iv_i=0.$$Since $\beta$ is a basis for $\mathsf V$, we have $b_i=0$ for all $i$. Hence $S$ is linearly independent. Notice that this argument also shows that $\mathsf T(v_{k+1}),\ldots,\mathsf T(v_n)$ are distinct; therefore $\operatorname{rank}(\mathsf T)=n-k$.

Here's the definition of linear dependence.

Definition. A subset $S$ of a vector space $\mathsf V$ is called linearly dependent if there exist a finite number of distinct vectors $u_1,u_2,\ldots,u_n$ in $S$ and scalars $a_1,a_2,\ldots,a_n$, not all zero, such that $$a_1u_1+a_2u_2+\cdots+a_nu_n=0.$$In this case we also say that the vectors of $S$ are linearly dependent.

I don't understand in the proof how they claim that $\mathsf T(v_{k+1}),\ldots,\mathsf T(v_n)$ are distinct. Suppose on the contrary they are not and for some $i,j$, $\mathsf T(v_{i})=\mathsf T(v_{j})$. We want to contradict linear independence, i.e. show linear dependence. The problem is that linear dependence in the book is defined to be a nontrivial representation of the zero vector for distinct vectors in $S$. It seems like I can circumvent showing linear dependence of $S$ by choosing distinct vectors in $S$. I am perplexed.

 

[psie]

I figured it out with some help. Use $\mathsf T(v_{i})=\mathsf T(v_{j})$ and notice that $v_i-v_j$ is thus in the null space of $\mathsf T$. Derive a contradiction for the set $\beta$.