Subsets of finite sets of linearly independent vectors are linearly independent Larson Edwards Falvo - Elementary Linear Algebra





I don't understand the part in red. Please explain what exactly is the contradiction here.
(WOLOG?) S can be $v_1, v_2, ..., v_k, v_{k+1}$ and so $\sum_{i=1}^{k+1} c_iv_i = 0 \to c_i = 0$
So what if $\sum_{i=1}^{k} c_iv_i = 0, c_j \ne 0$?
WOLOG, suppose it is $c_1$. Then $v_1$ can be written as a linear combination of $v_2, ..., v_k$. Then I have no idea what to do.
 A: Suppose $T=\{\mathbf{v}_1,\mathbf{v}_2\} \subset \{\mathbf{v}_1,\mathbf{v}_2, \mathbf{v}_3\} = S.$
If $T$ is a linearly dependent set of vectors, then there are scalars $c_1,c_2$ such that $c_1\mathbf{v}_1 + c_2\mathbf{v}_2 = \mathbf 0$ and $c_1,c_2$ are not both $0$.
Therefore there are scalars $c_1,c_2,c_3$ (where $c_3=0$) such that $c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{c}_3 = \mathbf 0$ (the two scalars $c_1,c_2$ are still the same scalars they were above).
Therefore $S$ would not be linearly independent.
A: Let $T=\{v_1,\dots,v_k\}$ and $S=\{v_1,\dots,v_k,v_{k+1},\dots,v_n\}$; suppose $T$ is linearly dependent. If $c_1,\dots,c_k$ are scalars, not all zero, such that $c_1v_1+\dots+c_kv_k=0$, set $c_{k+1}=\dots=c_n=0$; then $c_1,\dots,c_n$ are scalars, not all zero, such that
$$
c_1v_1+\dots+c_kv_k+c_{k+1}v_{k+1}+\dots+c_nv_n=0
$$
and therefore $S$ is linearly dependent.
(This is a proof by contrapositive, rather than by contradiction.)
Alternative way: suppose $S$ is linearly independent and suppose $c_1v_1+\dots+c_kv_k=0$; set $c_{k+1}=\dots=c_n=0$ so that
$$
c_1v_1+\dots+c_kv_k+c_{k+1}v_{k+1}+\dots+c_nv_n=0
$$
By linear independence of $S$ you get $c_1=\dots=c_k=0$.

The “without loss of generality” can be justified by the fact that $T$ can be obtained from $S$ by successively removing one element; thus you just need to prove the statement when $T$ is obtained from $S$ by removing one element. However, I don't think this makes the proof clearer: the above one is just applying the definitions. I find the book's suggestions at least misleading.
