I know this question has answers here but they involve linear transformations which I haven't reached upto yet. There was a proof by contradiction there as well but I had a different idea in mind:
Let $S=\{v_1,v_2,...,v_n\}$ where $v$ is a vector. Assume that $T$ is linearly dependent. So there exists scalars $t_1,t_2,...,t_m$, not all zero such that the relation of linear dependence on $T$ becomes $$t_1v_1+t_2v_2+...+t_mv_m=0 \tag{1}$$ where $m<n$ (i.e. choose any number and combination of vectors from $S$ but not all vectors from $S$). Now add $0\times\sum v_i$, where $v_i$ denotes those vectors from $S$ not chosen in $(1)$, to $(1).$ Hence, $$t_1v_1+t_2v_2+...+t_mv_m+0\sum v_i=0 \tag{2}$$ But this is the relation of linear dependence on $S$ which suggests that there exists scalars not all zero such that $(2)$ holds true. This contradicts our assumption that $S$ was linearly independent. Hence $T$ is linearly independent. $\Box$
I am self-teaching Linear Algebra and theres no solution for this problem in my book, so I'm looking to see if my english/correctness of this proof flows smoothly.
Is this proof of contradiction done correctly? Any flaws or is it fine?