Show that any set of vectors containing a linearly dependent subset is again linearly
dependent.
I think you're supposed to show this by contradiction, but not sure how.
I tried:
Let V be a linearly independent set of vectors {$v_1,…v_n$} such that $a_1v_1+…+a_nv_n$=0 and $a_1=…=a_n=0$. Then, V is a subset of {$v_1,…,v_{n+1}$} and $a_1v_1+…+a_{n+1}v_{n+1}=0$ where $a_1=…=a_{n+1}=0$. Therefore, the set {$v_1,…,v_{n+1}$} is again linearly independent.
Best Answer
Suppose $\{x_1,\ldots, x_m\}$ is dependent. Then there are coefficients $a_1,\ldots, a_m$, not all zero, such that $$a_1x_1+\cdots+a_mx_m=0$$ Now, let $n\ge m$, and consider the set $\{x_1,\ldots, x_m,x_{m+1},\ldots, x_n\}$. Now, take $b_1=a_1, b_2=a_2,\ldots, b_m=a_m$, and $b_{m+1}=b_{m+2}=\cdots=b_n=0$. Since the $a_i$'s were not all zero, the $b_i$'s are not all zero either. We have $$b_1x_1+\cdots+b_nx_n=a_1x_1+\cdots a_mx_m=0$$ Hence $\{x_1,\ldots,x_n\}$ is dependent as well.