Question

Suppose v1,v2, vm is linearly independent in V and w ∈ V . Prove that dim span(v+w,v2+w,..vm+w)≥ m-1

Answer 1

Let {$v_1,v_2,...,v_m$ } is linearly independent in V and $w\in W.$

Suppose $(v_1-w,v_2-w,...,v_m-w)$ is Linearly dependent. Then there esists scalars $a_1,a_2,...,a_m$ , not all zero such that

 $a_1(v_1-w),a_2(v_2-w)..,a_m(v_m-w)$

Rearranging the equation-

$a_1v_1+...+a_mv_m=(a_1+...+a_m)w$

If $a_1+...+a_m$ were zero, then the equation above would contradict the linear independence of {$v_1,v_2,...,v_m$ }. Thus $a_1+....+a_m\neq 0.$

Thus divide both sides of the equation by $(a_1+...+a_m)$ showing that $w\in span(v_1,v_2,..,v_m)$

$\Rightarrow \text{dim span}(v_1-w,v_2-w,...,v_m-w)\ge m-1.$