Question

Suppose v₁v₂..v_mis linearly independent in V and w element of V. Show that v_1,v_2,...v_m;w is linearly independent if and only if w = not element of span(v_1,v_2,...v_m).

Answer 1

Let $v_1,v_2,...,v_m,$ w is linearly dependent and let

$w\in span(v_1,v_2,...v_m)$

$w=\alpha_1v_1+\alpha_2v_2+....+\alpha_mv_m=0,$ Where at least one of $\alpha_i\neq 0.$

$\Rightarrow \alpha_1v_1+\alpha_2v_2+....+\alpha_mv_m-w=0$

$\Rightarrow \alpha_1v_1+\alpha_2v_2+....+\alpha_mv_m+\alpha_{m+1}w=0,\text{ Where }\alpha_{m+1}=-1\neq 0$

So, ($v_1,v_2,...v_m$ ), w are linearly independent which is contradiction.

Hence, $w\in span((v_1,v_2,...v_m)$

Conversely: Let $w \in span(v_1,v_2,..,v_m)$

TO show: $x_1,x_2...,x_m , w$ is linearly independent

then, $\alpha_1,\alpha_2,.....,\alpha_{m+1}\neq 0$

Such that,

 $\alpha_1v_1+\alpha_2v_2+....+\alpha_mv_m+\alpha{m+1}w=0$

 say $\alpha_{m+1}\neq 0.$

$\alpha_{m+1} w=-\alpha_1v_1-\alpha_2v_2.... -\alpha_mv_m$

$w=-\alpha_1\alpha^{-1}_{m+1}v_1-\alpha_2\alpha^{-1}_{m+1}2v_2.... -\alpha_m\alpha^{-1}_{m+1}v_m$

$w\in span(v_1,v_2,...,v_m),$ Which is contradiction.

Hence , $v_1,v_2,...,v_m , w$ is linearly independent.