linearly independent subset of a vector space

Question

Suppose ${w_1, w_2, w_3, w_4}$ is a linearly independent subset of a vector space W.

Prove that the set ${w_1 − w_2, w_2 − w_3, w_3 − w_4, w_4}$ is also linearly independent.

I have an intuition that this can be solved using an augmented matrix but I am not sure. Can someone please help me do it?

Answer 1

$${w_1, w_2, w_3, w_4} -Linear \ independent \ set $$ $$\Rightarrow {a_1,a_2,a_3,a_4} \in F_n - scalars$$ $$\sum_{i=1}^4 a_iw_i = a_1w_1+a_2w_2+a_3w_3+a_4w_4 = 0 \Rightarrow a_i = 0 (independence \ criteria)$$ $${w_1 − w_2, w_2 − w_3, w_3 − w_4, w_4}$$ $$Let \ {b_1,b_2,b_3,b_4} \in F_n - \ scalars $$ $$\Rightarrow b_1(w_1-w_2) + b_2(w2-w3)+b_3(w3-w4)+b_4(w4) = 0$$

$$\Rightarrow Calculate \ and \ get: b_1w_1 + (b_2-b_1)w_2 + (b_3-b_2)w_3 + (b_4-b_3)w_4 = 0$$ $$But \ we \ showed \ that: a_1w_1+a_2w_2+a_3w_3+a_4w_4 = 0$$ $$\Rightarrow b_1 = a_1 = 0 \Rightarrow b_1 = 0$$ $$\Rightarrow (b_2-b_1) = a_2 = 0 \Rightarrow b_2=b_1 = 0$$ $$\Rightarrow (b_3-b_2) = a_3 = 0 \Rightarrow b_3=b_2 = 0 \Rightarrow b_3 = 0$$ $$\Rightarrow (b_4-b_3) = a_4 = 0 \Rightarrow b_4=b_3 = 0 \Rightarrow b_4=0$$ So for $b_1(w_1-w_2) + b_2(w_2-w_3)+b_3(w_3-w_4)+b_4(w_4) = 0$ all the coefficients are $0$'s $\Rightarrow $ $${w_1 − w_2, w_2 − w_3, w_3 − w_4, w_4} - \ an \ independent \ set$$