Proving a list is linearly independent.

Question

Section 2.A #6 Linear Algebra Done Right.

Suppose v1, v2, v3, v4 is linearly independent in V.
Prove that the list: v1-v2, v2-v3, v3-v4, v4 is also linearly independent.

My answer is different from the one I have found on the internet and I just want to make sure that it is valid.

Basically, I just show that because v1, v2, v3, v4 can be written as a linear combination of v1-v2, v2-v3, v3-v4, v4 it is thus in it's span. Thus the span(v1, v2, v3, v4) is a subset of the span(v1-v2, v2-v3, v3-v4, v4)

I then do the converse.

Finally, because each span is a subset of one and the other, they are thus equal. By transition, v1-v2, v2-v3, v3-v4, v4 is also linearly independent.

Could that have worked as a valid answer as well?

Thank you

Answer 1

You don't need to show the converse: since $\operatorname{span}\{v_1-v_2, v_2-v_3, v_3-v_4, v_4\}\subseteq\operatorname{span}\{v_1, v_2, v_3, v_4\}$, once you have shown the converse inclusion, you have a generating set with the same cardinality as a basis, hence a basis itself.

A different way to prove the statement is by considering the matrix $$ \begin{bmatrix} 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 1 \end{bmatrix} $$ which clearly has rank $4$ (a unitriangular matrix is invertible; or you can easily compute its RREF). Since the coordinate vectors of $v_1-v_2$, $v_2-v_3$, $v_3-v_4$ and $v_4$ form a linearly independent list, then also the vectors form a linearly independent list.

Answer 2

Since $v_1$,$v_2$,$v_3$ and $v_4$ are linearly independent, the two system are equivalent since they span the same subspace and there are 4 elements in both sets.

Thus also $v_1-v_2, v_2-v_3, v_3-v_4, v_4$ are linearly independent.

I think that you don't need to show the converse.