Proof that if $v_1,. . .v_n$ spans $V$, then $v_1 -v_2,. . .,v_{n-1} - v_n, v_n$ also spans $V$. I saw the following problem in Linear Algebra Done Right and thought if the general version of it would hold.
Problem: Suppose that $v_1,.  .  .,v_4$. spans $V$. Prove that $v_1 - v_2,v_2-v_3,v_3-v_4,v_4$ also spans $V$.
I would like if someone could confirm if the following proof is correct.
Q: Suppose that $v_1,.  .  .,v_n$ spans $V$. Prove that $v_1-v_2,v_2-v_3,.  .  .,v_{n-1}-v_n,v_n$ also spans $V$ for $n\geq 2$.
Proof: It suffices to show that there exists scalars such that $a_1(v_1-v_2)+.  .  .+a_{n-1}(v_{n-1}-v_n)+a_nv_n = c_1v_1+.  .  .+c_nv_n$.
Let
$a_1 = c_1$
$a_2 =c_1+c_2$
.
.
.
$a_{n-1}=c_1+.  .  .+c_{n-1}$
$a_n=c_1+. . .+c_n$
Then, $a_1(v_1-v_2)+.  .  .+a_{n-1}(v_{n-1}-v_n)+a_nv_n=c_1v_1+.  .  .+c_nv_n$.
Proving the desired result.
Is this reasoning correct?
 A: Let me denote the span of a set $S$ by $L(S)$ . So let us denote the set $\{v_{1}-v_{2},...,v_{n-1}-v_{n},v_{n}\}=T$ and $S=\{v_{1},...,v_{n}\}$
Two equivalent definitons of span:-
1:- The span of a set $S$ is the set of all finite linear combinations of the elements of $S$ .
2:- The span of a set $S$ is the smallest subspace $L(S)$ of $V$ such that $S\subseteq L(S)$ . By smallest we mean the intersection of all subspaces of $V$ that contain $S$.
Now notice that $(v_{n-1}-v_{n})+(v_{n})=v_{n-1}\in L(T)$
Then again $v_{n-2}-v_{n-1}+v_{n-1}=v_{n-2}\in L(T)$ .
Use a simple induction to prove that $\{v_{1},...,v_{n}\}\in L(T)$
This directly gives you that $L(\{v_{1},...,v_{n}\})\subseteq L(L(T))=L(T)$
Also $T\subset L(S)\implies L(T)\subseteq L(L(S))=L(S)$
Thus $L(T)=L(S)$ .
A: There is another way to tackle the problem. Notice that there is an $nxn$ matrix $A$ which is 
which satisfies

It is easy to notice that $A$ is invertible and therefore any
linear combination of the one side can be written as a linear
combination of the other side and vice versa. That is if a vector is a linear combination of v's is

which is a linear combination of $v_{i-1}-v_{i}$'s and we can do the same the inverse way, and that proves the result!
