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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?

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  • $\begingroup$ How are you quantifying $a_i$'s and $c_i$'s? As stated your proof is incomplete at best, incorrect at worst. A much simpler solution is to show that $v_{n-1}\in W:=\text{span}\{v_1-v_2,\dots, v_{n-1}-v_n,v_n\}$. Then, show $v_{n-2}\in W$, then show $v_{n-3}\in W$ and so on (a rigorous proof uses induction). Why does this complete the proof? $\endgroup$
    – peek-a-boo
    Jul 5, 2022 at 7:56
  • $\begingroup$ @peek-a-boo If we let $F$ denote the field or real or complex numbers. Then there exists scalars $a_1,. . .,a_n \in F$ such that for any $c_1,. . .,c_n \in F$, the claim holds. $\endgroup$
    – Seeker
    Jul 5, 2022 at 7:59
  • $\begingroup$ Nope, still incorrect. $\endgroup$
    – peek-a-boo
    Jul 5, 2022 at 7:59
  • $\begingroup$ @peek-a-boo Oh I think I got what you mean. I’ll give that a try. $\endgroup$
    – Seeker
    Jul 5, 2022 at 8:00
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    $\begingroup$ @Mateo OP is essentially asking a 'check my work' question, and in my experience these are better discussed in comments. Even if I write up my comments as an answer, OP would still end up asking the same follow-up questions for clarification, so I don't see any benefit to me prematurely writing up an answer. $\endgroup$
    – peek-a-boo
    Jul 5, 2022 at 8:09

2 Answers 2

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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 $L(T)=L(S)$ .

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There is another way to tackle the problem. Notice that there is an $nxn$ matrix $A$ which is enter image description here

which satisfies enter image description here

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

enter image description here

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!

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  • $\begingroup$ It isn't mentioned that $V$ is $n$-dimensional. Also, why bother showing independence (which here is the incorrect way to go anyway) when proving spanning is much simpler? $\endgroup$
    – peek-a-boo
    Jul 5, 2022 at 10:52
  • $\begingroup$ You are right, I did not notice that the given vectors were not necessarily independent. I will make a correction! $\endgroup$
    – user1054388
    Jul 5, 2022 at 11:14

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