# 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?

• 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? Jul 5, 2022 at 7:56
• @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. Jul 5, 2022 at 7:59
• Nope, still incorrect. Jul 5, 2022 at 7:59
• @peek-a-boo Oh I think I got what you mean. I’ll give that a try. Jul 5, 2022 at 8:00
• @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. Jul 5, 2022 at 8:09

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

There is another way to tackle the problem. Notice that there is an $$nxn$$ matrix $$A$$ which is 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!

• 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? Jul 5, 2022 at 10:52
• You are right, I did not notice that the given vectors were not necessarily independent. I will make a correction!
– user1054388
Jul 5, 2022 at 11:14