# Show that a given list spans V

• I
• Mayhem
In summary, the conversation discusses the use of the definition of span to prove that a given list of vectors span a vector space. It is shown that by rearranging the terms in the span, the list can indeed span the vector space as the coefficients can be arbitrary elements in the underlying field. However, it is important to consider cases where the new vectors may not span the vector space, as in the case where the list is linearly dependent. The argument for the coefficients should show that for any given set of coefficients, there exist other coefficients that can be used to form the original list, thus proving that the new list spans the vector space.

#### Mayhem

TL;DR Summary
Problem text:

Suppose ##v_1, v_2, v_3, v_4## spans ##V##. Prove that the list $$v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4$$ Also spans ##V##
Using the definition of the span, we obtain $$\mathrm{span}\left(v_1,v_2,v_3,v_4\right) = \left \{ a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4 : a_1, a_2, a_3, a_4 \in \mathbb{F}\right \}$$

Suppose $$\mathrm{span}\left ( v_1-v_2,v_2-v_3,v_3-v_4,v_4 \right ) = \left \{a_1(v_1-v_2) + a_2(v_2-v_3) + a_3(v_3-v_4) + a_4v_4 : a_1, a_2, a_3, a_4 \in \mathbb{F}\right \}$$

The terms can be rearranged as follows: $$\mathrm{span}\left ( v_1-v_2,v_2-v_3,v_3-v_4,v_4 \right ) = \left \{a_1v_1-a_1v_2 + a_2v_2-a_2v_3 + a_3v_3-a_3v_4 + a_4v_4 : a_1, a_2, a_3, a_4 \in \mathbb{F}\right \}$$ $$\Updownarrow$$ $$\mathrm{span}\left ( v_1-v_2,v_2-v_3,v_3-v_4,v_4 \right ) = \left \{a_1v_1-(a_1+a_2)v_2-(a_2+a_3)v_3-(a_3+a_4)v_4 : a_1, a_2, a_3, a_4 \in \mathbb{F}\right \}$$

Which shows that the list indeed spans ##V## since the coefficients are arbitrary elements of ##\mathbb{F}##.

Is this correct?

I think it's nearly fine, save the sign errors in the last line. I guess you might also need to justify briefly why the last set does indeed include every vector in ##V## i.e. why the coefficients ##a_1##, ##a_1 - a_2##, ##a_2 - a_3## and ##a_3 - a_4## can each take on any possible number in ##\mathbb{F}## for suitable choice of the numbers ##a_1, a_2, a_3, a_4 \in \mathbb{F}##. But that's pretty easy to do!

You might also think about it like, take any ##v \in V##, then you have for some ##\alpha##, ##\beta##, ##\gamma##, ##\delta## ##\in \mathbb{F}##\begin{align*} v &= \alpha v_1 + \beta v_2 + \gamma v_3 + \delta v_4 \\ &= \alpha (v_1 - v_2) + (\alpha + \beta)(v_2 - v_3) + (\alpha + \beta + \gamma)(v_3 - v_4) + (\alpha + \beta + \gamma + \delta)v_4 \end{align*}and because ##v## is arbitrary, by construction we showed that the new list also spans ##V##

Mayhem
etotheipi said:
I think it's nearly fine, save the sign errors in the last line. I guess you might also need to justify briefly why the last set does indeed include every vector in ##V## i.e. why the coefficients ##a_1##, ##a_1 - a_2##, ##a_2 - a_3## and ##a_3 - a_4## can each take on any possible number in ##\mathbb{F}## for suitable choice of the numbers ##a_1, a_2, a_3, a_4 \in \mathbb{F}##. But that's pretty easy to do!

You might also think about it like, take any ##v \in V##, then you have for some ##\alpha##, ##\beta##, ##\gamma##, ##\delta## ##\in \mathbb{F}##\begin{align*} v &= \alpha v_1 + \beta v_2 + \gamma v_3 + \delta v_4 \\ &= \alpha (v_1 - v_2) + (\alpha + \beta)(v_2 - v_3) + (\alpha + \beta + \gamma)(v_3 - v_4) + (\alpha + \beta + \gamma + \delta)v_4 \end{align*}and because ##v## is arbitrary, by construction we showed that the new list also spans ##V##
Yes, I should probably have factored in the minus sign.

I considered arguing for the coefficients but thought it was pretty trivial.

etotheipi
Mayhem said:
I considered arguing for the coefficients but thought it was pretty trivial.

You should give the argument. You might consider it trivial, but it's the point of the question.

Another approach is to check that ##v_1,v_2,v_3,v_4## are all in ##\text{Span}(v_1-v_2,v_2-v_3,v_3-v_4,v_4)## and hence any linear combination of them (which by assumption is an arbitrary vector in ##V##) is too.

Last edited:
Infrared said:
You should give the argument. You might consider it trivial, but it's the point of the question.

Another approach is to check that ##v_1,v_2,v_3,v_4## are all in ##\text{Span}(v_1,v_2,v_3,v_4)## and hence any linear combination of them (which by assumption is an arbitrary vector in ##V##) is too.
Is there more to it other than arguing why the "new" coefficients are also in F?

Mayhem said:
Is there more to it other than arguing why the "new" coefficients are also in F?
Yes.
For example, consider a similar problem: Prove or disprove that if the list ##v_1,v_2,v_3## spans ##V##, then so does the list ##v_1-v_2,v_2-v_3,v_3-v_1.## The structure of your argument would say that they do, but this isn't the case: these three new vectors are always linearly dependent (their sum is zero), so if ##v_1,v_2,v_3## are independent, these new vectors cannot span.

In terms of coefficients, in your approach you need to show that for any ##(b_1,b_2,b_3,b_4)\in F^4##, there exist ##a_1,...,a_4\in F## such that ##a_1=b_1, a_2-a_1=b_2##, etc.

Mayhem
Infrared said:
Yes.
For example, consider a similar problem: Prove or disprove that if the list ##v_1,v_2,v_3## spans ##V##, then so does the list ##v_1-v_2,v_2-v_3,v_3-v_1.## The structure of your argument would say that they do, but this isn't the case: these three new vectors are always linearly dependent (their sum is zero), so if ##v_1,v_2,v_3## are independent, these new vectors cannot span.

In terms of coefficients, in your approach you need to show that for any ##(b_1,b_2,b_3,b_4)\in F^4##, there exist ##a_1,...,a_4\in F## such that ##a_1=b_1, a_2-a_1=b_2##, etc.
It is less trivial than I thought then. Could you maybe help me in the right direction?

As I suggested above (with a corrected typo):

Can you show that ##v_1\in\text{Span}(v_1-v_2,v_2-v_3,v_3-v_4,v_4)##? And the same for ##v_2,v_3,v_4##? Do you see why this tells you that ##v_1-v_2,v_2-v_3,v_3-v_4,v_4## is a spanning list?

Infrared said:
As I suggested above (with a corrected typo):

Can you show that ##v_1\in\text{Span}(v_1-v_2,v_2-v_3,v_3-v_4,v_4)##? And the same for ##v_2,v_3,v_4##? Do you see why this tells you that ##v_1-v_2,v_2-v_3,v_3-v_4,v_4## is a spanning list?
Can it be done by simply showing that the given span, written as its linear combination, contains ##av_1## + other stuff?

I don't know what you mean by that. Can you post your attempt?

I can show that ##v_1 \in \text{span}(v_1 - v_2, v_2-v_3, v_3-v_4, v_4)## for appropriate ##a_1, a_2, a_3, a_4 \in \mathbb{F}## which satisfy ##v_1 = a_1(v_1-v_2) + a_2(v_2-v_3) + a_3(v_3-v_4) + a_4v_4##. Is this correct?

Since the RHS is nothing but ##\text{span}(v_1 - v_2, v_2-v_3, v_3-v_4, v_4)##, this would by extension show that ##(v_1,v_2,v_3,v_4) \in \text{span}(v_1 - v_2, v_2-v_3, v_3-v_4, v_4)## if verified for the other vectors.

Is this a step in the right direction?

Looking into it some more, if I can write every vector in ##V## as a linear combination of vectors in some set ##S## (in this case ##S = {v_1-v_2, v_2-v_3, v_3-v_4, v_4}##, then ##S## spans ##V##. So the question is that if I can write ##v_1, v_2, v_3, v_4## as linear combination of vectors in ##S##, then I have shown that ##S## spans ##V##.

Is this right?

Yes, that's a good approach (and what I suggested in post 4).

Infrared said:
Yes, that's a good approach (and what I suggested in post 4).
Ah.

So if ##S## spans ##V##, there must exist ##b_1, b_2, b_3, b_4 \in \mathbb{F}## which satisfy $$v_1 = b_1(v_1-v_2)+b_2(v_2-v_3)+b_3(v_3-v_4)+b_4v_4$$These can be found by expanding and refactoring the above as such: $$v_1 = b_1v_1+(b_2-b_1)v_2 + (b_3-b_2)v_3 +(b_4-b_3)v_4$$ Let ##c_1 = b_1, c_2 = b_2-b_1, c_3 =b_3-b_2, c_4 = b_4-b_3##, such that we obtain: $$v_1 = c_1v_1+c_2v_2 + c_3v_3 +c_4v_4$$ Then it should be easily to see that ##b_2 = b_1, b_3 = b_2, b_4 = b_3## and lastly ##b_1 = 1## give us ##v_1 = v_1## which is what we wanted. Then the last step whould be to choose ##b_1, b_2, b_3, b_4 \in \mathbb{F}## such that ##c_1, c_3, c_4 = 0## and ##c_2 = 1## for ##v_2##, and so on for the remaining vectors. Correct?