# Homework Help: Linear Algebra - Spans

1. Oct 24, 2011

### bugatti79

1. The problem statement, all variables and given/known data

Let V be a real vector space and {b_1,b_2,b_3,b_4} a linearly independent set of vectors in V

3. The attempt at a solution

Show that the span $(b_1,b_2,b_3,b_4)=span(b_1-b_3,b_2-b_1,b_3,b_4-b_2)$

If I equate the LHS and RHS as

$\alpha_1b_1=+\alpha_1b_1-\alpha_3b_3$ implies $\alpha_3=0$

$\alpha_2b_2=\alpha_2b_2-\alpha_1b_1$ implies $\alpha_1=0$

$\alpha_3b_3=\alpha_3b_3$ but $\alpha_3=0$

$\alpha_4b_4=\alpha_4b_4-\alpha_2b_2$ implies $\alpha_2=0$

This correct? What about $\alpha_4$?

2. Oct 24, 2011

### Staff: Mentor

No, this isn't correct. Let v be an arbitrary vector in $span(b_1-b_3,b_2-b_1,b_3,b_4-b_2)$, which means that v is some linear combination of these vectors. Show that this same vector is a linear combination of $(b_1,b_2,b_3,b_4)$.

Now go the other way. Let u be an arbitrary vector in $span(b_1, b_2, b_3, b_4)$. Show that u is also a linear combination of the vectors in the second set.

3. Oct 25, 2011

### bugatti79

$$(b_1,b_2,b_3,b_4)=\alpha_1\left ( b_1 - b_3 \right )+\alpha_2(b_2-b_1)+\alpha_3(b_3)+\alpha_4(b_4-b_2)$$

Is this the correct start?

thanks

4. Oct 25, 2011

### Staff: Mentor

No. Let v = <v1, v2, v3, v4> be an arbitrary vector in $span(b_1-b_3,b_2-b_1,b_3,b_4-b_2)$.

5. Oct 25, 2011

### bugatti79

Not sure what to do...?

$$(v_1,v_2,v_3,v_4)=\alpha_1\left ( b_1 - b_3 \right )+\alpha_2(b_2-b_1)+\alpha_3(b_3)+\alpha_4(b_4-b_2)$$

6. Oct 25, 2011

### Staff: Mentor

Show that v is a linear combination of $(b_1,b_2,b_3,b_4)$.

Now go the other way. Let u be an arbitrary vector in $span(b_1, b_2, b_3, b_4)$. Show that u is also a linear combination of the vectors in the second set.

7. Oct 25, 2011

### bugatti79

but how do I show this? I realise that a vector v is a linear combination of the vectors b_1, b_2, b_3 and b_4 if it can be expressed in the form

$$v=\alpha_1b_1 +\alpha_2b_2+\alpha_3b_3+\alpha_4b_4 +....+\alpha_nb_n$$

Dont know how to proceed.. thanks

8. Oct 25, 2011

### Staff: Mentor

Work with the right side of the equation in post #6.

9. Oct 26, 2011

### bugatti79

$$V=\left \{ v_1,v_2,v_3,v_4 \right \}=b_1(\alpha_1-\alpha_2)+b_2(\alpha_2-\alpha_4)+b_3(\alpha_3-\alpha_1)+b_4(\alpha_4)$$

$$U=\left \{ u_1,u_2,u_3,u_4 \right \}=\alpha_1b_1+\alpha_2b_2+\alpha_3b_3+\alpha_4b_4$$

I dont know how one would be a linear combination of the other? Thanks

10. Oct 26, 2011

### Staff: Mentor

$$=(\alpha_1-\alpha_2) b_1 +(\alpha_2-\alpha_4)b_2 + (\alpha_3-\alpha_1)b_3+ (\alpha_4)b_4$$

Doesn't this show that v (not V) is a linear combination of b1, b2, b3, and b4?

11. Oct 26, 2011

### bugatti79

Sorry I don get it..Arent we to show that

if $$v=(v_1,v_2...v_n)$$ and $$u=(u_1,u_2...u_n)$$ then span v = span u if and only if v is a linear combination of those in u and vice versa?

12. Oct 26, 2011

### bugatti79

OK, I think I see what is required. I now need to express the vectors b_1, b_2, b_3 and b_4 as linear combinations of $$(b_1-b_3), (b_2-b_1), (b_3)$$ and $$(b_4-b_2)$$

Hmmmm....how is that done?!

13. Oct 26, 2011

### Staff: Mentor

From post #2
Then u = ?

14. Oct 26, 2011

### bugatti79

$$u=\alpha_1(b_1-b_3)+\alpha_2(b_2-b_1)+\alpha_3b_3+\alpha_4(b_4-b_2)$$

I dont know why its coming up like this...I didnt use the strike tags at all......
Mod Note: Fixed LaTeX.

Last edited by a moderator: Oct 26, 2011
15. Oct 26, 2011

### Staff: Mentor

OK, so now you have proved what you needed to per your post #1.

16. Oct 26, 2011

### bugatti79

Thank you Mark,

I am slowly learning :-)

17. Oct 27, 2011

### bugatti79

Actually, should the vector u be represented by beta scalars and v by the alpha scalars?

18. Oct 27, 2011

### Staff: Mentor

I don't think that matters. What matters is that for a given vector will have a different set of coordinates for the two bases.

19. Oct 27, 2011

### bugatti79

Ok, the last section

Is the span $$\left \{ b_1+b_2,b_2+b_3,b_3 \right \}=\left \{ b_1,b_2,b_3,b_4 \right \}$$

Let $$v=\left \{ v_1,v_2,v_3,v_4 \right \}$$ span $$\left \{ b_1+b_2,b_2+b_3,b_3 \right \}$$

rearranging

$$v=\alpha_1b_1+(\alpha_1+\alpha_2)b_2+(\alpha_2+\alpha_3)b_3$$

Let $$u=\left \{ u_1,u_2,u_3,u_4 \right \}$$ span $$\left \{ b_1,b_2,b_3,b_4 \right \}$$

$$u=\alpha_1b_1+\alpha2b_2+\alpha_3b_3+alpha_4b_4$$

Hence v does not span u because v does not contain $$b_4$$
How do I state this correctly?

20. Oct 27, 2011

### Staff: Mentor

I don't know why you are still asking. You have already proved the two parts of this. The second part was in post #14.

21. Oct 27, 2011

### bugatti79

No I dont think so. Its another question, ie its a different span we are asked to check....?

22. Oct 27, 2011

### Staff: Mentor

OK, I didn't realize this was a different question.

Are b1, b2, b3, and b4 linearly independent?

If so, span{b1, b2, b3, and b4} couldn't possibly be the same as the span of the three vectors you listed on the left side. More specifically, the group of three vectors doesn't include b4.

23. Oct 27, 2011

### bugatti79

Yes,

THey are linearly independant. Ok, thanks for the clarification!!
bugatti

24. Oct 27, 2011

### Staff: Mentor

Better: Is Span{b1 + b2, b2 + b3, b3} = Span{b1, b2, b3, b4}?
Let v = <v1, v2, v3, v4>. What you have written makes no sense. A vector (v) doesn't span a set of other vectors. A set of vectors spans a subspace of some vector space.

If V is the vector space (or subspace) that is spanned by {b1 + b2, b2 + b3, b3}, then v can be written as some specific linear combination of these vectors.

Clearly there's going to be a problem, because your set of vectors doesn't include b4.

25. Oct 27, 2011

### bugatti79

ok, thanks for clarification of the nonsense I was writing :-)