# Linear Algebra - Spans

## Homework Statement

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

## 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 $\alpha_4$?

Mark44
Mentor

## Homework Statement

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

## 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 $\alpha_4$?
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.

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.

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

Is this the correct start?

thanks

Mark44
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.

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

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

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

Not sure what to do...?

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

Mark44
Mentor
Not sure what to do...?

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

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

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

Mark44
Mentor
Work with the right side of the equation in post #6.

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

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

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

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

Mark44
Mentor
$$V=\left \{ v_1,v_2,v_3,v_4 \right \}=b_1(\alpha_1-\alpha_2)+b_2(\alpha_2-\alpha_4)+b_3(\alpha_3-\alpha_1)+b_4(\alpha_4)$$
$$=(\alpha_1-\alpha_2) b_1 +(\alpha_2-\alpha_4)b_2 + (\alpha_3-\alpha_1)b_3+ (\alpha_4)b_4$$

Doesn't this show that v (not V) is a linear combination of b1, b2, b3, and b4?
$$U=\left \{ u_1,u_2,u_3,u_4 \right \}=\alpha_1b_1+\alpha_2b_2+\alpha_3b_3+\alpha_4b_4$$

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

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

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

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?

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

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

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?!

Mark44
Mentor
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?!

From post #2
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.

Then u = ?

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

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:
Mark44
Mentor
OK, so now you have proved what you needed to per your post #1.

Thank you Mark,

I am slowly learning :-)

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

Mark44
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.

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?

Mark44
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.

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

Mark44
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.

Yes,

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

Mark44
Mentor
Ok, the last section

Is the span $$\left \{ b_1+b_2,b_2+b_3,b_3 \right \}=\left \{ b_1,b_2,b_3,b_4 \right \}$$
Better: Is Span{b1 + b2, b2 + b3, b3} = Span{b1, b2, b3, b4}?
Let $$v=\left \{ v_1,v_2,v_3,v_4 \right \}$$ span $$\left \{ b_1+b_2,b_2+b_3,b_3 \right \}$$
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.
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?

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