Proving Linear Independence in Real Vector Spaces

[bugatti79]

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]

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.

 

[bugatti79]

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]

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 $span(b_1-b_3,b_2-b_1,b_3,b_4-b_2)$.

 

[bugatti79]

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]

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.

 

[bugatti79]

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

but how do I show this? I realize 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]

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

 

[bugatti79]

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 don't know how one would be a linear combination of the other? Thanks

 

[Mark44]

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

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

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 don't know how one would be a linear combination of the other? Thanks

 

[bugatti79]

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

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?

 

[bugatti79]

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

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]

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 = ?

 

[bugatti79]

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

I don't know why its coming up like this...I didnt use the strike tags at all...
Mod Note: Fixed LaTeX.

 

[Mark44]

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

 

[bugatti79]

Thank you Mark,

I am slowly learning :-)

 

[bugatti79]

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

 

[Mark44]

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.

 

[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?

 

[Mark44]

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.

 

[bugatti79]

No I don't think so. Its another question, ie its a different span we are asked to check...?

 

[Mark44]

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.

 

[bugatti79]

Yes,

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

 

[Mark44]

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?

 

[bugatti79]

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