Linear Algebra - Spans

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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 [itex](b_1,b_2,b_3,b_4)=span(b_1-b_3,b_2-b_1,b_3,b_4-b_2)[/itex]

If I equate the LHS and RHS as

[itex]\alpha_1b_1=+\alpha_1b_1-\alpha_3b_3[/itex] implies [itex]\alpha_3=0[/itex]

[itex]\alpha_2b_2=\alpha_2b_2-\alpha_1b_1[/itex] implies [itex]\alpha_1=0[/itex]

[itex]\alpha_3b_3=\alpha_3b_3[/itex] but [itex]\alpha_3=0[/itex]

[itex]\alpha_4b_4=\alpha_4b_4-\alpha_2b_2[/itex] implies [itex]\alpha_2=0[/itex]

This correct? What about [itex]\alpha_4[/itex]?
 

Answers and Replies

  • #2
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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 [itex](b_1,b_2,b_3,b_4)=span(b_1-b_3,b_2-b_1,b_3,b_4-b_2)[/itex]

If I equate the LHS and RHS as

[itex]\alpha_1b_1=+\alpha_1b_1-\alpha_3b_3[/itex] implies [itex]\alpha_3=0[/itex]

[itex]\alpha_2b_2=\alpha_2b_2-\alpha_1b_1[/itex] implies [itex]\alpha_1=0[/itex]

[itex]\alpha_3b_3=\alpha_3b_3[/itex] but [itex]\alpha_3=0[/itex]

[itex]\alpha_4b_4=\alpha_4b_4-\alpha_2b_2[/itex] implies [itex]\alpha_2=0[/itex]

This correct? What about [itex]\alpha_4[/itex]?
No, this isn't correct. Let v be an arbitrary vector in [itex]span(b_1-b_3,b_2-b_1,b_3,b_4-b_2)[/itex], which means that v is some linear combination of these vectors. Show that this same vector is a linear combination of [itex](b_1,b_2,b_3,b_4)[/itex].

Now go the other way. Let u be an arbitrary vector in [itex]span(b_1, b_2, b_3, b_4)[/itex]. Show that u is also a linear combination of the vectors in the second set.
 
  • #3
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No, this isn't correct. Let v be an arbitrary vector in [itex]span(b_1-b_3,b_2-b_1,b_3,b_4-b_2)[/itex], which means that v is some linear combination of these vectors. Show that this same vector is a linear combination of [itex](b_1,b_2,b_3,b_4)[/itex].

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

Is this the correct start?

thanks
 
  • #4
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No, this isn't correct. Let v be an arbitrary vector in [itex]span(b_1-b_3,b_2-b_1,b_3,b_4-b_2)[/itex], which means that v is some linear combination of these vectors. Show that this same vector is a linear combination of [itex](b_1,b_2,b_3,b_4)[/itex].

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

Is this the correct start?
No. Let v = <v1, v2, v3, v4> be an arbitrary vector in [itex]span(b_1-b_3,b_2-b_1,b_3,b_4-b_2)[/itex].
 
  • #5
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No. Let v = <v1, v2, v3, v4> be an arbitrary vector in [itex]span(b_1-b_3,b_2-b_1,b_3,b_4-b_2)[/itex].
Not sure what to do...?

[tex](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)[/tex]
 
  • #6
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Not sure what to do...?

[tex](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)[/tex]
Show that v is a linear combination of [itex](b_1,b_2,b_3,b_4)[/itex].

Now go the other way. Let u be an arbitrary vector in [itex]span(b_1, b_2, b_3, b_4)[/itex]. Show that u is also a linear combination of the vectors in the second set.
 
  • #7
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Show that v is a linear combination of [itex](b_1,b_2,b_3,b_4)[/itex].
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

[tex]v=\alpha_1b_1 +\alpha_2b_2+\alpha_3b_3+\alpha_4b_4
+....+\alpha_nb_n[/tex]

Dont know how to proceed.. thanks
 
  • #8
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Work with the right side of the equation in post #6.
 
  • #9
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Work with the right side of the equation in post #6.
[tex]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)[/tex]

[tex]U=\left \{ u_1,u_2,u_3,u_4 \right \}=\alpha_1b_1+\alpha_2b_2+\alpha_3b_3+\alpha_4b_4[/tex]

I dont know how one would be a linear combination of the other? Thanks
 
  • #10
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[tex]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)[/tex]
[tex]=(\alpha_1-\alpha_2) b_1 +(\alpha_2-\alpha_4)b_2 + (\alpha_3-\alpha_1)b_3+ (\alpha_4)b_4 [/tex]

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

I dont know how one would be a linear combination of the other? Thanks
 
  • #11
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[tex]=(\alpha_1-\alpha_2) b_1 +(\alpha_2-\alpha_4)b_2 + (\alpha_3-\alpha_1)b_3+ (\alpha_4)b_4 [/tex]

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 [tex]v=(v_1,v_2...v_n)[/tex] and [tex]u=(u_1,u_2...u_n)[/tex] then span v = span u if and only if v is a linear combination of those in u and vice versa?
 
  • #12
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[tex]=(\alpha_1-\alpha_2) b_1 +(\alpha_2-\alpha_4)b_2 + (\alpha_3-\alpha_1)b_3+ (\alpha_4)b_4 [/tex]

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 [tex](b_1-b_3), (b_2-b_1), (b_3)[/tex] and [tex](b_4-b_2)[/tex]

Hmmmm....how is that done?!
 
  • #13
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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 [tex](b_1-b_3), (b_2-b_1), (b_3)[/tex] and [tex](b_4-b_2)[/tex]

Hmmmm....how is that done?!
From post #2
Now go the other way. Let u be an arbitrary vector in [itex]span(b_1, b_2, b_3, b_4)[/itex]. Show that u is also a linear combination of the vectors in the second set.
Then u = ?
 
  • #14
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[tex]u=\alpha_1(b_1-b_3)+\alpha_2(b_2-b_1)+\alpha_3b_3+\alpha_4(b_4-b_2)[/tex]

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:
  • #15
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OK, so now you have proved what you needed to per your post #1.
 
  • #16
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Thank you Mark,

I am slowly learning :-)
 
  • #17
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Actually, should the vector u be represented by beta scalars and v by the alpha scalars?
 
  • #18
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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
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Ok, the last section

Is the span [tex]\left \{ b_1+b_2,b_2+b_3,b_3 \right \}=\left \{ b_1,b_2,b_3,b_4 \right \}[/tex]

Let [tex]v=\left \{ v_1,v_2,v_3,v_4 \right \}[/tex] span [tex] \left \{ b_1+b_2,b_2+b_3,b_3 \right \}[/tex]

rearranging

[tex]v=\alpha_1b_1+(\alpha_1+\alpha_2)b_2+(\alpha_2+\alpha_3)b_3[/tex]

Let [tex]u=\left \{ u_1,u_2,u_3,u_4 \right \}[/tex] span [tex] \left \{ b_1,b_2,b_3,b_4 \right \}[/tex]

[tex]u=\alpha_1b_1+\alpha2b_2+\alpha_3b_3+alpha_4b_4[/tex]

Hence v does not span u because v does not contain [tex]b_4[/tex]
How do I state this correctly?
 
  • #20
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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
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No I dont think so. Its another question, ie its a different span we are asked to check....?
 
  • #22
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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
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Yes,

THey are linearly independant. Ok, thanks for the clarification!!
bugatti
 
  • #24
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Ok, the last section

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

[tex]v=\alpha_1b_1+(\alpha_1+\alpha_2)b_2+(\alpha_2+\alpha_3)b_3[/tex]

Let [tex]u=\left \{ u_1,u_2,u_3,u_4 \right \}[/tex] span [tex] \left \{ b_1,b_2,b_3,b_4 \right \}[/tex]

[tex]u=\alpha_1b_1+\alpha2b_2+\alpha_3b_3+alpha_4b_4[/tex]

Hence v does not span u because v does not contain [tex]b_4[/tex]
How do I state this correctly?
 
  • #25
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ok, thanks for clarification of the nonsense I was writing :-)
 

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