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Linear Algebra - Spans

  1. Oct 24, 2011 #1
    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 [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]?
     
  2. jcsd
  3. Oct 24, 2011 #2

    Mark44

    Staff: Mentor

    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.
     
  4. Oct 25, 2011 #3
    [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
     
  5. Oct 25, 2011 #4

    Mark44

    Staff: Mentor

    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].
     
  6. Oct 25, 2011 #5
    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]
     
  7. Oct 25, 2011 #6

    Mark44

    Staff: Mentor

    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.
     
  8. Oct 25, 2011 #7
    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
     
  9. Oct 25, 2011 #8

    Mark44

    Staff: Mentor

    Work with the right side of the equation in post #6.
     
  10. Oct 26, 2011 #9
    [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
     
  11. Oct 26, 2011 #10

    Mark44

    Staff: Mentor

    [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?
     
  12. Oct 26, 2011 #11
    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?
     
  13. Oct 26, 2011 #12
    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?!
     
  14. Oct 26, 2011 #13

    Mark44

    Staff: Mentor

    From post #2
    Then u = ?
     
  15. Oct 26, 2011 #14
    [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: Oct 26, 2011
  16. Oct 26, 2011 #15

    Mark44

    Staff: Mentor

    OK, so now you have proved what you needed to per your post #1.
     
  17. Oct 26, 2011 #16
    Thank you Mark,

    I am slowly learning :-)
     
  18. Oct 27, 2011 #17
    Actually, should the vector u be represented by beta scalars and v by the alpha scalars?
     
  19. Oct 27, 2011 #18

    Mark44

    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.
     
  20. Oct 27, 2011 #19
    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?
     
  21. Oct 27, 2011 #20

    Mark44

    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.
     
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