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Basis of a Vector Space

  1. Mar 16, 2015 #1
    1. The problem statement, all variables and given/known data
    Let V be a vector space, and suppose that [itex]\vec{v_1}[/itex], [itex]\vec{v_2}[/itex], ... [itex]\vec{v_n}[/itex] is a basis of V. Let [itex]c\in\mathbb R[/itex] be a scalar, and define [itex]\vec{w}[/itex] = [itex]\vec{v_1} + c\vec{v_2}[/itex]. Prove that [itex]\vec{w}, \vec{v_2}, ... , \vec{v_n}[/itex] is also a basis of V.

    2. Relevant equations
    If two of the following conditions hold, the third holds automatically, and the list is a basis for a vector space V.
    1. If dimV = the number of vectors in the list.
    2. If the list of vectors spans V.
    3. If the list of vectors is linearly independent.


    3. The attempt at a solution
    So, I feel like I did this right, but I have no way to check because we don't use a textbook for the class and I couldn't find it online anywhere. I did the following.

    Since [itex]\vec{v_1}[/itex], [itex]\vec{v_2}[/itex], ... [itex]\vec{v_n}[/itex] is a basis of V, [itex]\vec{v_1}[/itex], [itex]\vec{v_2}[/itex], ... [itex]\vec{v_n}[/itex] are linearly independent and span V. Also, dimV = n, because n vectors form the basis of V.
    [itex]\vec{w}, \vec{v_2}, ... , \vec{v_n}[/itex] is also a list of n vectors, so it will be sufficient to show that this list either spans V or is linearly independent in order for it to be a basis of V.
    Since [itex]\vec{v_1}[/itex], [itex]\vec{v_2}[/itex], ... [itex]\vec{v_n}[/itex] span V, that means there exists some scalars ai (where i is between 1 and n inclusive) such that
    a1v1 + a2v2 + ... + anvn = [itex]\vec{x}[/itex], where [itex]\vec{x}[/itex] is in V.
    So, for [itex]\vec{w}, \vec{v_2}, ... , \vec{v_n}[/itex] to span V, there must exist some scalars bi such that b1w + b2v2 + ... + bnvn = [itex]\vec{x}[/itex].

    Then I basically expand this, and set it equal to the previous expression with all the a's. I show that in order for them to be equal, choose b3 = a3, b4 = a4, ... bn = an.

    Then all we have to show is that
    b1v1 + (b1c + b2)v2 = a1v1 + a2v2.
    Choose b1 = a1, and b2 = a2 - a1c.

    Then a1w+ (a2 - a1c)v2 + a3v3 + ... + anvn = [itex]\vec{x}[/itex], so it spans V.

    Thus, since the dimension if n and it spans V, it also forms a basis of V.

    Is this right, though?

     
  2. jcsd
  3. Mar 16, 2015 #2
    Looks ok to me.... :rolleyes:
     
  4. Mar 17, 2015 #3

    Ray Vickson

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    There is a theorem that a set of ##n## linearly-independent vectors in an ##n##-dimensional space ##V## does, indeed, span ##V##. If you know and are allowed to use that theorem, you can shorten your argument a bit: just show that ##\vec{w}, \vec{v}_2, \ldots, \vec{v}_n## are linearly independent.
     
  5. Mar 17, 2015 #4

    HallsofIvy

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    The OP said in his first post
    "
    If two of the following conditions hold, the third holds automatically, and the list is a basis for a vector space V.
    1. If dimV = the number of vectors in the list.
    2. If the list of vectors spans V.
    3. If the list of vectors is linearly independent. "
    so, yes, he does know that.
     
  6. Mar 17, 2015 #5

    Ray Vickson

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    I missed the opening sentence.
     
  7. Mar 17, 2015 #6
    I tried it like that first, but I couldn't actually figure out how to show that ##\vec{w}, \vec{v}_2, \ldots, \vec{v}_n## are linearly independent. Can I have a hint for how to do that?

    Also, I'm a girl btw. :P
     
  8. Mar 17, 2015 #7

    Ray Vickson

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    The message you are quoting above said nothing about your gender!

    Anyway, you want to know if there are ##c_1, c_2, \ldots, c_n##, not all zero, such that ##0 = c_1 w + c_2 v_2 + \cdots + c_n v_n = 0##. Take it from there.
     
  9. Mar 17, 2015 #8
    I think I got it.
    ##0 = c_1 w + c_2 v_2 + \cdots + c_n v_n = 0##
    Let ##c_3, ... c_n = 0##, and now we'll show that ##c_1w + c_2v_2## = 0.
    ##c_1(v_1+cv_2) + c_2v_2 = 0##
    ##c_1v_1+c_1cv_2 + c_2v_2 = 0##
    ##c_1v_1 + (c_1c+c_2)v_2 = 0##
    ##c_1## must be 0, because it's the only way to get rid of ##v_1##, so then ##c_2## must also be 0.
    Thus call the constants must be zero, so it's linearly independent?
     
  10. Mar 17, 2015 #9

    Mark44

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    I don't think you have it. In the equation #c_1v_1 + c_2v_2 + \cdots + c_nv_n = 0##, the fact that all of the constants ci = 0 means absolutely nothing. This happens when the vectors are linearly independent and when they are linearly dependent. The important fact for a set of linearly independent vectors is that there is only a single solution in the constants with all of them being zero.

    ##c_1 w + c_2 v_2 + \cdots + c_n v_n = 0##
    ##\Rightarrow c_1(v_1 + cv_2) + c_2 v_2 + \cdots + c_n v_n = 0##

    I would get an equation with only the vectors v1, v2, ..., vn in it, and then use the fact that this set of vectors is linearly independent.
     
  11. Mar 17, 2015 #10

    Mark44

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    You quoted Ray Vickson, but it was HallsofIvy who said "so, yes, he does know that."
     
  12. Mar 17, 2015 #11
    So, do you mean that I should just say
    ##c_1(v_1 + cv_2) + c_2 v_2 + \cdots + c_n v_n = 0##
    ##c_1v_1 + (c_1c+c_2)v_2 + ... + c_nv_n = 0 ## is linearly independent because ##v_1, v_2, ... v_n## are linearly independent?

    Also, I know HallsofIvy said it, I just didn't go back and quote him. I figured he'd figure it out lol.
     
  13. Mar 17, 2015 #12

    Dick

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    In the second form of the sum all of the coefficients of the v's must be zero since the v's are linearly independent. What does that tell you about ##c_1, c_3, ...##?
     
    Last edited: Mar 17, 2015
  14. Mar 17, 2015 #13

    Mark44

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    You can't describe an equation as "linearly independent". You're trying to show that w, v2, ..., vn are a linearly independent set.
    Well, you confused Ray Vickson, whom you quoted, and who thought you were talking about him.
     
  15. Mar 17, 2015 #14
    That they're 0. So the only way to make the second sum 0 is if ##c_2## is also 0. And since the second sum is equivalent to the first sum, all the coefficients must still be 0, so w, ##v_2, ..., v_n## are linearly independent?

    Sorry Ray Vickson! Thought you were joking, not confused.
     
  16. Mar 17, 2015 #15

    Mark44

    Staff: Mentor

    You're still missing an important concept. Given any vectors v1, v2, ..., vn in the same space, you can always write this equation: ##c_1v_1 + c_2v_2 + \cdot + c_nv_n = 0##. The vectors could be linearly independent or linearly dependent.

    If the vectors are linearly independent, then there is one and only one solution, the trivial solution: c1 = c2 = ... = cn = 0. If the vectors are linearly dependent, then there is also a solution for which at least one of the constants is nonzero.
     
  17. Mar 17, 2015 #16

    Dick

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    Pretty much. Since ##c_1 c + c_2=0## and ##c_1=0## the whole sum reduces to ##c_2 v_2=0##. Since ##v_2## is nonzero (why?) that tells you ##c_2## must also be zero.
     
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