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Linear Algebra: Orthogonal basis ERG HELP!

  1. Jul 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Consider the vector V= [1 2 3 4]' in R4, find a basis of the subspace of R4 consisting of all vectors perpendicular to V.


    2. Relevant equations
    I mean, I'm just completely stumped by this one. I know that in R2, any V can be broken down to VParallel + VPerp, which represents a unique solution, but how does this apply to R4.

    I'm interested in the process to finding the basis, more so than the answer. What steps would you use in general, to find the basis of a vector, that satisfy the conditions that the basis must be perpendicular to the vector.

    Thanks,
     
    Last edited: Jul 14, 2010
  2. jcsd
  3. Jul 14, 2010 #2

    vela

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    If x is some vector in R4, what equation must it satisfy if it is perpendicular to V?
     
  4. Jul 14, 2010 #3
    Do you know this theorem?

    If A := {v1,...,vn} is an orthonormal set in an n-dim inner product space V, then A can be extended to an orthonormal basis for V { v1,.........., vn, v n+1 ,.............. vz }

    and if W = Span(A) , then B : = { v n+1 ,.............. vz } is an orthonormal basis for A perp
     
  5. Jul 14, 2010 #4

    HallsofIvy

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    That is true in any dimension. In this case, since [1 2 3 4] itself, being a single vector spans a 1 dimensional subspace of R4. R4 can be written as as a direct sum of that and a 3 dimensional subspace. I recommend vela's suggestion: What must be true of [w x y z] in order that it be perpendicular to [1 2 3 4]? Specifically, what does "perpendicular" mean in a general Rn?

    Use that equation to write one of w, x, y, or z in terms of the other three.

     
  6. Jul 14, 2010 #5
    X dot V = 0 right?
     
  7. Jul 14, 2010 #6

    vela

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    Yup!
     
  8. Jul 14, 2010 #7
    Okay, so you mean that:

    [w x y z]' dot [1 2 3 4]' = [0 0 0 0]
     
  9. Jul 14, 2010 #8
    whoooaaa wait a minute, hold up. what wizvuze said...

    can you explain that to me? I'm not sure where that comes from...?
     
  10. Jul 14, 2010 #9

    Mark44

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    The dot product does not produce a vector; it produces a number (scalar). So [w x y z]' dot [1 2 3 4]' = 0. Now work with that.
     
  11. Jul 14, 2010 #10

    Mark44

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    You don't need an orthonormal basis. Any old basis for your three-dimensional subspace of R4 will do just fine. Heed the advice from HallsOfIvy.
     
  12. Jul 14, 2010 #11
    whoops. I was trying to be smart and let that slip.
     
  13. Jul 14, 2010 #12
    okay, so if i wrote X in terms of y,z,w (e.g. x= -2y-3z-4w) and substituted that into X, which then becomes [-2y-3z-4w 2y 3z 4w], then that is basis consisting of the span of all perpendicular vectors to [1 2 3 4]??? In other words, if I took the dot product of those, then it would be zero???
     
  14. Jul 14, 2010 #13

    vela

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    A basis is a set of linearly independent vectors. In this case, you need to specify three linearly independent such that any linear combination of those vectors is perpendicular to V. So far you have

    [tex]\vec{x}=\begin{pmatrix}-2y-3z-4w\\y\\z\\w\end{pmatrix}[/tex]

    You want to rewrite that as a linear combination of three vectors. Those three vectors will form the basis you're trying to find.
     
    Last edited: Jul 14, 2010
  15. Jul 14, 2010 #14

    vela

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    No, those aren't correct. If you multiply any of those by V, you won't get 0.

    Start by writing x as a sum of three vectors, one involving just y, one involving just z, and one involving just w. Then pull the variables out of each one so you end up with

    x = y(vector 1)+z(vector 2)+w(vector 3)

    Those three vectors are a basis for the subspace.
     
  16. Jul 14, 2010 #15
    So then X= y[-2 2 0 0] + z[-3 0 3 0] + w[ -4 0 0 4] ?

    and each of those vectors represents the basis?
     
  17. Jul 14, 2010 #16
    * mean the three vectors are a basis for R3, the subspace of R4 that are all perpendicular to V
     
  18. Jul 14, 2010 #17

    vela

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    A basis is a set of linearly independent vectors that span a vector space. In this problem, you have a three-dimensional space, so a basis consists of a set of three vectors. Each of the vectors isn't a basis; the three vectors together are a basis.
     
  19. Jul 14, 2010 #18

    Mark44

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    The three vectors might be a basis for a three-dimensional subspace of R4. (I say "might be" because I didn't check your work.)
    Since the vectors are in R4, they can't possibly have anything to do with R3.
     
  20. Jul 14, 2010 #19

    Mark44

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    No, you have an error. None of these vectors is perpendicular to <1, 2, 3, 4>.

    Edit: I didn't realize that vela had already spotted the error in post 14. Look carefully at what vela is showing in post 13.
     
    Last edited: Jul 14, 2010
  21. Jul 14, 2010 #20
    so, um, i don't understand then...yes i see that they're not perpendicular...i followed all of the outlined steps...what happened?
     
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