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Finding a basis

  1. Aug 29, 2011 #1
    I'm moving on to my next section of work and i come across this example:
    Consider the homogeneous system
    x + 2y − z + u + 2v = 0
    x + y + 2z − 3u + v = 0

    It asks for a basis to be found for the solution space S of this system. And also what is the dimension of S.

    I know this might be basic but I can't remember how to form a matrix from that system. I believe I will know how to answer the question once i have the matrix. Can anyone help?
     
  2. jcsd
  3. Aug 29, 2011 #2

    micromass

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    Let me do an example. From the system

    [tex]\left\{\begin{array}{c} 2x-y+z=0\\ 3x-2y+z=0\\ 5x+9y-z=0\end{array}\right.[/tex]

    yields the matrix

    [tex]\left(\begin{array}{cccc} 2 & -1 & 1 & 0\\ 3 & -2 & 1 & 0\\ 5 & 9 & -1 & 0 \end{array}\right)[/tex]

    Did you see what I did?? Can you do the same thing?
     
  4. Aug 29, 2011 #3

    HallsofIvy

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    Another way of looking at it:
    If we subtract the second equation from the first, we get y- 3z+ 4u+ v= 0. Solving for y,
    y= 3z- 4u- v. Any vector, (x, y, z, u, v) satisfying those equations can be written as (x, 3z-4u-v, z, u, v)= (x, 0, 0, 0)+ (0, 3z, z, 0, 0)+ (0, -4u, 0, u, 0)+ (0, -v, 0, 0, v). Can you get the basis for the solution space from that? What is the dimension?
     
  5. Aug 30, 2011 #4
    Well I got the matrix
    (1 2 -1 1 2 0)
    (1 1 2 -3 1 0)


    Then I used Gaussian elimination and I'm not sure if I did it correctly but I did end up with
    (1 0)
    (0 1)

    which is my basis and the dimension is 2?
     
  6. Aug 30, 2011 #5
    I think what I did was wrong. If I follow what HallsofIvy said; is this how I'm supposed to do it:
    (x, 3z-4u-v, z, u, v)= (x, 0, 0, 0)+ (0, 3z, z, 0, 0)+ (0, -4u, 0, u, 0)+ (0, -v, 0, 0, v)
    = x(1, 0, 0, 0) + z(0, 3, 1, 0, 0) + u(0, -4, 0, 1, 0) + v(0, -1, 0, 0, 1)
    i.e v1=(1, 0, 0, 0) v2=(0, 3, 1, 0, 0) v3=(0, -4, 0, 1, 0) v4=(0, -1, 0, 0, 1)
    which are linearly independent, thus S={v1, v2, v3, v4} is the basis and the dimension=4
     
  7. Aug 30, 2011 #6
    double post o_O
     
  8. Aug 30, 2011 #7

    Char. Limit

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    I'm pretty sure that you did it incorrectly, as Gaussian elimination doesn't usually take a 5x2 matrix and make it 2x2.
     
  9. Aug 30, 2011 #8
    Char.Limit can you check if my post after that is correct?
     
  10. Aug 30, 2011 #9

    Char. Limit

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    Yeah, that looks right to me.
     
  11. Aug 30, 2011 #10
    I would like to pose a question here that i am not so clear about, if someone would please help.


    All Basis Vectors should be orthogonal to each other . Is that right ?

    Thank You.
     
  12. Aug 30, 2011 #11

    HallsofIvy

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    No, that is not correct. All we require of a basis is that it span the space and its vectors be independent.

    Of course, it's nice if the basis is orthogonal- for any vector v, the coefficient in its expansion in an orthogonal basis [itex]\{v_i\}[/itex] is just [itex]<v, v_i>/<v_i, v_i>[/itex], where < , > is the inner product. It's even better if the basis is orthonormal. In that case the coefficient is just [itex]<v, v_i>[/itex].

    Note, by the way, that in order to talk about "orthogonal" and/or "normal", a vector space must have an inner product and a vector space, in general, does not have to have an inner product.
     
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