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Basis for the nullspace of this matrix

  1. Oct 20, 2009 #1
    this is apparently "really simple", but I just don't know how to do it from the examples I have and I feel like a moron...

    what's the basis for the nullspace of this matrix

    [ 2 3 1]
    [ 5 2 1]
    [ 1 7 2]
    [ 6 -2 0]
     
  2. jcsd
  3. Oct 20, 2009 #2
    Re: Basis

    I'm not sure if your notation is a list of vectors in the matrix or if that is the actual matrix so I can't be more exact for now than to remind you that the nullspace is the solution space of the system Ax = 0, so start off by solving for the coefficient matrix A augmented with 0. Remember also that the basis of a space is the set of linearly independent vectors that spans the space in question.
     
  4. Oct 20, 2009 #3

    HallsofIvy

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    Re: Basis

    The nullspace is the set of vectors, <x, y, z> such that
    [tex]\begin{bmatrix}2 & 3 & 1 \\ 5 & 2 & 1\\ 1 & 7 & 2\\6 & -2 & 0\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}[/tex]
    That is the same as the four equations 2x+ 3y+ z= 0, 5x+ 2y+ z= 0, x+ 7y+ 2z= 0, and 6x- 2y= 0. If we subtract the first equation from the second, we get 3x- y= 0 or y= 3x. Putting that into 6x- 2y= 0 we get 6x- 2(3x)= 6x- 6x= 0 which is satisfied for all x.

    If we subtract 2 times the first equation from the third we get -3x+ y= 0 or y= 3x again. Those equations all reduce to y= 3x which cannot be reduced further. Putting y= 3x into the first equation we get 2x+ 3(3x)+ z= 0 so z= -11x. Putting y= 3x into the second equation we get 5x+ 2(3x)+ z= 0 so z= -11x. Putting y= 3x into the third equation, we get x+ 7(3x)+ 2z= 0 so 2z= -22x and z= -11x.

    That is, any solution of that equation, any vector in the kernel, can be written as <x, y, z>= <x, 3x, -11x>= x <1, 3, -11>.
     
    Last edited: Oct 20, 2009
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