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Null Space of a Matrix and Its Iterates

  1. Aug 7, 2013 #1
    This might seem like a stupid question but would the null space of a matrix and its, say Gaussian elimination transforms, have the same null space. I guess, I am asking if this is valid:

    Let x be in N(A). Let A[itex]_{m}[/itex] be some iteration of A through elimination matrices, i.e. A[itex]_{m}[/itex] = E[itex]_{1}[/itex]E[itex]_{2}[/itex]...E[itex]_{m}[/itex]A. Is N(A) = N(A[itex]_{m}[/itex])?

    Seems like an obvious answer with a sorta obvious proof involving expanding A[itex]_{m}[/itex] to the elimination matrices multiplied by the original A and showing that since x is in A's nullspace, you just have elimination matrices being multiplied by the zero vector. Is this correct? And if it is, can it apply to other decompositions such as QR via Householder? Can it apply to any arbitrary iterate, 1st, 2nd, etc.?

    Thanks for any input.
     
  2. jcsd
  3. Aug 7, 2013 #2
    The null-space is the kernel of a matrix if I'm not mistaken, and yes, the original matrix and the transformed matrix have the same kernel. As for the proof, it's fairly simple, you just let the original matrix represent a system of linear equations, and then prove that any of the three elementary transformations don't change the solution of the system, which also means the kernel remains the same.
     
  4. Aug 7, 2013 #3

    HallsofIvy

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    Each of the [itex]E_i[/itex] is an "elementary matrix" corresponding to some row operation. That is, it is the matrix we get by applying that row operation to the identity matrix. Every row operation has an inverse operation and so every elementary matrix is invertible.
    (There are three kinds of row operations:
    1) Multiply a row by some non-zero number, a. The inverse is to multiply that same row by 1/a.
    2) Swap two rows. The inverse is the same- swap those same two rows.
    3) Add a multiple, a, of row i to row j. the inverse is to add -a times row ix to row j.)

    If Ax= 0, then, of course, [itex]E_1E_2...E_nAx= 0[/itex]. And, because the [itex]E_i[/itex] matrices are invertible, if Ax is NOT 0, then neither is [itex]E_1E_2...E_nAx= 0[/itex].
     
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