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Nullspace of mxn matrix

  1. Nov 22, 2009 #1
    Does anyone know how to approach this problem?

    Let A be an m×n matrix of rank m, where m<n. Pick a point x in R^n, and let x∗ be the point in the nullspace of A closest to x. Write a formula for x∗ in terms of x and A.

    What exactly is the significance of the point x* in the nullspace of A?
     
  2. jcsd
  3. Nov 23, 2009 #2
    i think this question has something to do with the orthogonality of subspaces and is analogous to saying the shortest distance between a point and an axis is just a perpendicular line...
     
  4. Nov 23, 2009 #3
    >> What exactly is the significance of the point x* in the nullspace of A?

    So, Ax* = 0 which is basically what it means by x* being in the nullspace of A. x* is a vector which multiplies A to take it to 0.

    I think it has something to do with the least square solution for linear systems.
     
  5. Nov 23, 2009 #4
    I think you should think of the solution in terms of the projection of x onto the nullspace of A, which would be x*.

    This is simply given by A*(A'A)-1*A'*x. Sorry, I am not fluent with latex.

    A' = transpose of A.
    -1 = inverse
    * = multiplication

    Refer to the Strang book on least square and projection of vectors onto the column space of a a matrix.
     
  6. Nov 23, 2009 #5
    We know that the row space is in [tex]R^{n}[/tex] and that it is orthogonal to the null space.

    Imagine that we have a 2x3 matrix with rank 2. It's row space would be a plane in [tex]R^{3}[/tex], and it's null space a line perpendicular to that plane. If we pick a point [tex]x[/tex] in that plane, wouldn't the point closest to the null space be that same point x?

    I'm eagerly awaiting someone to help us out :)
     
  7. Nov 23, 2009 #6
    ^^ i think you're on the right track, dafe.

    i think an analogy to this problem is, what is the shortest distance between a point in the xy plane and the x axis. it's just a line perpendicular to the x-axis.

    so in the context of the problem i asked, it's like...the dot product between the point x and...the row space? something like that...
     
  8. Nov 23, 2009 #7
    ok, some help:

    1) set this up with a variable (btw, better to call the starting point x0 rather than x, since it is fixed), an objective function and a set of equality constraints. Write them down. You should replace the objective function by something nicer: for example no square roots, since to apply Lagrange you need a differentiable function.
    2) show geometrically in terms of the level sets of the objective function and the tangent space of the constraints that the vector x0 - x* is perpendicular to the null space.
    3) form the Lagrangian. solve.
     
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