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Projection into the left null space

  1. Mar 14, 2007 #1
    1. The problem statement, all variables and given/known data

    I am trying to find the matrix M that projects a vector b into the left nullspace of A, aka the nullspace of A transpose.

    2. Relevant equations

    A = matrix
    A ^ T = A transpose
    A ^ -1 = inverse of A
    e = b - A x (hat)
    e = b-p

    I know that the matrix P that projects the vector b into the collumn space of A is P = A(A ^T*A)^-1 A^T. Col space is orthogonal to the left nullspace

    3. The attempt at a solution

    Since Col space is orth to left null, I was thinking of just find a matrix that, when doted with P is equal to zero (the definition of orthogonality); but thats what they want us to do in part b

    Also, since we can get the left nullspace from the column space, i was thinking we could just apply that to P in order to get M (as in find the left null space of P) but the problem is that A is not given

    Third idea; the error e used in finding P is in the left nullspace. so if i could somehow make it only have a component in the left nullspace, none in the column space, i could somehow find P.

    So i have plenty of ideas, but no idea how to implement them. any help would be GREATLY appreciated, as this pset is due in 3 hours!
  2. jcsd
  3. Mar 14, 2007 #2
    heh, it is actually very simple... and you have basically solved it... yes use that error term...

    a vector consists of components in Col A and components in Col A perp... so v=x+y, x in Col A and y in Col A prep. just subtract y from x and there you go!

    x=Px + y, so y=x-Px or just (I-P)x

    edit: fixed the typo.
    Last edited: Mar 14, 2007
  4. Mar 14, 2007 #3


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    This doesn't look right. If A has a non-trivial nullspace, then (ATA)-1 doesn't exist, so your formula for the matrix which projects to column space of A doesn't make sense when A is singular. On the other hand, if A is non-singular, then A(ATA)-1AT = I, so your formula just gives the identity function, but it's obvious that the identity function is what projects the column space of A when A is non-singular.
  5. Mar 14, 2007 #4
    I think he meant that A is the matrix whose columns span a vector space. thus A is nonsingular (though not necessarily a square matrix). the P matrix is the projection operator of a vector onto Col A. of course, if A is a square matrix.. there is nothing to project, thus P becomes identity.

    edit: sry I meant the columns of A consists of a set of basis for Col A.
    Last edited: Mar 14, 2007
  6. Mar 14, 2007 #5


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    Every matrix's columns span a vector space. The columns of the zero matrix span the vector space {0}. What you need is that if A is mxn, then rank(A) = n, or something like that. If we're allowed to assume this, then you've already given away the answer, so there's nothing left for the original poster to do. But nothing warrants this assumption in the first place.
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