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Householder matrix

  1. Oct 12, 2009 #1
    a) Show that the householder matrix H=I-2ww* is unitary ( where ||w||=1).

    b) Given a vector x in C^n (C: complex numbers) and an integer k with 1<k<n, derive a formula for a Householder matrix with the property that (Hx)_i = 0 for i> k. Be sure to choose the signs so that the formula is numerically stable.

    For a), I think H is a unitary matrix if H H* = Identity matrix I. Now I guess, the thing is to just multiply H and H* and see if I get the identity matrix. Now the question is
    (I-2ww*)*= (I-2w*w)? Or is it rather (I-2ww*)*= (I-2ww*) ?

    For b)I don't know how to even start that one. Any help is much appreciated.
     
  2. jcsd
  3. Oct 12, 2009 #2

    Dick

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    * is Hermitian conjugate, right? Then (a(b*))*=((b*)*)(a*)=b(a*). Apply that to the case a=b=v. Not sure about 'numerical stability' issues in the second one.
     
  4. Oct 14, 2009 #3
    I assume ww* is a square matrix right? and (I-2ww*)*= (I-2ww*)
    Now HH*= (I-2ww*)(I-2ww*)*= (I-2ww*)(I-2ww*)= I-4ww*+ 4(ww*)(ww*)
    I suppose -4ww*+ 4(ww*)(ww*) will cancel out but what's the reason? I mean, we have that ||w||=1, but how do we use this info?
     
  5. Oct 14, 2009 #4

    Dick

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    How might (w*)(w) be related to ||w||?
     
  6. Oct 15, 2009 #5
    I wonder because I thought (w*)(w) was a square matrix, I am not sure, is it like the determinant?
     
  7. Oct 15, 2009 #6

    Dick

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    w* is a row vector. w is a column vector. What's the form of the product?
     
  8. Oct 18, 2009 #7
    Say w* is an 1xn vector and w is an nx1, I believe ww*is of the form nxn which is a square matrix. now w*w should be a number (of the form 1x1)., so is ||w||=square root of w*w ?
     
  9. Oct 18, 2009 #8

    Dick

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    Sure. w*w=||w||=1.
     
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