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A little proof

  1. Feb 25, 2008 #1
    I need help with this proof.
    We have u and v of size n*1. It is giving that I of size n*n.
    A = I + u*v^Transpose

    Proof that if u^T*v is not = -1
    then A is reverseble and that A is
    A^-1 = I - (1 / (1+u^T*V))*uv^T
  2. jcsd
  3. Feb 25, 2008 #2


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    The obvious thing to do is to look at
    [tex](I+uv^T)(I- \frac{1}{1+ u^Tv}uv^T)[/tex]
    [tex](I- \frac{1}{1+ u^Tv}uv^T)(I+ uv^T)[/tex]

    What do you get when you multiply those out?

    (Are you certain that one of those [itex]uv^T[/itex] is not supposed to be [itex]vu^T[/itex]?)
    Last edited by a moderator: Feb 25, 2008
  4. Feb 26, 2008 #3
    I did multiply the equation out and find out that ((v^T)* u)^T = u^T * v
    But I was wondering if this was enough to show out the proof.
  5. Feb 26, 2008 #4


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    No, that is not the point at all!
    The products of both
    [tex](I+uv^T)(I- \frac{1}{1+ u^Tv}uv^T)[/tex]
    [tex](I- \frac{1}{1+ u^Tv}uv^T)(I+ uv^T)[/tex]
    should be the identity matrix!
  6. Feb 27, 2008 #5
    Yes thank you I think that I got it now.
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