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Homework Help: Nonsingular perturbed identity matrix

  1. Jan 19, 2009 #1
    1. The problem statement, all variables and given/known data
    If A = I + uv*, where u and v are m vectors and A is known to be nonsingular, show that the inverse of A = I + [tex]\alpha[/tex]uv* where [tex]\alpha[/tex] is a scalar value


    2. Relevant equations



    3. The attempt at a solution
    Since A is nonsingular, we know the rank of A is m.
    Since both u and v are vectors of dimension m, we know that uv* is a square matrix

    From the above statements, we conclude that A is an orthogonal matrix which imples:

    A[tex]^{-1}[/tex]
    = A [tex]^{*}[/tex]
    = (I + uv*)*
    = I + (uv*)*
    = I + u*v
    = I + [tex]\alpha[\tex], since u*v is a scalar value

    As you can see, I am arriving at the wrong conclusion. What am I missing? Thank you!!
     
  2. jcsd
  3. Jan 19, 2009 #2

    Hurkyl

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    I note two problems:

    (1) I have no idea how you conclude A should be orthogonal.
    (2) (AB)* = B*A*
     
  4. Jan 19, 2009 #3
    (1) - I think one of the definition of an Orthogonal matrix is a square matrix, whose columns are all linearly independent so that they make the basis for an m space. One of the properties of a nonsingular matrix is that it has the maximum rank (in this case m since it is a square matrix). These statements led me to conclude our matrix A is orthogonal!

    (2) - That is a correct property...did I miss something?
     
  5. Jan 19, 2009 #4

    Hurkyl

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    You forgot a key word in that characterization: they have to make an orthonormal basis.

    (Lemme switch to S and T instead of A and B, to eliminate confusion)

    Yes. The property is (ST)*=T*S*, but you used it in your calculation as (ST)*=S*T*.
     
  6. Jan 19, 2009 #5
    Wow! Thank you I missed that! I was assuming (AB)* = A*B* all along - even after I read your post! Thank you.

    If A is not an orthogonal matrix, then I guess I am back to square one :(
     
  7. Jan 19, 2009 #6

    Hurkyl

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    Simple is often best, even if it's just to get ideas. Rather than looking for a clever argument, why not test directly whether or not this alledged inverse to A really is?
     
  8. Jan 19, 2009 #7

    Dick

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    The matrices are inverses if (I+uv*)(I+alpha uv*)x=x for any vector x. Expand it out and try to figure out a value of alpha that will make that true.
     
  9. Jan 19, 2009 #8
    Thank you! Figured it out :)

    (I + uv*) (I + alpha uv*)x = x
    (I + alpha Iuv* + Iuv* + alpha uv*uv*)x = x
    v*u is a scalar and let it be beta
    (I + alpha Iuv* + Iuv* + alpha beta uv*)x = x
    [I + Iuv*(alpha + 1 + alpha beta)]x = x
    this implies (alpha + 1 + alpha beta) = 0!
    alpha = -1/(1+beta)!!!
     
  10. Jan 19, 2009 #9

    Hurkyl

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    Note that x was really just a placebo. You could have gone straight to the definition of inverse, and put an I on the right hand side, without any of the x's, and done exactly the same thing.
     
  11. Jan 19, 2009 #10
    I like the expression (placebo). Yup, thanks Hurkyl. Thanks so much.
     
  12. Jan 19, 2009 #11
    If beta = -1 (which would imply A is singular), what would the nullspace of A be? I got only as far as writing the equation as:

    (I+uv*)x = 0
    x + uv*x = 0....

    Do you have any suggestions?
     
  13. Jan 19, 2009 #12

    Hurkyl

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    Well, if it's going to be at all easy to figure out, it's probably going to be some expression involving u or v. (I+uv*)x = 0 is a linear equation, so I would start by plugging in a few things, and see if I can combine them to make 0.
     
  14. Jan 19, 2009 #13

    Dick

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    Sure you don't want to take a crack at this without a hint? It's pretty easy. Remember beta=v*u=(-1).
     
  15. Jan 19, 2009 #14
    Thanks a lot for your help guys.
    Dick, I think it could be the pressure:), but I still cannot see it. Here is what I did

    v*x is some constant beta,
    x + uv*x = x + beta u
    => x = -beta u
    Therefore, I concluded x is a scalar multiple of u.

    This is what I came up with, but it looks dodgy because I somewhat defined beta with x itself..???
     
  16. Jan 19, 2009 #15
    OH!!!! I see it now :) if x = u, then

    x + uv*x = u + uv*u = u + -u = 0

    You are right Dick, it was easy! :)

    Thanks a bunch!!!
     
  17. Jan 19, 2009 #16

    Hurkyl

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    I think you confused yourself, because you now have two variables named beta! One of them was the inner product of u with v, and the other is the inner product of x with v.

    But no matter; you can test directly if u is a right null-vector or not (I now see you've done that). Of course, you'll have to clean up your proof if you want to claim that the right nullspace is one-dimensional.

    (p.s. as an alternative, there is a rank-counting argument that tells you that A cannot be less than n-1 dimensional. Can you see it? Hint: it might be easier to think instead of the equation I = A - uv*)
     
    Last edited: Jan 19, 2009
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