1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Nonsingular perturbed identity matrix
  1. Identity matrix (Replies: 1)

Loading...