A little proof

  • Thread starter John Smith
  • Start date
  • #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
 

Answers and Replies

  • #2
HallsofIvy
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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]
and
[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]?)
 
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  • #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.
 
  • #4
HallsofIvy
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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]
and
[tex](I- \frac{1}{1+ u^Tv}uv^T)(I+ uv^T)[/tex]
should be the identity matrix!
 
  • #5
Yes thank you I think that I got it now.
 

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