Proving Reversibility of A with u & v of Size n*1

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Discussion Overview

The discussion revolves around proving the reversibility of a matrix A defined as A = I + u*v^T, where u and v are column vectors of size n*1 and I is the identity matrix of size n*n. Participants are exploring the conditions under which A is invertible, specifically when u^T*v is not equal to -1, and the proposed form of the inverse.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks help with proving that A is reversible under the condition that u^T*v is not equal to -1 and provides a proposed inverse.
  • Another participant suggests multiplying the matrices (I + uv^T) and (I - (1 / (1 + u^T*v))uv^T) to verify the proof.
  • A participant confirms that they have computed the multiplication but questions if their result is sufficient for the proof.
  • Another participant emphasizes that the goal is to show that the products of the two matrices yield the identity matrix, which is essential for proving reversibility.
  • A later reply indicates that the original poster believes they have understood the proof after the discussion.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and clarity regarding the proof. There is no consensus on the sufficiency of the calculations presented, and the discussion includes corrections and clarifications about the multiplication of matrices.

Contextual Notes

Participants have not resolved the specific steps needed to demonstrate that the product of the matrices equals the identity matrix, and there are questions about the notation used in the expressions.

John Smith
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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
 
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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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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.
 
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!
 
Yes thank you I think that I got it now.
 

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