- #1

- 6

- 0

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**

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter John Smith
- Start date

- #1

- 6

- 0

We have

Proof that if

then A is reverseble and that A is

- #2

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 964

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]?)

[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]?)

Last edited by a moderator:

- #3

- 6

- 0

But I was wondering if this was enough to show out the proof.

- #4

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 964

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

- 6

- 0

Yes thank you I think that I got it now.

Share: