Stuck on a proof

bump

Hey reinout-g!


Let's see...

Suppose we define s = b' inv(A) b, which is a scalar.
It's important that it is a scalar, because it means we can reorder any multiplications of s with a matrix or a vector.

Then the question is whether: inv( A + b b' ) = inv(A) - inv(A) b b' inv(A) / (1 + s).

If we multiply both sides with (A + b b'), which is presumably invertable, we get:

I = (inv(A) - inv(A) b b' inv(A) / (1 + s)) (A + b b')

Working it out, we get:

I = (inv(A) - inv(A) b b' inv(A) / (1 + s)) (A + b b')
= I + inv(A)b b' - inv(A) b b' / (1 + s) - inv(A) b b' inv(A) b b' / (1 + s)
= I + (inv(A)b b' (1 + s) - inv(A) b b' - inv(A) b s b') / (1 + s)
= I + (inv(A)b b' + s inv(A)b b' - inv(A) b b' - s inv(A) b b') / (1 + s)
= I

(Sorry if it looks a bit cumbersome. )

This reasoning can be applied in reverse... so I think we got it!
Yay!

I like Serena:

Hey reinout-g!


Let's see...

Suppose we define s = b' inv(A) b, which is a scalar.
It's important that it is a scalar, because it means we can reorder any multiplications of s with a matrix or a vector.

Then the question is whether: inv( A + b b' ) = inv(A) - inv(A) b b' inv(A) / (1 + s).

If we multiply both sides with (A + b b'), which is presumably invertable, we get:

I = (inv(A) - inv(A) b b' inv(A) / (1 + s)) (A + b b')

Working it out, we get:

I = (inv(A) - inv(A) b b' inv(A) / (1 + s)) (A + b b')
= I + inv(A)b b' - inv(A) b b' / (1 + s) - inv(A) b b' inv(A) b b' / (1 + s)
= I + (inv(A)b b' (1 + s) - inv(A) b b' - inv(A) b s b') / (1 + s)
= I + (inv(A)b b' + s inv(A)b b' - inv(A) b b' - s inv(A) b b') / (1 + s)
= I

(Sorry if it looks a bit cumbersome. )

This reasoning can be applied in reverse... so I think we got it!
Yay!

Wow Serena!

I can't imagine how long it must have took you to get that into 'inv(A)' notation!
Let alone the proof itself.
I finally deciphered it and do understand what you did there.
The idea of taking s for b' inv(A) b really helped met out.
Also I appreciated how you chose to multiply with (A + b b') since it makes it way easier to see where you're going in the proof.

Thanks a lot!

reinout-g:

Wow Serena!

I can't imagine how long it must have took you to get that into 'inv(A)' notation!
Let alone the proof itself.
I finally deciphered it and do understand what you did there.
The idea of taking s for b' inv(A) b really helped met out.
Also I appreciated how you chose to multiply with (A + b b') since it makes it way easier to see where you're going in the proof.

Thanks a lot!

Thanks!

I have to admit that it looked much easier to understand when I wrote it on paper.