# Thread: would like help on formulating proof

1. ## would like help on formulating proof

Dear all,

I have an exercise which asks me to prove the following;

Let A be a non-singular matrix with transpose $\displaystyle A^T$ and inverse $\displaystyle A^{-1}$
Let b be a vector with transpose $\displaystyle b^T$.

Prove that for $\displaystyle b^T A^{-1} b$ =//not equal to//= $\displaystyle -1$

$\displaystyle (A + b b^T)^{-1} = A^{-1} - (A^{-1} b b^T A^{-1})/(1+b^T A^{-1} b)$

Regards,
Reinout

2. ## Re: would like help on formulating proof

Originally Posted by reinoutg
Dear all,

I have an exercise which asks me to prove the following;

Let A be a non-singular matrix with transpose A' and inverse inv(A)
Let b be a vector with transpose b'.

Prove that for b' A b =//not equal to//= -1

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

I tried to give the easiest notation possible and hope you can help me.

Regards,
Reinout

p.s. I attached an image to clarify the equation.
there's a problem in your expression.

you have $\displaystyle 1+b A^{-1}b$ in the denominator.

$\displaystyle b$ is $n\times 1$

$\displaystyle A^{-1}$ is $n\times n$

$\displaystyle$ so $A^{-1}b$ is $n\times 1$ and cannot be multiplied on the left by a $n \times 1$ vector.

do you maybe mean $\displaystyle 1+b^T A^{-1} b$ ?

3. ## Re: would like help on formulating proof

I had not noticed this yet, but it must be a typo in the exercise. This indeed makes it either $\displaystyle 1+b^T A^{-1} b$ or $\displaystyle 1+b A^{-1} b^T$.
I am also dubious about the $\displaystyle b^TAb$ is not equal to -1 since obviously it is there to make the denominator be unequal to 0.
Therefore I suppose this is another typo which is supposed to be $\displaystyle b^TA^{-1}b$.