# would like help on formulating proof

• Jan 27th 2014, 04:37 AM
reinoutg
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 $A^T$ and inverse $A^{-1}$
Let b be a vector with transpose $b^T$.

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

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

Regards,
Reinout

• Jan 27th 2014, 04:52 AM
romsek
Re: would like help on formulating proof
Quote:

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 $1+b A^{-1}b$ in the denominator.

$b is n\times 1$

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

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

do you maybe mean $1+b^T A^{-1} b$ ?
• Jan 27th 2014, 05:28 AM
reinoutg
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 $1+b^T A^{-1} b$ or $1+b A^{-1} b^T$.
I am also dubious about the $b^TAb$ 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 $b^TA^{-1}b$.