Matrix Algebra proof

Hendrik

New member
Joined
Nov 28, 2014
Messages
8
Hi,

I'm not sure if this is the right forum to ask this question. If not please let me know.
My problem: I don't know how to prove the following statement:

Let A be nonsingular and b'A-1b≠-1, then

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

Here b is a vector. I do not even know how to start.

I really hope someone can help me out!
 
A is a square invertible matrix, b is a column vector, b' is its transpose and is a row vector. Thus b'A-1b is a scalar so let the scalar a be defined by
a = 1 + b'A-1b
What you are asked to show is
(A + b b') (a A-1 - A-1bb'A-1) = a I
where I is the identity matrix. So start manipulating the left hand expression
(A + b b') (a A-1 - A-1 b b' A-1)
= (A + b b') A-1 ( a - b b' A-1)
= (AA-1 + b b'A-1) ( a - b b'A-1)
= ...

Is that enough?
 
Thanks! At first I thought that with those tips I could easily complete the proof. However,
I still havent found out what the last steps should be in order to get aI=aI.
I see how you got to:

aI = (AA-1 + b b'A-1) ( a - b b'A-1)

then

aI = (I + b b'A-1) ( a - b b'A-1)
aI = aI -bb'A-1+abb'A-1-bb'A-1bb'A-1 So this last part should equal zero
aI = aI + (-1+a-bb'A-1)bb'A-1

But how to continue? I tried writing out a as its original form, but I wasnt
able to show that the last part equals zero. Can you help me out?
 
Thanks! At first I thought that with those tips I could easily complete the proof. However,
I still havent found out what the last steps should be in order to get aI=aI.
I see how you got to:

aI = (AA-1 + b b'A-1) ( a - b b'A-1)

then

aI = (I + b b'A-1) ( a - b b'A-1)
aI = aI -bb'A-1+abb'A-1-bb'A-1bb'A-1 So this last part should equal zero
aI = aI + (-1+a-bb'A-1)bb'A-1

But how to continue? I tried writing out a as its original form, but I wasnt
able to show that the last part equals zero. Can you help me out?
aI = aI -bb'A-1+abb'A-1-b [ b'A-1b ] b'A-1
 
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