1. ## identity proof

hey guys,

i need help in proving this identity. i've been working on it for 3 days now

some notation:
A is a matrix;
A^T is A transpose;
A^-1 is A inverse;
v is a vector;
v^-1 is v inverse;
v^T is v transpose;

Assume A is a non singular matrix and v is a vector with comparable dimension. Prove:

(A + vv^T)^-1 = A^-1 - [ (A^-1v)(v^TA^-1) / (1 + v^TA^-1v) ]

Need help please. or at least some tips. Thanks!

2. Originally Posted by appsmail12
hey guys,

i need help in proving this identity. i've been working on it for 3 days now

some notation:
A is a matrix;
A^T is A transpose;
A^-1 is A inverse;
v is a vector;
v^-1 is v inverse;
v^T is v transpose;

Assume A is a non singular matrix and v is a vector with comparable dimension. Prove:

(A + vv^T)^-1 = A^-1 - [ (A^-1v)(v^TA^-1) / (1 + v^TA^-1v) ]

Need help please. or at least some tips. Thanks!
What work do you have from your three days of effort?

3. Multiply out

$\left[A^{-1}-\dfrac{A^{-1}\mathbf{v}\mathbf{v}^{T}A^{-1}}{1+\mathbf{v}^{T}A^{-1}\mathbf{v}}\right]\left(A+\mathbf{v}\mathbf{v}^{T}\right),$ and

$\left(A+\mathbf{v}\mathbf{v}^{T}\right)\left[A^{-1}-\dfrac{A^{-1}\mathbf{v}\mathbf{v}^{T}A^{-1}}{1+\mathbf{v}^{T}A^{-1}\mathbf{v}}\right],$

and show that you get $I$ both times. Keep in mind that the product $\mathbf{v}^{T}A^{-1}\mathbf{v}$ is a scalar, and thus commutes with any multiplication.

How does that work for you?

4. Solved it. Thanks Ackbeet!

5. You're welcome! Have a good one.