# Householder matrix Proof

1. Oct 11, 2009

### Chris Rorres

I'm working on trying to figure this proof out but its proving to be quite difficult does anyone have any insight?

Let u and w be vectors in (all real numbers)^n, and let I denote the (n × n) identity matrix. Let A= I + u(w^T), and assume that (w^T)u doesn’t equal -1 (notice that (w^T)u produces a scalar). Prove that
A^-1= I–au(w^T), where a = 1/(1+(w^T)u)

Last edited: Oct 11, 2009
2. Oct 11, 2009

### Office_Shredder

Staff Emeritus
The definition of the inverse of A is B so that AB=BA = Identity

So if B=I-auwT, what should you do to check that B is the inverse of A?