Chris Rorres
Oct11-09, 03:24 PM
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)
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)