How can the Sherman-Morrison formula simplify matrix inversion?

[nikozm]

Hi,

I am trying to derive the Sherman-Morrison formula for the following expression:
a*x*(b*AA^H+I)^-1*y, where a, b are non-negative scalar values. Also, vectors and matrices are represented by lowercase bold typeface and uppercase bold typeface letters, respectively. Also, (.)^H and (.)^-1 are the Hermitian transposition and inverse operation, respectively.

Any help would be useful.

 

[Hawkeye18]

You need to give more details.

Sherman-Morrison formula deals with the inversion of a rank one perturbation of an invertible matrix $A$. Assuming that $A^{-1}$ is already computed, the formula gives you simple algebraic expression for the inverse of the perturbation; simple means that it does not involve any other inversions, only (matrix) additions and multiplications.

What do you want here? You should be more specific.
Is your matrix $A$ a "tall" one? What inverse you assume to be known?