Before diving into the Sherman-Morrison formula, Meyer in Matrix Analysis and Applied Linear Algebra, pg. 124, starts with
$$
(I+cd^T)^{-1} = I – \frac{cd^T}{1+dc^T}
$$
where $c,d$ are vectors, and says "it's straightforward to verify by direct multiplication".
I took that to mean multiply right (in essence, both) sides by $(I+cd^T)$, and see if I could reach the identity matrix on RHS, so
$$
I + cd^T – \frac{cd^T (I + cd^T)}{1+d^Tc}
$$
$$
= I + cd^T – \frac{I cd^T (1+cd^T)}{1+d^Tc}
$$
$$
= I + cd^T – cd^T = I
$$
It seemed to work.
I am wondering though how one could derive this equality by just starting from $(I+cd^T)^{-1}$. How did the mathematician go about finding that?
Best Answer
Well, $I + \alpha cd^T$ with $\alpha \in \mathbb{R}$ is the general form of a matrix that fixes vectors orthogonal to $d$ by multiplication on the left and fixes vectors orthogonal to $c^T$ by multiplication on the right. If $I + cd^T$ has an inverse, then its inverse also fixes these subspaces by multiplication and so is in this form. Now, one can solve for $\alpha$.
$$(I + cd^T) (I + \alpha cd^T) = I + cd^T + \alpha cd^T + \alpha cd^Tcd^T \\ = I + (1 + \alpha + \alpha d^T c) cd^T$$
For this to be identity, $1 + \alpha + \alpha d^T c$ must be zero, so $$\alpha = \frac{-1}{1 + d^Tc}$$