Given $A \in \mathbb{R}^{n \times n}$ non-singular and $a,b \in \mathbb{R}^n$, let $\overline{A} = A + ab^T$. If $\overline{A}$ is nonsingular then show that:
\begin{gather*} \overline{A}^{-1} = A^{-1} - \frac{A^{-1} ab^T A^{-1}}{1 + b^TA^{-1}a} \tag{A.27} \\ \end{gather*}
Attempt to alebraically show that $\overline{A} \cdot \overline{A}^{-1} = I$:
\begin{gather*} \overline{A} \cdot \overline{A}^{-1} = (A + ab^T) A^{-1} - (A + ab^T) \frac{A^{-1} ab^T A^{-1}}{1 + b^TA^{-1}a} \\ = I + a b^T A^{-1} - \frac{ab^T A^{-1} + (ab^T A^{-1})^2}{1 + b^TA^{-1}a} \\ = I + a b^T A^{-1} - \frac{ab^T A^{-1} (I + ab^T A^{-1})}{1 + b^TA^{-1}a} \\ = \left(I + \frac{1}{1 + b^TA^{-1}a} ab^T A^{-1} \right) (I + ab^T A^{-1}) \\ % = \frac{(1 + b^TA^{-1}a) (I + a b^T A^{-1}) - ab^T A^{-1} (I + ab^T A^{-1})}{1 + b^TA^{-1}a} \\ % = \frac{(I + b^TA^{-1}a I - ab^T A^{-1}) (I + a b^T A^{-1})}{1 + b^TA^{-1}a} \\ \end{gather*}
from here, I'm stuck. The algebra isn't giving me $I$.
I believe the key is to note that $$ab^T A^{-1} + ab^T A^{-1}ab^TA^{-1} = a(1 + b^TA^{-1}a)b^TA^{-1}$$ so then you get \begin{align*} \overline{A} \cdot \overline{A}^{-1} &= (A + ab^T) A^{-1} - (A + ab^T) \frac{A^{-1} ab^T A^{-1}}{1 + b^TA^{-1}a} \\ &= I + a b^T A^{-1} - \frac{ab^T A^{-1} + ab^T A^{-1}ab^TA^{-1}}{1 + b^TA^{-1}a} \\ &= I + ab^TA^{-1} - \frac{a(1 + b^TA^{-1}a)b^TA^{-1}}{1 + b^TA^{-1}a} \end{align*} And then you are done since the quantity $(1 + b^TA^{-1}a)$ is just a scalar.