A formula vaguely similar to Sherman and Morrison's If $V$ is a $p\times n$ matrix and $u$ is an $n$-vector, how can I prove the following equality?
$$
\frac{u^TV^T(VV^T)^{-1}Vu}{1-u^TV^T(VV^T)^{-1}Vu} = u^TV^T[V(I-uu^T)V^T]^{-1}Vu.
$$
My line of attack is to recognize the bottom denominator is a scalar, and tried to bring it into the inverse on top. However, I cannot get equivalence, does anyone know how?
 A: Consider the Formula (1) and (2) in the Woodbury matrix identity. Applied to the block matrix $M = \begin{pmatrix} 1 & u^t V^t \\ V u & VV^t\end{pmatrix}$ they give us two possible ways to calculate the $(1, 1)$-entry of $M^{-1}$:
$$(M^{-1})_{1, 1} = (1 - u^t V^t(VV^t)^{-1} Vu)^{-1}$$
$$(M^{-1})_{1, 1} = 1 + u^tV^t[VV^t - Vuu^tV^t]^{-1}Vu = 1 + u^tV^t[V(I - uu^t)V^t]^{-1}Vu$$
Now this yields your equation:
$$\begin{align*}
\frac{u^t V^t(VV^t)^{-1} Vu}{1 - u^t V^t(VV^t)^{-1} Vu} &= [1 - u^t V^t(VV^t)^{-1} Vu]^{-1} - 1 = (M^{-1})_{1, 1} - 1 \\
& = u^tV^t[V(I - uu^t)V^t]^{-1}Vu
\end{align*}$$
(Assuming that all occuring terms are well-defined).
A: This is simple consequence of Sherman-Morrison-Woodbury formula.
By a direct application the Sherman-Morrison-Woodbury formula we have $$
\begin{align}
{\begin{pmatrix} V(1 - uu^T) V^T \end{pmatrix}}^{-1} &= (VV^T-(Vu)(Vu)^T)^{-1}\\
&= (VV^T)^{-1} + \dfrac{(VV^T)^{-1}(Vu)(Vu)^T(VV^T)^{-1}}{1 - u^TV^T(VV^T)^{-1}Vu}
\end{align}
$$
Multiplying the LHS and RHS above first on the left by $(Vu)^T$ and then on the right by $Vu$ we get $u^TV^T{\begin{pmatrix} V(1 - uu^T) V^T \end{pmatrix}}^{-1}Vu = c + \dfrac{c^2}{1-c}$ where $c = u^TV^T(VV^T)^{-1}Vu.$ 
This simplifies to $\dfrac{c}{1-c}$ and the answer follows.
