TITLE: Proof relating inverse to determinant
QUESTION [13 upvotes]: I'm reading a paper regarding the consistency of a statistical estimator, and the author claimed that the following is an identity:
$$ \mathbf{x}^\top (\Sigma + \mathbf{x}\mathbf{x}^\top)^{-1}\mathbf{x} = 1- \frac{\det (\Sigma)}{\det (\Sigma+\mathbf{x}\mathbf{x}^\top)}$$
Here $\mathbf{x}$ is a vector and $\Sigma$ is a covariance matrix. Apart from that, I don't think the author has specified any more assumptions/constraints.  
While superficially it appears that this result might be provable via the Sherman-Morrison formula, I haven't been able to work it out. 
Any help/pointers would be appreciated !
Thanks.

REPLY [4 votes]: Whenever a leading/trailing principal submatrix is invertible, the determinant of a matrix is the product of the determinants of that submatrix and of its Schur complement in the enclosing matrix. Let $M=\pmatrix{1&x^T\\ x&\Sigma+xx^T}$. If $\Sigma+xx^T$ is invertible, by considering the Schur complements of respectively $1$ and $\Sigma+xx^T$ in $M$ respectively, we see that
$$
\det(\Sigma+xx^T)\det\left(1-x^T(\Sigma+xx^T)^{-1}x\right)
=\det(M)
=\det(1)\det(\Sigma).
$$
Now the result follows immediately.