Matrix determinant lemma derivation While reading this wikipedia article on the determinant lemma, I stumbled upon this expression (in a proof section):
\begin{equation}
\begin{pmatrix} \mathbf{I} & 0 \\ \mathbf{v}^\mathrm{T} & 1 \end{pmatrix}
\begin{pmatrix} \mathbf{I}+\mathbf{uv}^\mathrm{T} & \mathbf{u} \\ 0 & 1 \end{pmatrix}
\begin{pmatrix} \mathbf{I} & 0 \\ -\mathbf{v}^\mathrm{T} & 1 \end{pmatrix} =
\begin{pmatrix} \mathbf{I} & \mathbf{u} \\ 0 & 1 + \mathbf{v}^\mathrm{T}\mathbf{u} \end{pmatrix}.
\end{equation}
Although I see that this equation "works", I'm interested in HOW this thing was invented. For example, why we have $u$ term in a central block matrix of the left side?
UPD
A little clarification of the question above.
Let
\begin{equation}L = \begin{pmatrix} \mathbf{I} & 0 \\ \mathbf{v}^\mathrm{T} & 1 \end{pmatrix} \end{equation}
I see that
\begin{equation}
L^{-1}=
\begin{pmatrix} \mathbf{I} & 0 \\ -\mathbf{v}^\mathrm{T} & 1 \end{pmatrix} 
\end{equation}
and hence the first equation looks like 
\begin{equation}
L\begin{pmatrix}\mathbf{I + uv^T} & \mathbf{u} \\ 0 & 1\end{pmatrix}L^{-1} = 
\begin{pmatrix} \mathbf{I} & \mathbf{u} \\ 0 & 1 + \mathbf{v}^\mathrm{T}\mathbf{u} \end{pmatrix}.
\end{equation}
I see that $\det(L) = \det(L^{-1}) = 1 $. Hence determinants of RHS and LHS are equal as well.
What I do not understand is how we jumped from simple $\begin{pmatrix}\mathbf{I + uv^T} & \mathbf{0} \\ 0 & 1\end{pmatrix}$ or $\begin{pmatrix}\mathbf{I} & \mathbf{u} \\ \mathbf{-v}^T & 1\end{pmatrix}$ to $\begin{pmatrix}\mathbf{I + uv^T} & \mathbf{u} \\ 0 & 1\end{pmatrix}$ for a central part of LHS.
Thank you.
 A: This is just Gaussian elimination. We begin with $\pmatrix{I+uv^T&0\\ 0&1}$ and we want to eliminate the summand $uv^T$ in the first sub-block using row or column additions (so that the determinant is preserved):
\begin{align*}
\pmatrix{I+uv^T&0\\ 0&1}\pmatrix{I&0\\ -v^T&1}&=\pmatrix{I+uv^T&0\\ -v^T&1},\\
\pmatrix{I&u\\ 0&1}\pmatrix{I+uv^T&0\\ -v^T&1}&=\pmatrix{I&u\\ -v^T&1},\\
\pmatrix{I&0\\ v^T&1}\pmatrix{I&u\\ -v^T&1}&=\pmatrix{I&u\\ 0&1+v^Tu}.
\end{align*}
Put them together, we get
$$
\pmatrix{I&0\\ v^T&1}\pmatrix{I&u\\ 0&1}\pmatrix{I+uv^T&0\\ 0&1}\pmatrix{I&0\\ -v^T&1}=\pmatrix{I&u\\ 0&1+v^Tu}.
$$
If you want, you may perform a further column addition to make the RHS block-diagonal:
$$
\pmatrix{I&0\\ v^T&1}\pmatrix{I&u\\ 0&1}\pmatrix{I+uv^T&0\\ 0&1}\pmatrix{I&0\\ -v^T&1}\pmatrix{I&-u\\ 0&1}=\pmatrix{I&0\\ 0&1+v^Tu}.
$$
Sherman–Morrison–Woodbury formula and Schur complement are also derived in the same spirit.
A: I have no idea how this was invented and what was the original motivation, but let me outline a different proof for the formula $\det(I + uv^T) = 1 + v^T u$ which I think is much less magical and more natural.
Set 
$$ B = uv^T = \begin{pmatrix} u_1 v_1 & \dots & u_1 v_n \\
u_2 v_1 & \dots & u_2 v_n \\
\vdots & \ddots & \vdots \\
u_n v_1 & \dots  & u_n v_n \end{pmatrix} \in M_n(\mathbb{F}). $$
Let us start by finding the characteristic polynomial $p_B(x) = \det(B - xI)$ of $B$. Since $B$ has $\operatorname{rank}(B) \leq 1$, we know that it has at least $n - 1$ eigenvectors associated to the eigenvalue $0$. Since the sum of all eigenvalues must be $\operatorname{tr}(B) = v^T u$, we see that
$$ p_B(x) = \det(B - xI) = (-1)^{n} x^{n-1}(x - v^T u). $$
Now plug in $x = (-1)$ and deduce the required formula:
$$ p_B(1) = \det(B - (-1)I) = \det(B + I) = \det(I + uv^T) = (-1)^n (-1)^{n-1} (-1 - v^T u) = 1 + v^T u.$$
