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.
Best Answer
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.