Prove $(\mathbf{A}+\mathbf{u}\mathbf{v}^{\text{T}})^{-1}=\cdots$ Problem:
Assuming $\mathbf{A}$ and $\mathbf{A}+\mathbf{u}\mathbf{v}^{\text{T}}$ are nonsingular, prove
\begin{equation}
(\mathbf{A}+\mathbf{u}\mathbf{v}^{\text{T}})^{-1}=\mathbf{A}^{-1}-\frac{\mathbf{A}^{-1}\mathbf{uv}^{\text{T}}\mathbf{A}^{-1}}{(1+\mathbf{v^{\text{T}}\mathbf{A}^{-1}\mathbf{u}})}
\end{equation}
My Attempt
Since $\mathbf{A}+\mathbf{u}\mathbf{v}^{\text{T}}$ is nonsingular, then the inverse of its inverse exists. And using the fact that
\begin{equation}
(\mathbf{A}+\mathbf{u}\mathbf{v}^{\text{T}})^{-1}(\mathbf{A}+\mathbf{u}\mathbf{v}^{\text{T}})=\mathbf{I},
\end{equation}
then we want to show that
\begin{equation}
\left[\mathbf{A}^{-1}-\frac{\mathbf{A}^{-1}\mathbf{uv}^{\text{T}}\mathbf{A}^{-1}}{(1+\mathbf{v^{\text{T}}\mathbf{A}^{-1}\mathbf{u}})}\right](\mathbf{A}+\mathbf{u}\mathbf{v}^{\text{T}})=\mathbf{I}.
\end{equation}
Simpliying this leads to the following
\begin{align}
\mathbf{A}^{-1}(\mathbf{A}+\mathbf{u}\mathbf{v}^{\text{T}})-\frac{\mathbf{A}^{-1}\mathbf{uv}^{\text{T}}\mathbf{A}^{-1}}{(1+\mathbf{v}^{\text{T}}\mathbf{A}^{-1}\mathbf{u})}(\mathbf{A}+\mathbf{u}\mathbf{v}^{\text{T}})&\overset{?}{=}\mathbf{I}\\
\mathbf{I}+\mathbf{A}^{-1}\mathbf{uv}^{\text{T}}-\frac{\mathbf{A}^{-1}\mathbf{uv}^{\text{T}}+\mathbf{A}^{-1}\mathbf{uv}^{\text{T}}\mathbf{A}^{-1}\mathbf{uv}^{\text{T}}}{(1+\mathbf{v}^{\text{T}}\mathbf{A}^{-1}\mathbf{u})}&\overset{?}{=}\mathbf{I}\\
\frac{(1+\mathbf{v}^{\text{T}}\mathbf{A}^{-1}\mathbf{u})(\mathbf{I}+\mathbf{A}^{-1}\mathbf{uv}^{\text{T}})-\mathbf{A}^{-1}\mathbf{uv}^{\text{T}}(\mathbf{I}+\mathbf{A}^{-1}\mathbf{uv}^{\text{T}})}{(1+\mathbf{v}^{\text{T}}\mathbf{A}^{-1}\mathbf{u})}&\overset{?}{=}\mathbf{I}
\end{align}
For brevity, let the scalar $(1+\mathbf{v}^{\text{T}}\mathbf{A}^{-1}\mathbf{u})$ factor be $\alpha$. Then
\begin{equation}
\frac{\alpha(\mathbf{I}+\mathbf{A}^{-1}\mathbf{uv}^{\text{T}})-\mathbf{A}^{-1}\mathbf{uv}^{\text{T}}(\mathbf{I}+\mathbf{A}^{-1}\mathbf{uv}^{\text{T}})}{\alpha}\overset{?}{=}\mathbf{I}.
\end{equation}
I am stuck here.
 A: You are almost there. Just notice $\mathbf{v}^T \mathbf{A}^{-1} \mathbf{u}$ is a scalar, so it can be moved from the middle into front:
$$\frac{\mathbf{A}^{-1}\mathbf{uv}^{\text{T}}+\mathbf{A}^{-1}\mathbf{uv}^{\text{T}}\mathbf{A}^{-1}\mathbf{uv}^{\text{T}}}{(1+\mathbf{v}^{\text{T}}\mathbf{A}^{-1}\mathbf{u})} = \frac{\mathbf{A}^{-1}\mathbf{uv}^{\text{T}}+(\mathbf{v}^{\text{T}}\mathbf{A}^{-1}\mathbf{u})\mathbf{A}^{-1}\mathbf{u}\mathbf{v}^{\text{T}}}{(1+\mathbf{v}^{\text{T}}\mathbf{A}^{-1}\mathbf{u})} = \frac{(1+\mathbf{v}^{\text{T}}\mathbf{A}^{-1}\mathbf{u})\mathbf{A}^{-1}\mathbf{u}\mathbf{v}^{\text{T}}}{1+\mathbf{v}^{\text{T}}\mathbf{A}^{-1}\mathbf{u}} =\mathbf{A}^{-1}\mathbf{u}\mathbf{v}^{\text{T}}$$
A: As a side note, I prefer rewriting the equality as
$$
(I+xv^T)^{-1} = I - \frac{xv^T}{1+v^Tx}.
$$
where $x=A^{-1}u$. The merit of doing so is that, we immediately see why the inverse of the rank-1 update of a matrix is a rank-1 update of the inverse: by Cayley-Hamilton theorem, $(I+xv^T)^{-1}$ is a polynomial in $I+xv^T$. Therefore, $(I+xv^T)^{-1}$ must be in the form of $aI+bxv^T$ because every nonnegative integer power of $I+xv^T$ is of this form. But $a$ must be equal to $1$ because $(I+0v^T)^{-1}=I$. Therefore $(I+xv^T)^{-1}=I+bxv^T$, which is a rank-1 update of $I^{-1}=I$.
We can also determine the coefficient $b$ easily. Expand the LHS of the equation $(I+xv^T)(I+bxv^T)=I$, we get $I+(1+b+bv^Tx)xv^T=I$. Hence $b=-1/(1+v^Tx)$.
