We are to find under which conditions for $u$ and $v$ the rank-one update of the identity $A = I - uv^{T}$, is invertible.
Apparenetly the condition is simply that $1 + v^T u \neq 0$. I can see that we must require that $Ax \neq 0$ for all $x$ which leads me to $x \neq (v^Tx)u$.
This is where I get stuck. The solution somehow concludes that is becomes sufficient to check that $Au \neq 0$. What is the intuition about this?
The necessary and sufficient condition for $I-uv^T$ to be invertible is not $1+v^Tu\ne0$, but $1-v^Tu\ne0$, because \begin{align} &I-uv^T\text{ is singular}\\ &\iff \exists x\ne0 \text{ such that } (I-uv^T)x=0\\ &\iff \exists x\ne0 \text{ such that } x=u(v^Tx)\tag{1}\\ &\iff \exists y\ne0 \text{ such that } y=u(v^Ty) \text{ and } v^Ty=1\tag{2}\\ &\iff v^Tu=1\tag{3},\\ &\iff 1-v^Tu=0,\\ \end{align} where we take $y=\frac{1}{v^Tx}x$ in proving that $(1)\Rightarrow(2)$ and we take $y=u$in proving that $(3)\Rightarrow(2)$.
Since $Au=(I-uv^T)u=u-u(v^Tu)=(1-v^Tu)u$, when $Au\ne0$, we must have $1-v^Tu\ne0$. Hence the condition $Au\ne0$ is sufficient.