Inverse of $v \cdot v^\top$ Let's say I have a vector $v$. Now I want to calculate $(v\cdot v^\top)^{-1}$. Is there a known formula to solve this more directly than simply calculating it directly? Maybe something similar to the Sherman–Morrison formula 
$$(A+uv^\top)^{-1}=A^{-1}-{A^{-1}uv^\top A^{-1} \over 1+v^\top A^{-1}u}$$
which might be
$$(A+vv^\top)^{-1}=A^{-1}-{A^{-1}vv^\top A^{-1} \over 1+v^\top A^{-1}v}$$
for my case. However, I am still left with $A$ and the formula does not allow to be $A$ to be zero if I am not mistaken.
Or maybe there is a very efficient method to calculate $(v\cdot v^\top)^{-1}$ for such a special case?
 A: Note: $\mbox{rank}(vw^T)\leq 1$ for any $v,w\in\mathbb F^n$. So if $n\geq 2$, then $vw^T$ is not invertible.
A: $vv^\top$ is not invertible unless $v$ is a nonzero scalar, in which case $vv^\top=v^2$, so $(vv^\top)^{-1}=1/v^2$.
When $v$ is a vector, we can find another vector $u \neq \mathbf{0}$ to be orthogonal to $v$: $v^\top u=0$. Then $(vv^\top)u=v(v^\top u)= \mathbf{0}$, so $vv^\top$ is not invertible. 
One way you can find such a $u$ is to try performing Gram-Schmidt orthogonalization on the columns of the identity matrix. Here is an example: say $v=(1,1)^\top$. I will try Gram-Schmidt with the columns of the identity: $\mathbf{e}_1=(1,0)^\top$ and $\mathbf{e}_2=(0,1)^\top$. We try $\mathbf{e}_1$ first:
$$u= \mathbf{e}_1- \frac{\mathbf{e}_1^\top v}{v^\top v}v= \begin{pmatrix}1 \\ 0 \end{pmatrix}- \frac{1}{2}\begin{pmatrix}1 \\ 1 \end{pmatrix}= \frac{1}{2}\begin{pmatrix}1 \\ -1 \end{pmatrix}$$
Since this $u \neq \mathbf{0}_2$, we don't need to consider $\mathbf{e}_2$. And it's clear $v^\top u=0$. This is how in general you can always find $u \neq \mathbf{0}$ such that $v^\top u=0$.
