A simple matrix inverse problem

Closed form inverse of $I-\sigma uu^T$ for $\sigma \neq 1$.

We are interested in the matrix $A = I - \sigma uu^T$ where $u\in \mathbb R^n$ such that $||u||=1$.

From an answer of this post, we can easily see that $A=\text{diag} ( 1-\sigma, 1, ..., 1)$ in the basis $\{ v_1=u, v_2,..., v_n\}$.

We thus can easily see that if $\sigma \neq 1$, then $A$ is invertible. An interesting question is that

What is the closed form inverse of $A$?

This question is not very complicated and not simple since the construction of $v_2, ..., v_n$ is not obvious.

So this note is to provide a closed form inverse of $A$, which is
$$A^{-1} = I + \frac{\sigma}{1-\sigma ||u||^2} uu^T.$$

We can easily check that $A.A^{-1} = I$. But the interesting point is that how do we construct such formula for $A^{-1}$. The trick is as follows.

First, let
$$y = Ax = x - \sigma \langle u, x\rangle u.$$
Now, if we can recover $x$ form $y$, we will get the inverse of $A$.
Consider
$$\langle u, y \rangle= \langle u, x \rangle - \sigma \langle u, x\rangle ||u||^2.$$
Now we get
$$\langle u, x\rangle= \frac{\langle u, y\rangle}{1- \sigma ||u||^2}.$$
Substitute this expression to the first expression of $y$, we have
$$y = x - \frac{\sigma}{1-\sigma ||u||^2} uu^T y$$
or
$$x = y + \frac{\sigma}{1-\sigma ||u||^2} uu^T y=A^{-1} y.$$

We are done.