Does anyone know how to calculate the inverse matrix of $I+\mu\mu^{\top}$?

Question

As stated above, I am stuck in calculating the inverse matrix of $I+\mu\mu^{\top}$, where $\mu$ is a unit vector, i.e. $\vert\mu\vert=1$ and $\vert\cdot\vert$ is $L^{2}$-norm.

Answer 1

Apply the Sherman-Morrison formula $$\eqalign{ (I+\mu\mu^\top)^{-1} &= I^{-1} - \frac{I^{-1}\mu\mu^\top I^{-1}}{1+\mu^\top I^{-1}\mu} \cr &= I - \frac{\mu\mu^\top}{1+\mu^\top\mu} \cr &= I - \frac{1}{2}\mu\mu^\top \cr }$$

Answer 2

Let $u\in L(\mathbb{R}^n)$ representing $A$ in the canonique base, thats means $u:x\to x+(\mu,x)\mu$ where $(\cdot,\cdot)$ is the inner product. Indeed $$AX=X+\mu\mu^TX=X+(\mu^TX)\mu$$

Analysis : Let $y\in\mathbb{R}^n$, we want to solve $y=x+(\mu,x)\mu$.

Take the inner product with $\mu$ : $(\mu,y)=(\mu,x)(1+|\mu|^2)=2(\mu,x)$.

Then $x=y-(\mu,x)\mu=y-\frac{1}{2}(\mu,y)\mu$.

Synthesis : $A^{-1}=I-\frac{1}{2}\mu\mu^T$.