 Solution

The vector triple product identity gives $n\times (x\times n)=x-(n\cdot x)n$. Decompose $x=x_{\perp }+x_{\parallel }$, where $x_{\parallel }=(n\cdot x)n$ and $x_{\perp }\cdot n=0$. ThenThe linear map acts as the identity on $n^{\perp }$ and as multiplication by $\alpha$ along $n$. Consequently it is invertible exactly when $\alpha \ne 0$, withApplying $\Phi$ to this expression recovers the perpendicular and parallel components of $y$ separately.