If $\phi : M = \mathbb{R}^n \setminus \{0\} \to \mathbb{R}^n$ is defined by $\phi(x) = x + \frac{x}{|x|}$, find $\phi^{-1}$

Question

If $\phi : M = \mathbb{R}^n \setminus \{0\} \to \phi[M] \subseteq \mathbb{R}^n$ is defined by $\phi(x) = x + \frac{x}{|x|}$, find $\phi^{-1}$

I'm not sure how to go about finding $\phi^{-1}$ here. $\phi$ is clearly injective, and bijective on $M$, so $\phi^{-1}$ must exist, but I'm not entirely sure how to go about algebraically finding the inverse $\phi^{-1}(x)$.

How can I go about doing so? If it helps the boundary of $\phi[M]$ is $S^{n-1}$.

Answer 1

Geometrically, the function $\phi$ adds a unit vector to a point $x$ with the direction of this unit vector being in the same direction as $x$, and the result is a vector in the same direction as $x$. The inverse simply subtracts this unit vector. Hence, $$\phi^{-1}:\Bbb R^n\setminus \bar B_1(0)\to M,\phi^{-1}(y)=y-\frac{y}{\lvert y\rvert},$$ where $\bar B_1(0)$ is the unit closed ball centred at the origin.

Algebraically, the function $\phi$ can be seen as multiplying $x$ by the scalar $1+\frac{1}{\lvert x\rvert}$. To find the inverse, we find a scalar $c$ in terms of $y:=(1+\frac{1}{\lvert x\rvert})x$ such that $c(1+\frac{1}{\lvert x\rvert})=1$. Of course $c=\frac{\lvert x\rvert}{\lvert x\rvert+1},\lvert y\rvert=\lvert x\rvert+1$. So $c=\frac{\lvert y\rvert-1}{\lvert y\rvert}=1-\frac{1}{\lvert y\rvert}$. The inverse function $\phi^{-1}$ is just to multiply a vector $y$ by $c$ (note that $c$ depends on $y$).

Answer 2

Hint :

$\phi$ takes a point $x$, and extends it by a unit length of $x$. Thus the inverse $\phi^{-1}$ is such that it takes a point and remove that extension.

Answer 3

$R :=\phi (x) = x + \dfrac{x}{|x|} =$

$ (1+\dfrac{1}{|x|})x.$

Need to express $x$ in terms of $R$.

$|R| = |x|(1+ \dfrac{1}{|x|}) = |x| +1.$

$|x|= |R| -1.$

$1+\dfrac{1}{|x|} = \dfrac{|x|+1}{|x|} =\dfrac{ |R|}{|R|-1}$.

$x = \phi^{-1}(R) = \dfrac{|R|-1}{|R|}R.$

Answer 4

With

$\phi(x) = x + \dfrac{x}{\vert x \vert}, \tag 1$

we may immediately intuit that $\phi$ simply extends any vector $x \in \Bbb R^n \setminus \{ 0 \}$ by a unit vector in it's own direction; indeed we have

$\phi(x) = x + \dfrac{x}{\vert x \vert} = \dfrac{\vert x \vert x}{\vert x \vert} + \dfrac{x}{\vert x \vert} = \dfrac{(\vert x \vert + 1)x}{\vert x \vert }, \tag 2$

whence

$\vert \phi(x) \vert = (\vert x \vert + 1) \dfrac{\vert x \vert}{\vert x \vert} = \vert x \vert + 1. \tag 3$

In the light of (3), we may with equal ease intuit that

$\psi(y) = \phi^{-1}(y) \tag 4$

must then subtract from any $y$ a similar unit vector, thusly:

$\psi(y) = y - \dfrac{y}{\vert y \vert}; \tag 5$

we have:

$\psi(\phi(x)) = \phi(x) - \dfrac{\phi(x)}{\vert \phi(x) \vert } = \dfrac{(\vert \phi(x) \vert - 1) \phi(x)}{\vert \phi(x) \vert} = \dfrac{\vert x \vert \phi(x)}{\vert \phi(x) \vert }, \tag 6$

by virtue of (3). Combining (2) and (3) we see that

$\dfrac{\phi(x)}{\vert \phi(x) \vert } = \dfrac{x}{\vert x \vert}, \tag 7$

and using this in (6) we see that

$\psi(\phi(x)) = \phi(x) - \dfrac{\phi(x)}{\vert \phi(x) \vert } = \dfrac{\vert x \vert \phi(x)}{\vert \phi(x) \vert } = \dfrac{x \vert x \vert}{\vert x \vert} = x. \tag 8$

In a similar manner we may compute

$\phi(\psi(y)) = \psi(y) + \dfrac{\psi(y)}{\vert \psi(y) \vert}; \tag{9}$

from (5),

$\psi(y) = \dfrac{(\vert y \vert - 1)y}{\vert y \vert}, \tag{10}$

which follows in a manner similar to the way (2) follows from (1); and as (3) follows from (2) so we have

$\vert \psi(y) \vert = \vert y \vert - 1; \tag {11}$

therefore, combining (5), (9), (10) and (11),

$\phi(\psi(y)) = y - \dfrac{y}{\vert y \vert} + \dfrac{y}{\vert y \vert} = y. \tag{12}$

Bearing in mind their respective domains, we thus see that $\phi$ and $\psi$ are inverses of one another. Very clever diffeonorphisms 'twixt $\Bbb R^n \setminus \{0\}$ and $\Bbb R^n \setminus \bar B_1(0)$, indeed.