How to show $f(x) = ||x||^2 \cdot x$ is one-to-one?

Question

In the book of Analysis on Manifolds by Munkres, at page 70, as a part of the question, it is asked to show that

$f: \mathbb{R}^n \to \mathbb{R}^n $ defined by $$f(x) = ||x||^2 \cdot x$$ is one-to-one on $B(0,1)$.

However, I'm having trouble showing this. Could you provide some hint please ?

Answer 1

Hint: If $f(x)=f(y)$, then $\bigl\|f(x)\bigr\|=\bigl\|f(y)\bigr\|$.

Answer 2

Hint: Linear independence gets you almost all the way. Recall that two vectors are linear dependent if and only if one is a multiple of the other. Also note that $f$ will never make two linearly independent vectors linearly dependent.

Answer 3

Here's what I came up with:

Note that

$f(x) = \Vert x \Vert^2 x \Longrightarrow f(x) \cdot f(x) = \Vert x \Vert^4 x \cdot x = \Vert x \Vert^4 \Vert x \Vert^2 = \Vert x \Vert^6; \tag 1$

then

$f(x) = f(y) \Longrightarrow f(x) \cdot f(x) = f(y) \cdot f(y) \Longrightarrow \Vert x \Vert^6 = \Vert y \Vert^6 \Longrightarrow \Vert x \Vert = \Vert y \Vert; \tag 2$

thus if

$\Vert x \Vert^2 x = f(x) = f(y) = \Vert y \Vert^2 y, \tag 3$

and

$\Vert x \Vert = \Vert y \Vert \ne 0, \tag 4$

we have, dividing (3) through by $\Vert x \Vert^2 = \Vert y \Vert^2$,

$x = y; \tag 5$

finally,

$\Vert x \Vert = \Vert y \Vert = 0 \Longrightarrow x = y = 0. \tag 6$