To prove that f is one-one where $ f (x)=x \langle x,x\rangle$

Question

Let $f$ be a function from $\mathbb {R^n}$ to $\mathbb {R^n}$ and $f (x)=x \langle x,x \rangle$. If I define the inner product as $$\langle x,y \rangle =\sum_{i=1}^{n} x_i y_i $$ then $f$ is one one but what if the inner product is defined some other way? Then how to prove that $f$ is one one?

Answer 1

Note that $f(x)=x\|x\|^2$. So, if $x\neq0$, $f(x)$ is equal to $x$ times a number greater than $0$. Therefore, $f(x)=f(y)$ implies that $x=\lambda y$ for some $\lambda>0$. But then $f(y)=f(x)=f(\lambda y)=\lambda^3 f(y)$. So, $\lambda^3=1$ and so $\lambda=1$. In other words, $x=y$.

Note that this proof does not depend on the definition of the definition of the inner product and therefore can be used for every possible choice of inner product.