An easy proof that an isometry preserving the zero vector is linear

Question

I want to show that for real inner product spaces $V$ and $W$, if $L:V\to W$ satisfies the following properties:$$\parallel L(\vec{x})-L(\vec{y})\parallel=\parallel \vec{x} -\vec{y}\parallel\\$$and $$L(\vec{0})=\vec{0},$$ then this map is linear. I am aware of the existence of the (more general) theorem of Mazur-Ulam, but I was wondering if there is more accessible proof, which is suitable for beginners in linear algebra. Thanks in advance!

Answer 1

Note that the assumptions imply that if $L(x)=0$, then $x=0$. Besides, $\|L(x)\|=\|x\|$ for all $x$.

Now, note that $\|a+b\|=\sqrt{\|a\|^2+\|b\|^2+2\langle a,\,b\rangle}$.

As a consequence, for all $x,y$, $\langle L(x),\, L(y) \rangle=\langle x,y \rangle$.

Therefore, for any scalars $\lambda_i$, for any vectors $x_i$, $$\|\sum_i{\lambda_iL(x_i)}\|=\|\sum_i{\lambda_ix_i}\|.$$

Then take $x_1=\alpha u+\beta v$, $x_2=u$, $x_3=v$, $\lambda_1=-1$, $\lambda_2=\alpha$, $\lambda_3=\beta$.