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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!

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    $\begingroup$ Is the scalar field $\mathbb{R}$ or $\mathbb{C}$? $\endgroup$ – user159517 Jun 6 at 20:34
  • $\begingroup$ Ah, we are assuming $V$ and $W$ are real vector spaces! Thanks. $\endgroup$ – EBP Jun 6 at 20:35
  • $\begingroup$ Are we also assuming that $L$ is surjective? $\endgroup$ – user159517 Jun 6 at 20:49
  • $\begingroup$ No, the assumptions I mentioned are all the assumptions we make. However, we could perhaps restrict the codomain to the image of $L$ in order to get some kind of surjectivity. $\endgroup$ – EBP Jun 6 at 20:51
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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 $x_1=\alpha u+\beta v$, $x_2=u$, $x_3=v$, $\lambda_1=-1$, $\lambda_2=\alpha$, $\lambda_3=\beta$.

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  • $\begingroup$ Thank you! While reading your comment I realised I made a mistake in the properties, I don't think this works for the current ones. $\endgroup$ – EBP Jun 6 at 21:13
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    $\begingroup$ Nevermind! They are equivalent. Thank you very much! $\endgroup$ – EBP Jun 6 at 21:39

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