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

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!

• Is the scalar field $\mathbb{R}$ or $\mathbb{C}$? – user159517 Jun 6 at 20:34
• Ah, we are assuming $V$ and $W$ are real vector spaces! Thanks. – EBP Jun 6 at 20:35
• Are we also assuming that $L$ is surjective? – user159517 Jun 6 at 20:49
• 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. – EBP Jun 6 at 20:51

## 1 Answer

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

• 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. – EBP Jun 6 at 21:13
• Nevermind! They are equivalent. Thank you very much! – EBP Jun 6 at 21:39