Are inner product-preserving maps always linear?

Question

Let $E,F$ be Pre-Hilbert spaces and $T: E \rightarrow F$ be a map that preserves the inner product, that is $\langle Tu , Tv \rangle = \langle u , v \rangle$ for all $u,v \in E$. Must it be true that $T$ is linear? If $T$ is surjective one has

$$\langle T(\lambda u+v), Tw\rangle = \langle \lambda u + v, w \rangle = \lambda \langle u, w \rangle + \langle v, w \rangle = \langle \lambda Tu, Tw \rangle + \langle Tv, Tw \rangle \iff \langle T(\lambda u + v) - \lambda T u - Tv, Tw\rangle = 0$$

Now since $T$ is surjective one can choose $Tw$ to be $ T(\lambda u + v) - \lambda T u + Tv$, and by positive definiteness the linearity follows. Can this somehow be extended if $T$ isn't surjective?

@AniruddhaDeshmukh I somehow thought that since any isometry is injective, it is also surjective because $E,F$ have the same dimension but that obviously already assume that $T$ is linear. Dumb mistake. — Jannik Pitt, Nov 02 '18 at 13:24
https://mathoverflow.net/questions/62380/when-do-0-preserving-isometries-have-to-be-linear — Jochen, Nov 02 '18 at 13:25
T is onto on its image (which is a pre-hilbert space) hence the proof for the surjective case applies and T is linear. — user126154, Nov 09 '18 at 10:59

score 10 · Accepted Answer · answered Nov 09 '18 at 10:11

Here is a short proof. $$\begin{split} \|T(\lambda u + w) - \lambda T(u) - T(w) \|^2 &= \langle T(\lambda u + w) - \lambda T(u) - T(w) , T(\lambda u + w) - \lambda T(u) - T(w) \rangle\\ &=\langle T(\lambda u + w) , T(\lambda u + w) \rangle + \text{more such terms}\\ &= \langle \lambda u + w , \lambda u + w\rangle + \text{more such terms}\\ &= \langle (\lambda u + w) -\lambda u-w, (\lambda u + w) -\lambda u-w\rangle= 0. \end{split}$$ In the second step, we apply linearity of the inner product to arrive at a sum of terms of the type $\langle T(a),T(b)\rangle=\langle a,b\rangle$. In the last step, we put everything back into one inner product.

Are inner product-preserving maps always linear?

1 Answers1