# Bijective map preserving inner products is linear

The question comes from Kaplansky's book Linear Algebra and Geometry on page 96 exercise 2

Let $V$ be a non-singular inner product space of characteristic $\neq2$. Let $T$ be a one-to-one map of $V$ onto itself, sending $0$ to $0$ and satisfying $(x-y, x-y) = (Tx - Ty, Tx - Ty)$ for all $x,y \in V$. Prove that $T$ is orthogonal.

That $T$ preserves inner products is easy, but I don't know how to prove that it is linear. Any help would be appreciated.

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## 1 Answer

Let $a, b \in F$, a field of char $\ne 2$ and $V$ be a $F$-vector space.

Noting that $T$ preserves inner products, we see $0=\langle ax+by,z \rangle - a \langle x, z \rangle - b \langle y, z \rangle =\langle T(ax+by)-aTx -bTy, Tz \rangle$ for any $x ,y, z \in V$.

As the inner product is non-singular and $T$ is onto, we have that $T(ax+by) -aTx-bTy = 0 \forall x,y \in V$ and $a,b \in F$. This proves the linearity of $T$.

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This looks like a very straightforward solution to me... is 'characteristic not two' really unnecessary?! – rschwieb Jun 27 '13 at 21:08
To prove that $T$ preserves inner products, you have to use $F$ has char $\ne 2$. – nsoum Jun 27 '13 at 21:10
Aha that's where it was hiding. Thanks! – rschwieb Jun 27 '13 at 22:59