Are inner product-preserving maps always linear?

Solution 1:

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.