A bijective "isometric" map $T:V \rightarrow V$ preserve inner product.

Question

I don't see how the first statement holds. More precisely,

Let $V$ be an inner product space over $\mathbb{C}$, $T:V \rightarrow V$ a bijective map such that $T(0)=0$ and $||Ty-Tx||= ||y-x||$. Then why is $\langle y,x \rangle = \langle Ty, Tx \rangle $?

I can prove that the real parts coincide, but not the imaginary parts since $T$ is not assumed to be linear in $\mathbb{C}$. Perhaps this is not even true, is there a counter example?

As I recall, a theorem by Ulam-Mazur guarantees that a surjective isometry is linear. — uniquesolution, Jan 23 '18 at 12:48
However, I guess that in Kaplansky's setting inner products are bilinear, unlike what analysts call inner products in the case of complex vector spaces. — Daniel Fischer, Jan 23 '18 at 13:23

score 2 · Accepted Answer · answered Jan 23 '18 at 13:25

Daniel Fisher's comment.

What if $V = \mathbb C$ is the one-dimensional space with inner product $\langle x,y\rangle = x\;\overline{y}$ . Then $\|x\| = |x|$. Let $T$ be complex conjugation, $T(x) = \overline{x}$. $$ \|Tx - Ty\| = |\overline{x}-\overline{y}| = |\overline{x-y}| = |x-y| = \|x-y\| $$ Now try $x= i, y=1$, so that $$ \langle y,x\rangle = y\;\overline{x} = 1\;\overline{i} = -i, \\ \langle Ty,Tx\rangle = \overline{y}\;\overline{\;\overline{x}\;} = \overline{1}\;i = i $$ are not equal.

A bijective "isometric" map $T:V \rightarrow V$ preserve inner product.

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