Unitary map in hilbert spaces, is anti isometric.

Question

If T is an anti-unitary map, prove that for all $\varphi$, $\psi\in H$, $<T\varphi,T\psi> =<\psi,\varphi>$ (Hint: Polarization.)

I have this:

$<T\varphi,T\psi>=\frac{1}{4}(|T\varphi+T\psi|^2-|T\varphi-T\psi|^2+i|T\varphi-iT\psi|^2-i|T\varphi+iT\psi|^2) =\frac{1}{4}(|T(\varphi+\psi)|^2-|T(\varphi-\psi)|^2+i|T(\varphi-\overline{i}\psi)|^2-i|T(\varphi+\overline{i}\psi)|^2) =\frac{1}{4}(|\varphi+\psi|^2-|\varphi-\psi|^2+i|\varphi+i\psi|^2-i|\varphi-i\psi|^2)$

but it is not, $<\psi,\varphi>$...

pd: I have already been able to do it. I just had to notice that $<\psi,\varphi>=\overline{<\varphi,\psi>}$

Thank you.

Answer 1

\begin{align*}\cdots &= \frac{1}{4}(|\varphi+\psi|^2-|(-1)(-\varphi+\psi)|^2+i|i(-i\varphi+\psi)|^2-i|(-i)(i\varphi+\psi)|^2)\\ &=\frac{1}{4}(|\psi+\varphi|^2-|\psi-\varphi|^2+i|\psi-i\varphi|^2-i|\psi+i\varphi)|^2)\\ &=\langle\psi,\varphi\rangle. \end{align*}