Prove $\langle u,v \rangle = uAv^* $ defines an inner product.

Question

Consider $V= \Bbb C^2 $ and $A=\begin{bmatrix}1&i\\-i&2\end{bmatrix}$.

Prove that $\langle u,v \rangle = uAv^* $ defines an inner product.

To prove the 3rd condition: $ \overline {\langle u,v \rangle} = \langle v,u \rangle $ where bar denotes conjugate,

the solution states that

$$\overline {\langle u,v \rangle} = \overline {uAv^*} = (uAv^*)^*.$$

I do not understand the step where the conjugate is removed and replaced with a complex conjugate instead.

May I know what is the logic behind this step?

Thank you!

Answer 1

$\langle u,v \rangle \in \Bbb{C}$, so taking transpose will leave it unchanged. Hence $$\overline {\langle u,v \rangle} = \overline {uAv^*} = \left( \overline {uAv^*} \right)^T = (uAv^*)^* = vA^*u^*= vAu^*= \langle {v,u} \rangle.$$