The Unitary group was defined as follows: $U_n = \{g \in GL_n(\mathbb{C}):<gu,gv>=<u,v> u,v\in \mathbb C^n\}$ Where $<u,v>=\sum_1^n u_i \bar v_i$
I'm trying to prove that a matrix $g$ is in the Unitary Group iff $g\bar{g}^t=I $
My attempt (for =>) is to take specific vectors in $\mathbb C^n$. For example if I take $u_i=v_i=(0,...0,1,0,..0)$ where the $1$ is at the $i$th place I get that $g_i \bar g_i^t=1$ for all $i$ and that if $i\neq j$ than $g_i \bar g_j^t=0$. Does it follow that $g\bar{g}^t=I$ ?
I don't know where to start proving the other direction. Or is it exactly the same going backwards?
Use the theorem about the existence of adjoint operator:
$$gg^t=I\implies\;\forall\,u\in \Bbb C^n\;,\;\;\langle u,\,u\rangle=\langle u,\,g^*gu\rangle=\langle gu,\,gu\rangle$$
and we're done, since in unitary linear spaces, $\;g^*=\overline g^t=\overline{g^t}\;$
The other direction: for all $\;u\in\Bbb C^n\;$ ,we have
$$\langle u,\,u\rangle=\langle gu,\,gu\rangle=\langle u,\,g^*gu\rangle\implies g^*g=I$$