Let $T:\mathcal{V}\to\mathcal{W}$ be a linear map and let it be bijective. How can I prove now that its conjugate transpose, $A^*$ is bijective too?
My idea was that if $T$ is bijective then there exists a $T^{-1}$, so we know that $\det{A}\ne 0$, where $T(v)=Av$. Now we know that $\det{A^*}=\overline{\det{A}}\ne 0$, so we know $T^*$ has also an inverse $T^{*-1}$ and that means it should also be bijective?
Just use the fact that $(U\circ T)^*=\operatorname{Id}^*=\operatorname{Id}$, but also $\,(U\circ T)^*= T^*\circ U^*$.
This proves $T^*$ and $U^*$ are inverse to each other.