Hermitian form induces an isomorphism between $V$ and $V^*$? Let $H(-,-):V \times V \rightarrow \mathbb{C}$ be an Hermitian form, linear in the first factor. My question is, how can one show that this induces an isomorphism from $V$ to it's dual space $V^*$? Here's what I've tried so far:
We must check that $H(-,v):V \rightarrow \mathbb{C}$ given by $H(-,v)(u)=H(u,v)$ is linear. But since $H$ is linear in the first factor, this is clear and so $H(-,v)$ takes values in $V^*=\{ \text{ linear maps } \,  V \rightarrow \mathbb{C} \, \}$.
Now consider $\Phi_H : V \rightarrow V^* $ given by $\Phi_H : v \mapsto H(-,v)$. We need that $\Phi_H$ is a vector space isomorphism. Checking the condition on vector addition is fine, but it's the scalar multiplication condition which seems to go wrong for me:
\begin{equation} 
\Phi_H(cv) = H(-,cv) = \bar{c}H(-,v) \neq c\Phi_H(v)
\end{equation}
Any advice on where I'm going wrong would be much appreciated. Thank you!

There's nothing wrong in your derivation. Actually, what the Hermitian form $H$ induces is a conjugate linear bijection, not a vector space isomorphism. However, in some sense we can say that $V$ and $V^*$ are "the same". In an infinite-dimensional Hilbert space $H$, we have analogous one-to-one correspondence between $H$ and $H^*$, i.e. the famous Riesz Representation Theorem.
