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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!

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Actually, $\Phi_H$ is a vector space isomorphism if you start to interpret the situation differently. That is, a Hermitian form is really a bilinear map $V\times \overline V \to \mathbb C$, where the scalar multiplication in $\overline V$ is given by: $$\lambda \cdot v := \bar \lambda \cdot_V v$$

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  • $\begingroup$ Of course, how did I miss that?! Thank you for your answer! $\endgroup$ – mathphys Nov 3 '17 at 21:41
  • $\begingroup$ I guess you missed that because it is rarely pointed out, I think. $\endgroup$ – Stefan Perko Nov 3 '17 at 21:44
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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.

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