# Existence of a conjugate linear map

This is an extension of theorem 6.8 of Friedberg Linear Algebra.

Let $$V$$ be a finite dimensional complex inner product space. I know for every linear functional $$f$$ on $$V$$, $$\exists$$ a unique vector $$w \in V$$ such that $$f(v) = \langle v, w\rangle$$. [This is theorem 6.8 in Friedberg]

Now from this, I want to know how to prove the existence of a conjugate linear map, $$\phi_V : V^* \rightarrow V$$ by $$\phi_V(f) = w$$, where $$V^*$$ is a dual space of $$V$$.

The existence of $$\phi_v$$ is a direct consequence of the theorem of representation you mention.
The only thing to prove is that $$\phi_V$$ is linear. And this is immediate using the uniqueness of representation as stated by theorem 6.8 you mention.
• can you explain more explicitly why the existence of $\phi_v$ is a direct consequence of theorem 6.8 of Friedberg? Jan 24, 2022 at 7:44
• Remember that an element of $V^*$ is a linear map from $V$ to $k$ the base field of $V$. So can be represented with an element $w$ as stated by theorem 6.8. Jan 24, 2022 at 8:16