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 $V$.


1 Answer 1


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.

  • $\begingroup$ can you explain more explicitly why the existence of $\phi_v$ is a direct consequence of theorem 6.8 of Friedberg? $\endgroup$
    – phy_math
    Jan 24, 2022 at 7:44
  • $\begingroup$ 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. $\endgroup$ Jan 24, 2022 at 8:16
  • $\begingroup$ conjugate linear :) $\endgroup$ Jan 24, 2022 at 19:53

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