Given that the complex numbers $\mathbb{C}$ are considered to be canonically isomorphic to $\mathbb{R^2}$, I am wondering if the dual space of $\mathbb{C}$ is the same es that of $\mathbb{R}$, i.e. the vector space of linear functionals. I never heard about dual space of $\mathbb{C}$ until today, so I couldnt find much information in the litterature. Also:

  • How is the dual of $\mathbb{C}$ defined, what is a dual basis of $\mathbb{C}$ ?
  • What is the field $K$ in which the linear functionals get their values ($\mathbb{C}$ or $\mathbb{R}$) ?

Many thanks.

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    $\begingroup$ The dual of any vector space $V$ over a field $K$ is the space of linear functionals $f:V\to K$. $\endgroup$
    – A. Goodier
    Mar 5, 2018 at 17:50
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    $\begingroup$ For a finite dimensional vector space $\mathbb{F}^n$ (possibly barring some characteristic 2 issues - I'm a bit cloudy on it right now and fairly distraught, so I'm not sure about this), the dual space is $\mathbb{F}^n$. $\endgroup$ Mar 5, 2018 at 17:51
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    $\begingroup$ @CameronWilliams Why should characteristic $2$ be special here? It isn't. $\endgroup$
    – egreg
    Mar 5, 2018 at 18:17
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    $\begingroup$ The Question has to tell Readers whether $\mathbb C$ is being considered a vector space over the field $\mathbb R$ or over the field $\mathbb C$. $\endgroup$
    – hardmath
    Mar 5, 2018 at 19:20
  • $\begingroup$ @egreg I didn't think it was but I was having a particularly bad day today so I couldn't think well haha. $\endgroup$ Mar 5, 2018 at 22:40

1 Answer 1


The answer to your questions depends upon your answer to this question: do you see $\mathbb C$ as a real vector space or as a complex vector space?

Since you mention $\mathbb{R}^2$, I'll assume that you see it as a real vector space. In that case:

  • A basis of $\mathbb{C}^*$ is $\{\alpha,\beta\}$, with $\alpha(z)=\operatorname{Re}z$ and $\beta(z)=\operatorname{Im}z$.
  • The functionals get their values in $\mathbb R$.

On the other hand, if you see $\mathbb C$ as a complex vector space, then you can take any map $\alpha\colon\mathbb{C}\longrightarrow\mathbb{C}$ of the type $\alpha(z)=az$ (as long as $a\neq0$) and $\{\alpha\}$ will be a basis of $\mathbb{C}^*$. For instance, you can take $\alpha=\operatorname{Id}$ (which corresponds to choosing $a=1$).

  • $\begingroup$ Many thanks. So in that case is then obvious that $\alpha (1)=1, \alpha (i)=0$ and $\beta (1)=0, \beta (i)=1.$ How about if one considered $\mathbb{C}$ as a complex vector space ? $\endgroup$
    – user249018
    Mar 5, 2018 at 17:59
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    $\begingroup$ @user249018 Indeed. Of course, a shorter answer to the first question would have been $\{\operatorname{Re},\operatorname{Im}\}$. $\endgroup$ Mar 5, 2018 at 18:01
  • $\begingroup$ Since {1} is the only basis vector of $\mathbb{C}$ considered as a complex vector space, one would expect a single dual basis vector, say $\theta$. I am wondering if $\theta (1)=1$ or $0$ ? $\endgroup$
    – user249018
    Mar 5, 2018 at 18:10
  • $\begingroup$ @user249018 Acutaly, for every $z\in\mathbb{C}\setminus\{0\}$, $\{z\}$ is a basis of $\mathbb C$, if we see it as a complex vector space. But, yes, then $\mathbb{C}^*$ will be $1$-dimensional. Please see the paragraph that I've added. $\endgroup$ Mar 5, 2018 at 18:15
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    $\begingroup$ @user249018 No. What I wrote is what I meant: for each $z\in\mathbb{C}\setminus\{0\}$, $\{z\}$ is a basis of $\mathbb C$, if we see it as a complex vector space. Why do you ask? Do you know some non-zero complex number $z$ such that $\{z\}$ is not a basis of $\mathbb C$? $\endgroup$ Mar 5, 2018 at 18:26

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