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Suppose we have a $f\colon \mathbb{C} \to \mathbb{C} $. I am confused when people say $f$ is $\mathbb{R}$-linear map or $\mathbb{C}$-linear map. What is the difference between these two ?

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    $\begingroup$ $a + ib \mapsto a - ib$ is $\mathbb R-$linear but not $\mathbb C-$ linear. $\endgroup$
    – user171326
    Feb 14 '15 at 11:17
  • $\begingroup$ i cant understand what u mean what are there definition of R-linear and $C$linear maps ? $\endgroup$
    – user203867
    Feb 14 '15 at 11:18
  • $\begingroup$ A $\mathbb R$ linear map is just a map $\lambda$ respecting addition such that $\lambda(\mu \cdot v) = \mu \cdot \lambda(v)$ for all $\mu \in \mathbb R$. The definition for $\mathbb C$ is the same. $\endgroup$
    – user171326
    Feb 14 '15 at 11:21
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Suppose $V$ is a vector space over two fields $F$ and $K$. Then a map $f:V\to V$ is said to be $F$-linear if $f$ is a linear transformation over the field $F$ when $V$ is regarded as a vector space over $F$. Also a map $f:V\to V$ is said to be $K$-linear if $f$ is a linear transformation over the field $K$ and in this case $V$ is regarded as a vector space over $K$.

Now clearly a $\mathbb C$-linear map $f:\mathbb C\to \mathbb C$ is $\mathbb R$-linear as $\mathbb R$ can be treated as a sub-field of $\mathbb C$. But the converse is not true i.e. a $\mathbb R$-linear map $f:\mathbb C \to\mathbb C$ need not be a $\mathbb C$-linear map (since $f(z)=\bar{z}$ is the required map which is $\mathbb R$-linear but not $\mathbb C$-linear). But you can prove the following:

a $\mathbb R$-linear map $f:\mathbb C\to \mathbb C$ is $\mathbb C$-linear if $f(i)=if(1)$.

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  • $\begingroup$ Can you explain why the lest theorem is correct? $\endgroup$ Aug 11 at 10:53
  • $\begingroup$ Just prove it. It is straightforward. $\endgroup$
    – user149418
    Aug 13 at 21:06
  • $\begingroup$ @user149418 would one just swap $i$ in the power series and 1 in for the other with an $i$ outside and see if they're equal or just start with one and show you can get to other $\endgroup$ 15 hours ago
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  1. A $\mathbb{C}$-linear map $f : \mathbb{C} \to \mathbb{C}$ needs to fulfill $f(x+y) = f(x) + f(y)$ for all $x,y \in \mathbb{C}$ and $f(ax) = af(x)$ for all $a,x \in \mathbb{C}$.

  2. A $\mathbb{R}$-linear map $f : \mathbb{C} \to \mathbb{C}$ needs to fulfill $f(x+y) = f(x) + f(y)$ for all $x,y \in \mathbb{C}$ and $f(ax) = af(x)$ for all $a \in \mathbb{R}$ and $x \in \mathbb{C}$.

So every $\mathbb{C}$-linear map as above is $\mathbb{R}$-linear, but not conversely. Consider for instance the complex conjugation $f(a+ib) = a-ib$, which is $\mathbb{R}$-linear, but not $\mathbb{C}$-linear.

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  • $\begingroup$ So, for R-linear map variable x,y are considered as complex??? $\endgroup$
    – Arashium
    Feb 14 '15 at 11:30
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    $\begingroup$ Yes, this is why is wrote $x,y \in \mathbb{C}$. If you are confused: Probably you know the definition of linear maps $f : V \to W$ for vector spaces $V$ and $W$ over a field $K$. In the case of an $\mathbb{R}$-linear map from $\mathbb{C}$ to itself, you consider $\mathbb{C}$ as an $\mathbb{R}$-vector space (of course $\mathbb{C}$ is also a field, but you ignore this fact here), i.e. in the notation above you would have $K=\mathbb{R}$ and $V=W=\mathbb{C}$. $\endgroup$
    – user44400
    Feb 14 '15 at 12:49

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