R-Linear and C-Linear Mapings .... ....

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In summary, the conversation discusses the definition and examples of $\mathbb{R}$-linear and $\mathbb{C}$-linear mappings of $\mathbb{C}$ into $\mathbb{C}$, with a focus on Chapter 0, Section 1.2 of Reinhold Remmert's book "Theory of Complex Functions." The definition of these mappings is given as satisfying the conditions of additivity and scalar multiplication, with the only difference being that $\mathbb{C}$-linear mappings must satisfy the scalar multiplication condition for all complex numbers while $\mathbb{R}$-linear mappings only need to satisfy
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R-Linear and C-Linear Mappings...

I am reading Reinhold Remmert's book "Theory of Complex Functions" ...I am focused on Chapter 0: Complex Numbers and Continuous Functions ... and in particular on Section 1.2:\(\displaystyle \mathbb{R}\)-linear and \(\displaystyle \mathbb{C}\)-linear mappings of \(\displaystyle \mathbb{C}\) into \(\displaystyle \mathbb{C}\) ... ... I need help in order to get a clear idea of the definition and nature of \(\displaystyle \mathbb{R}\)-linear and \(\displaystyle \mathbb{C}\)-linear mappings of \mathbb{C} into \mathbb{C} ... ...

Remmert's section on \(\displaystyle \mathbb{R}\)-linear and \(\displaystyle \mathbb{C}\)-linear mappings of \(\displaystyle \mathbb{C}\) into \(\displaystyle \mathbb{C}\) reads as follows:View attachment 8534
View attachment 8535It seems to me (this may be unfair and I just may be missing the point! ... ) that Remmert has not clearly defined \(\displaystyle \mathbb{R}\)-linear and \(\displaystyle \mathbb{C}\)-linear mappings ...

Can someone please give a first-principles definition of \(\displaystyle \mathbb{R}\)-linear and \(\displaystyle \mathbb{C}\)-linear mappings ... and some simple examples ...Hope someone can help ... ... help will be appreciated ...

Peter
 

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  • Remmert - 2 - R-linear and C-linear Mappings, Ch. 0, Section 1.2 ... PART 2 .png
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  • #2
Peter said:
It seems to me (this may be unfair and I just may be missing the point! ... ) that Remmert has not clearly defined \(\displaystyle \mathbb{R}\)-linear and \(\displaystyle \mathbb{C}\)-linear mappings ...

Can someone please give a first-principles definition of \(\displaystyle \mathbb{R}\)-linear and \(\displaystyle \mathbb{C}\)-linear mappings ... and some simple examples ...
If $V$ and $W$ are vector spaces over a field $\Bbb{F}$, then a map $f:V\to W$ is defined to be $\Bbb{F}$-linear if it satisfies the conditions \[(1)\text{ Additivity: }\ f(x+y) = f(x) + f(y) \text{ for all }x,y\in V,\] \[(2)\text{ Scalar multiplication: }\ f(\lambda x) = \lambda f(x) \text{ for all }x\in V,\;\lambda\in\Bbb{F}.\]

If we identify the spaces $\Bbb{R}^2$ and $\Bbb{C}$ in the usual way by the correspondence $(x,y) \leftrightarrow x+iy$, then we can regard $\Bbb{C}$ as either a (two-dimensional) vector space over $\Bbb{R}$ or a (one-dimensional) space over $\Bbb{C}$. With that identification, we can ask whether a given a map $f:\Bbb{C} \to \Bbb{C}$ is $\Bbb{R}$-linear or $\Bbb{C}$-linear. In each case the conditions (1) and (2) must be satisfied.

The condition (1) does not explicitly mention the underlying field. So the map $f$ will be $\Bbb{R}$-additive if and only if it is $\Bbb{C}$-additive. But the scalar multiplication condition (2) is different in the two cases. In fact, the condition $f(\lambda z) = \lambda f(z)$ only has to be satisfied for real $\lambda$ for the map to be $\Bbb{R}$-linear. But it has to satisfy the stronger condition (namely for all complex $\lambda$) for it to be $\Bbb{C}$-linear.

The simplest concrete example to illustrate this is the complex conjugation map $f: x+iy\mapsto x-iy$. This is certainly additive. It is $\Bbb{R}$-linear, because \[f(\lambda(x+iy)) = f(\lambda x+i\lambda y) = \lambda x-i\lambda y = \lambda (x-i y) = \lambda f(x+iy)\] for all real $\lambda$.

But that map is not $\Bbb{C}$-linear, because if $\lambda = \mu+i\nu$ then \[f(\lambda(x+iy)) = f((\mu+i\nu)(x+iy)) = f((\mu x - \nu y) + i(\nu x + \mu y)) = (\mu x - \nu y) - i(\nu x + \mu y).\] But \[\lambda f(x+iy) = (\mu+i\nu)(x-iy) = (\mu x + \nu y) + (\nu x -\mu y),\] which is clearly different from $f(\lambda(x+iy))$ whenever $\nu\ne0$.
 
  • #3
Opalg said:
If $V$ and $W$ are vector spaces over a field $\Bbb{F}$, then a map $f:V\to W$ is defined to be $\Bbb{F}$-linear if it satisfies the conditions \[(1)\text{ Additivity: }\ f(x+y) = f(x) + f(y) \text{ for all }x,y\in V,\] \[(2)\text{ Scalar multiplication: }\ f(\lambda x) = \lambda f(x) \text{ for all }x\in V,\;\lambda\in\Bbb{F}.\]

If we identify the spaces $\Bbb{R}^2$ and $\Bbb{C}$ in the usual way by the correspondence $(x,y) \leftrightarrow x+iy$, then we can regard $\Bbb{C}$ as either a (two-dimensional) vector space over $\Bbb{R}$ or a (one-dimensional) space over $\Bbb{C}$. With that identification, we can ask whether a given a map $f:\Bbb{C} \to \Bbb{C}$ is $\Bbb{R}$-linear or $\Bbb{C}$-linear. In each case the conditions (1) and (2) must be satisfied.

The condition (1) does not explicitly mention the underlying field. So the map $f$ will be $\Bbb{R}$-additive if and only if it is $\Bbb{C}$-additive. But the scalar multiplication condition (2) is different in the two cases. In fact, the condition $f(\lambda z) = \lambda f(z)$ only has to be satisfied for real $\lambda$ for the map to be $\Bbb{R}$-linear. But it has to satisfy the stronger condition (namely for all complex $\lambda$) for it to be $\Bbb{C}$-linear.

The simplest concrete example to illustrate this is the complex conjugation map $f: x+iy\mapsto x-iy$. This is certainly additive. It is $\Bbb{R}$-linear, because \[f(\lambda(x+iy)) = f(\lambda x+i\lambda y) = \lambda x-i\lambda y = \lambda (x-i y) = \lambda f(x+iy)\] for all real $\lambda$.

But that map is not $\Bbb{C}$-linear, because if $\lambda = \mu+i\nu$ then \[f(\lambda(x+iy)) = f((\mu+i\nu)(x+iy)) = f((\mu x - \nu y) + i(\nu x + \mu y)) = (\mu x - \nu y) - i(\nu x + \mu y).\] But \[\lambda f(x+iy) = (\mu+i\nu)(x-iy) = (\mu x + \nu y) + (\nu x -\mu y),\] which is clearly different from $f(\lambda(x+iy))$ whenever $\nu\ne0$.
Hi Opalg,

Thanks so much for your post ... it was most helpful!

But ... just some clarifications ...

Are \(\displaystyle \mathbb{R}\)-linear mappings of \(\displaystyle \mathbb{C}\) into \(\displaystyle \mathbb{C}\) restricted to the situation where \(\displaystyle \mathbb{C}\) is regarded as a vector space of \(\displaystyle \mathbb{R}^2\) over \(\displaystyle \mathbb{R}\) ... ... or can \(\displaystyle \mathbb{R}\)-linear mappings of \(\displaystyle \mathbb{C}\) into \(\displaystyle \mathbb{C}\) be legitimately regarded as occurring in \(\displaystyle \mathbb{C}\) regarded as a vector space of \(\displaystyle \mathbb{C}\) over \(\displaystyle \mathbb{C}\) ... ... ? (I am guessing that an \(\displaystyle \mathbb{R}\)-linear mapping can occur in both ... )Just for the record ... ... my understanding of an \(\displaystyle \mathbb{R}\)-linear mapping for \(\displaystyle \mathbb{C}\) regarded as a vector space of \(\displaystyle \mathbb{R}^2\) over \(\displaystyle \mathbb{R}\) is as follows:

... an \(\displaystyle \mathbb{R}\)-linear mapping for \(\displaystyle \mathbb{C}\) regarded as a vector space of \(\displaystyle \mathbb{R}^2\) over \(\displaystyle \mathbb{R}\) is a function/map \(\displaystyle f : \mathbb{R}^2 \to \mathbb{R}^2\) that satisfies the two conditions ...

(1) Additivity : \(\displaystyle f( \ (x,y) + (v,w) \ ) = f( \ (x,y) \ ) + f( \ (v,w) \ )\)

(2) Scalar Multiplication : \(\displaystyle f( \ \lambda (x,y) \ ) = \lambda f( \ (x,y) \ )\)Now ...for the conjugation map ... we consider ...

\(\displaystyle f : (x,y) \mapsto (x, -y)\)Now ... we have:

\(\displaystyle f( \ (x,y) + (v,w) \ ) = f( \ (x + v, y + w) \ ) = (x + v, -y - w) = (x,-y) + (v,-w) = f( \ (x,y) \ ) + f( \ (v,w) \ )\)

and

\(\displaystyle f( \ \lambda (x,y) \ ) = f( \ (\lambda x, \lambda y) \ ) = (\lambda x, - \lambda y) = \lambda (x, -y) = \lambda f( \ (x,y) \ ) \)
Is that correct?BUT ... given the above is a real vector space ...

... presumably we would not write \(\displaystyle \[f(\lambda(x+iy)) = f(\lambda x+i\lambda y) = \lambda x-i\lambda y = \lambda (x-i y) = \lambda f(x+iy)\]\) (for all real $\lambda$) because we are essentially dealing with a real vector space of \(\displaystyle \mathbb{R}^2\) over \(\displaystyle \mathbb{R}\)?

Indeed ... should \(\displaystyle i \) be in a real vector space of \(\displaystyle \mathbb{R}^2\) over \(\displaystyle \mathbb{R}\) ...
Can you comment?

Peter
 
  • #4
Peter said:
Are \(\displaystyle \mathbb{R}\)-linear mappings of \(\displaystyle \mathbb{C}\) into \(\displaystyle \mathbb{C}\) restricted to the situation where \(\displaystyle \mathbb{C}\) is regarded as a vector space of \(\displaystyle \mathbb{R}^2\) over \(\displaystyle \mathbb{R}\) ... ... or can \(\displaystyle \mathbb{R}\)-linear mappings of \(\displaystyle \mathbb{C}\) into \(\displaystyle \mathbb{C}\) be legitimately regarded as occurring in \(\displaystyle \mathbb{C}\) regarded as a vector space of \(\displaystyle \mathbb{C}\) over \(\displaystyle \mathbb{C}\) ... ... ?

(I am guessing that an \(\displaystyle \mathbb{R}\)-linear mapping can occur in both ... )
Yes. In fact, any vector space over $\Bbb{C}$ can be regarded as a vector space over $\Bbb{R}$, simply by restricting scalar multiplication to real scalars.

Peter said:
Now ...for the conjugation map ... we consider ...

\(\displaystyle f : (x,y) \mapsto (x, -y)\)

Now ... we have:

\(\displaystyle f( \ (x,y) + (v,w) \ ) = f( \ (x + v, y + w) \ ) = (x + v, -y - w) = (x,-y) + (v,-w) = f( \ (x,y) \ ) + f( \ (v,w) \ )\)

and

\(\displaystyle f( \ \lambda (x,y) \ ) = f( \ (\lambda x, \lambda y) \ ) = (\lambda x, - \lambda y) = \lambda (x, -y) = \lambda f( \ (x,y) \ ) \)

Is that correct?
Yes.

Peter said:
BUT ... given the above is a real vector space ...

... presumably we would not write \(\displaystyle \[f(\lambda(x+iy)) = f(\lambda x+i\lambda y) = \lambda x-i\lambda y = \lambda (x-i y) = \lambda f(x+iy)\]\) (for all real $\lambda$) because we are essentially dealing with a real vector space of \(\displaystyle \mathbb{R}^2\) over \(\displaystyle \mathbb{R}\)?

Indeed ... should \(\displaystyle i \) be in a real vector space of \(\displaystyle \mathbb{R}^2\) over \(\displaystyle \mathbb{R}\) ...
There is nothing wrong in considering $\Bbb{C}$ as a vector space over $\Bbb{R}$. it doesn't matter that the vectors are complex numbers, just so long as scalar multiplication is only done with real scalars.
 

FAQ: R-Linear and C-Linear Mapings .... ....

What is the difference between R-linear and C-linear mappings?

R-linear and C-linear mappings are both types of linear transformations, but they operate on different vector spaces. R-linear mappings are linear transformations that operate on real vector spaces, while C-linear mappings operate on complex vector spaces.

How do you determine if a mapping is R-linear or C-linear?

To determine if a mapping is R-linear or C-linear, you need to look at the vector space it operates on. If the vector space contains only real numbers, then the mapping is R-linear. If the vector space contains both real and imaginary numbers, then the mapping is C-linear.

What are some examples of R-linear and C-linear mappings?

An example of an R-linear mapping is the derivative operator, which maps a real-valued function to its derivative. An example of a C-linear mapping is the complex conjugate operator, which maps a complex number to its conjugate.

How are R-linear and C-linear mappings used in mathematics?

R-linear and C-linear mappings are used to study the properties of vector spaces and to solve problems in linear algebra. They are also used in fields such as physics and engineering to model real-world systems and phenomena.

What are the applications of R-linear and C-linear mappings?

R-linear and C-linear mappings have many applications in mathematics, physics, engineering, and computer science. They are used to solve systems of linear equations, perform transformations in geometry, and analyze the behavior of dynamic systems. They are also used in data analysis and machine learning algorithms.

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