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

[Math Amateur]

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

 

[Opalg]

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 $\nu\ne0$.

 

[Math Amateur]

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

 

[Opalg]

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.

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.

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.