MHB 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:$$ \mathbb{R}$$-linear and $$\mathbb{C}$$-linear mappings of $$\mathbb{C}$$ into $$\mathbb{C}$$ ... ... I need help in order to get a clear idea of the definition and nature of $$\mathbb{R}$$-linear and $$\mathbb{C}$$-linear mappings of \mathbb{C} into \mathbb{C} ... ...

Remmert's section on $$\mathbb{R}$$-linear and $$\mathbb{C}$$-linear mappings of $$\mathbb{C}$$ into $$\mathbb{C}$$ 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 $$\mathbb{R}$$-linear and $$\mathbb{C}$$-linear mappings ...

Can someone please give a first-principles definition of $$\mathbb{R}$$-linear and $$\mathbb{C}$$-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 $$\mathbb{R}$$-linear and $$\mathbb{C}$$-linear mappings ...

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

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

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

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

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

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

and

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

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

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

Peter

 

[Opalg]

Are $$\mathbb{R}$$-linear mappings of $$\mathbb{C}$$ into $$\mathbb{C}$$ restricted to the situation where $$\mathbb{C}$$ is regarded as a vector space of $$\mathbb{R}^2$$ over $$\mathbb{R}$$ ... ... or can $$\mathbb{R}$$-linear mappings of $$\mathbb{C}$$ into $$\mathbb{C}$$ be legitimately regarded as occurring in $$\mathbb{C}$$ regarded as a vector space of $$\mathbb{C}$$ over $$\mathbb{C}$$ ... ... ?

(I am guessing that an $$\mathbb{R}$$-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 ...

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

Now ... we have:

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

and

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

Is that correct?

Yes.

BUT ... given the above is a real vector space ...

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

Indeed ... should $$i $$ be in a real vector space of $$\mathbb{R}^2$$ over $$\mathbb{R}$$ ...

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.