Are Injective R-Linear Mappings in C Necessarily Surjective?

[Math Amateur]

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 1.4: Angle-Preserving Mappings ... ...

I need help in order to fully understand a remark of Remmert's regarding injective \(\displaystyle \mathbb{R}\)-linear mappings The relevant part of Remmert's section on Angle-Preserving Mappings reads as follows:View attachment 8550In the above text from Remmert we read the following:

" ... ... we look at \(\displaystyle \mathbb{R}\)-linear injective (consequently also bijective) mappings \(\displaystyle T : \mathbb{C}\to \mathbb{C}\) ... ... " Can someone please explain how/why exactly \(\displaystyle \mathbb{R}\)-linear injective mappings are necessarily surjective ... ... ?

Peter

 

[Opalg]

In the above text from Remmert we read the following:

" ... ... we look at \(\displaystyle \mathbb{R}\)-linear injective (consequently also bijective) mappings \(\displaystyle T : \mathbb{C}\to \mathbb{C}\) ... ... "

Can someone please explain how/why exactly \(\displaystyle \mathbb{R}\)-linear injective mappings are necessarily surjective ... ... ?

Considered as a vector space over $\Bbb{R}$, $\Bbb{C}$ is a two-dimensional space. It is a theorem from linear algebra (the rank-nullity theorem) that the rank plus the nullity of a linear map on a vector space equals the dimension of the space. In this case, if the mapping is injective then its nullity is zero, so its rank is equal to the dimension of the space. That is equivalent to saying that the map is surjective, and therefore bijective.

 

[math771]

Hi Peter,

I can try to explain this for you. First, let's define what \mathbb{R}-linear means. It means that the mapping T is linear, meaning it preserves addition and scalar multiplication, and it also maps real numbers to real numbers. So, it is a linear mapping that only operates on real numbers.

Now, let's look at the injective part. This means that the mapping T is one-to-one, meaning that for every element in the domain, there is a unique element in the range. This also means that no two distinct elements in the domain can be mapped to the same element in the range.

Combining these two properties, we can see that the mapping T must be surjective as well. This is because for every real number in the range, there must be a corresponding element in the domain that maps to it. Since T is one-to-one, there cannot be any "leftover" elements in the range that do not have a corresponding element in the domain. Therefore, T must be surjective in order to map every element in the range to a unique element in the domain.

I hope this helps clarify the remark for you. Let me know if you have any other questions.