R^n x R^m isomorphic to R^{n+m}

  • Thread starter yifli
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In summary, an isomorphism exists between sets when there is a bijection between them, and in this case, it is a linear mapping between two finite dimensional commutative vector spaces. This mapping is obtained by sending the basis of one space to the other, and is referred to as a natural isomorphism.
  • #1
yifli
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Here is how I prove it:

Let [tex]\theta_1[/tex](resp.,[tex]\theta_2[/tex]) be the injection from [tex]R^m[/tex](resp.,[tex]R^n[/tex]) to [tex]R^{m+n}[/tex]

Since injection is an isomorphism, [tex]\theta_1[/tex]+[tex]\theta_2[/tex] is the isomorphism

Is this correct?
 
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  • #2
When you say isomorphic, do you mean does there exist a bijection between the sets? Or are you talking about ring or group isomorphism? I ask this because an isomorphism between sets is a group theory topic, but you posted in the calculus section.
 
  • #3
And what exactly do you mean with [tex]\theta_1+\theta_2[/tex]?
 
  • #4
I mean bijection between the sets.
 
  • #5
sum of two functions. (f+g)(x) = f(x)+g(x)
 
  • #6
yifli said:
Here is how I prove it:

Let [tex]\theta_1[/tex](resp.,[tex]\theta_2[/tex]) be the injection from [tex]R^m[/tex](resp.,[tex]R^n[/tex]) to [tex]R^{m+n}[/tex]

Since injection is an isomorphism, [tex]\theta_1[/tex]+[tex]\theta_2[/tex] is the isomorphism

Is this correct?

To be more clear, injection [tex]\theta[/tex]is a linear mapping from [tex]V_i[/tex] to [tex]\prod{V_i}[/tex] such that [tex]\theta(\alpha_j)[/tex]=[tex](0,...0,\alpha_j , 0,... ,0)[/tex]
 
  • #7
Any 2 finite dimensional commutative vector spaces are naturally isomorphic with each other. A natural isomorphism is obtained by considering the map that sends the basis of one to the other
 

1. What does "isomorphic" mean in this context?

In mathematics, isomorphic refers to two structures that have the same structure or shape, even if their elements or components are different. In this context, it means that the vector spaces R^n x R^m and R^{n+m} have the same structure and can be considered equivalent.

2. Why is it important to know if R^n x R^m is isomorphic to R^{n+m}?

Knowing if two vector spaces are isomorphic is important because it allows us to understand if they have the same properties and behave in the same way. It also allows us to translate concepts and theorems from one space to another, making it easier to solve problems and make connections between different areas of mathematics.

3. How can we prove that R^n x R^m is isomorphic to R^{n+m}?

We can prove that two vector spaces are isomorphic by showing that there exists a linear transformation between them that is both one-to-one and onto. This means that the transformation preserves the structure of the original space and all elements in the new space can be mapped back to unique elements in the original space.

4. Can R^n x R^m be isomorphic to R^{n+k} for k ≠ m?

No, R^n x R^m cannot be isomorphic to R^{n+k} for k ≠ m. The dimension of a vector space is a fundamental property that determines its structure, and in this case, the dimensions of R^n x R^m and R^{n+k} are different. Therefore, they cannot be considered equivalent or isomorphic.

5. Are there any real-life applications of the concept of isomorphism in mathematics?

Yes, isomorphism has many real-life applications in mathematics, particularly in fields such as algebra, topology, and geometry. It is also used in computer science and physics. For example, isomorphism can help us understand the structure and properties of molecules in chemistry, and it is used in cryptography for secure communication.

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