Natural Isomorphism b/w Dual Spaces Tensor Prod & Multilinear Form Space

In summary, natural isomorphism is a fundamental concept in mathematics that refers to a type of isomorphism that exists independently of any specific choice of coordinates or basis. It is more restrictive than regular isomorphism, as it requires the isomorphism to be independent of any basis or coordinates. In the context of dual spaces, tensor products, and multilinear form spaces, natural isomorphism allows for a unique correspondence between these spaces. This concept is important in mathematics as it allows for the study of abstract structures and has practical applications in fields such as physics and engineering.
  • #1
caffeinemachine
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I am trying to prove the following.

Let be finite dimensional vector spaces over a field .
There is a natural isomorphism between and .

Define a map as

for all .
It can be seen that is a multilinear map.
By the universal property of tensor product, there exists a unique linear map such that .

We also know that

Thus we just need to show that to show that and are isomorphic.

My Problem: I want to show the triviality of the kernel in a basis free manner. But here I am stuck.

Can anybody help?
 
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  • #2
caffeinemachine said:
I am trying to prove the following.

Let be finite dimensional vector spaces over a field .
There is a natural isomorphism between and .

Define a map as

for all .
It can be seen that is a multilinear map.
By the universal property of tensor product, there exists a unique linear map such that .

We also know that

Thus we just need to show that to show that and are isomorphic.

My Problem: I want to show the triviality of the kernel in a basis free manner. But here I am stuck.

Can anybody help?
Instead of trying to show that is injective, I think it would be easier to show that it is surjective. This should somehow be equivalent to the fact that a matrix is a sum of rank matrices.
 
  • #3
Opalg said:
Instead of trying to show that is injective, I think it would be easier to show that it is surjective. This should somehow be equivalent to the fact that a matrix is a sum of rank matrices.
Hello Opalg,

Sorry for the late reply. I somehow forgot about this post.

I can show that is surjective by choosing a basis. I am getting more and more convinced that this cannot be done without choosing a basis.
 
  • #4
caffeinemachine said:
I am getting more and more convinced that this cannot be done without choosing a basis.
I tend to agree. For one thing, I believe that the result is false if the spaces are infinite-dimensional. So you somehow need to make use of the fact that the spaces are finite-dimensional, and the obvious way is to choose bases for them.
 

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