Prove that dim(V⊗W)=(dim V)(dim W)

In summary, the proof in the book defines the tensor product ##V \otimes W## as a set of bilinear maps ##f: V^* \times W^* \to k## and then shows that the set ##\{e_i \otimes f_j\}_{(i,j)}## is a basis for ##V \otimes W##, leading to the conclusion that ##\dim(V \otimes W) = \dim V \dim W##. The step ##(e_i \otimes f_j)(\phi, \psi) = \phi(e_i) \psi(f_j)## is simply part of the definition, and it is shown that this definition satisfies the properties of a tensor product
  • #1
Karl Karlsson
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TL;DR Summary
This proof was in my book.
(see image below)
Tensor product definition according to my book: $$V⊗W=\{f: V^*\times W^*\rightarrow k | \textrm {f is bilinear}\}$$ wher ##V^*## and ##W^*## are the dual spaces for V and W respectively.

I don't understand the step where they say ##(e_i⊗f_j)(φ,ψ) = φ(e_i)ψ(f_j)##. Why is this equality true? What definition has been used? My understanding for all of this is still quite basic.
This proof was in my book.
IMG_0775.jpg

Tensor product definition according to my book: $$V⊗W=\{f: V^*\times W^*\rightarrow k | \textrm {f is bilinear}\}$$ wher ##V^*## and ##W^*## are the dual spaces for V and W respectively.

I don't understand the step where they say ##(e_i⊗f_j)(φ,ψ) = φ(e_i)ψ(f_j)##. Why is this equality true? What definition has been used? My understanding for all of this is still quite basic.

Thanks in advance!
 
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  • #2
This is true because they define it that way. By definition, ##e_i \otimes f_j## is the bilinear map defined by $$e_i \otimes f_j: V^* \times W^* \to k: (\phi, \psi) \mapsto \psi(e_i) \psi(f_j)$$
It is just a definition. The only thing you should check is that this map is indeed ##k##-bilinear. We then obtain that ##e_i \otimes f_j \in V \otimes W## and it is then checked that ##\{e_i \otimes f_j\}_{(i,j)}## is a basis for ##V \otimes W## and since this basis has ##\dim V \dim W## amount of elements, you can conclude ##\dim(V \otimes W) = \dim V \dim W##.
 
  • #3
It should have been part of the definition. As it is written, it suggests that it cannot be done otherwise, which would require a proof. However, uniqueness of the tensor product is easier to prove in the language of categories, rather than using coordinates. The way it was done in the book is a mixture of both - neither done rigorously. The shortest way out of the dilemma is to incorporate it in the definition:
$$
T\in V\otimes W \Longleftrightarrow T=\sum_{\rho=1}^R v_{\rho} \otimes w_{\rho} \, : \,(X,Y)\longmapsto \sum_{\rho=1}^R v_{\rho}(X) \cdot w_{\rho}(Y) \in k
$$
for some ##v_{\rho}\in V^*,w_{\rho}\in W^*## and all ##X\in V, Y\in W.##
 
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1. What does "dim" refer to in the equation "dim(V⊗W)=(dim V)(dim W)"?

"Dim" refers to the dimension of a vector space, which is the number of linearly independent vectors that span the space.

2. Can you provide an example of a vector space V and W where the equation "dim(V⊗W)=(dim V)(dim W)" holds true?

Yes, let V be a vector space with dimension 3 and W be a vector space with dimension 4. Then, the tensor product V⊗W will have a dimension of 12, which is equal to (3)(4).

3. How does the tensor product of two vector spaces relate to their dimensions?

The tensor product of two vector spaces V and W is a new vector space that is formed by combining the elements of V and W in a specific way. The dimension of this new vector space is equal to the product of the dimensions of V and W.

4. Is the equation "dim(V⊗W)=(dim V)(dim W)" always true for any two vector spaces?

Yes, the equation is always true for any two vector spaces V and W. This is because the dimension of the tensor product V⊗W is determined by the number of elements in V and W, which is equal to the product of their dimensions.

5. What is the significance of the equation "dim(V⊗W)=(dim V)(dim W)" in mathematics and science?

The equation is significant because it shows that the dimension of the tensor product of two vector spaces is related to the dimensions of the individual vector spaces. This has applications in various fields such as quantum mechanics, where tensor products are used to describe composite systems, and in linear algebra, where tensor products are used to define multilinear maps.

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