Prove that dim(V⊗W)=(dim V)(dim W)

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SUMMARY

The discussion centers on the proof that the dimension of the tensor product of two vector spaces, denoted as dim(V⊗W), equals the product of their dimensions, dim V * dim W. The tensor product is defined as V⊗W = {f: V^* × W^* → k | f is bilinear}, where V^* and W^* are the dual spaces of V and W. The equality (e_i⊗f_j)(φ,ψ) = φ(e_i)ψ(f_j) is established through the definition of the bilinear map, confirming that the set {e_i⊗f_j} forms a basis for V⊗W, leading to the conclusion that dim(V⊗W) = dim V * dim W.

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  • Understanding of tensor products in linear algebra
  • Familiarity with dual spaces, specifically V^* and W^*
  • Knowledge of bilinear maps and their properties
  • Basic concepts of vector space dimension
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Karl Karlsson
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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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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##.
 
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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