After one more word of preliminary explanation we shall be ready to discuss the formal definition of tensor products. It turns out to be technically preferable to get at \(\mathcal{U} \otimes \mathcal{V}\) indirectly, by defining it as the dual of another space; we shall make tacit use of reflexivity to obtain \(\mathcal{U} \otimes \mathcal{V}\) itself. Since we have proved reflexivity for finite-dimensional spaces only, we shall restrict the definition to such spaces.
Definition 1. The tensor product \(\mathcal{U} \otimes \mathcal{V}\) of two finite-dimensional vector spaces \(\mathcal{U}\) and \(\mathcal{V}\) (over the same field) is the dual of the vector space of all bilinear forms on \(\mathcal{U} \oplus \mathcal{V}\) . For each pair of vectors \(x\) and \(y\) , with \(x\) in \(\mathcal{U}\) and \(y\) in \(\mathcal{V}\) , the tensor product \(z = x \otimes y\) is the element of \(\mathcal{U} \otimes \mathcal{V}\) defined by \(z(w) = w(x, y)\) for every bilinear form \(w\) .
This definition is one of the quickest rigorous approaches to the theory, but it does lead to some unpleasant technical complications later. Whatever its disadvantages, however, we observe that it obviously has the two desired properties: it is clear, namely, that dimension is multiplicative (see Section: Bilinear forms , Theorem 2, and Section: Dual bases , Theorem 2), and it is clear that \(x \otimes y\) depends linearly on each of its factors.
Another possible (and deservedly popular) definition of tensor product is by formal products. According to that definition \(\mathcal{U} \otimes \mathcal{V}\) is obtained by considering all symbols of the form \(\sum_i \alpha_i(x_i \otimes y_i)\) , and, within the set of such symbols, making the identifications demanded by the linearity of the vector operations and the bilinearity of tensor multiplication. (For the purist: in this definition \(x \otimes y\) stands merely for the ordered pair of \(x\) and \(y\) ; the multiplication sign is just a reminder of what to expect.) Neither definition is simple; we adopted the one we gave because it seemed more in keeping with the spirit of the rest of the book. The main disadvantage of our definition is that it does not readily extend to the most useful generalizations of finite-dimensional vector spaces, that is, to modules and to infinite-dimensional spaces.
For the present we prove only one theorem about tensor products. The theorem is a further justification of the product terminology, and, incidentally, it is a sharpening of the assertion that dimension is multiplicative.
Theorem 1. If \(\mathcal{X} = \{x_1, \ldots, x_n\}\) and \(\mathcal{Y} = \{y_1, \ldots, y_m\}\) are bases in \(\mathcal{U}\) and \(\mathcal{V}\) respectively, then the set \(\mathcal{Z}\) of vectors \(z_{ij} = x_i \otimes y_j\) ( \(i = 1, \ldots, n\) ; \(j = 1, \ldots, m\) ) is a basis in \(\mathcal{U} \otimes \mathcal{V}\) .
Proof. Let \(w_{pq}\) be the bilinear form on \(\mathcal{U} \oplus \mathcal{V}\) such that \(w_{pq}(x_i, y_j) = \delta_{ip} \delta_{jq}\) ( \(p = 1, \ldots, n\) ; \(j = 1, \ldots, m\) ); the existence of such bilinear forms, and the fact that they constitute a basis for all bilinear forms, follow from Section: Bilinear forms , Theorem 2. Let \(\{w'_{pq}\}\) be the dual basis in \(\mathcal{U} \otimes \mathcal{V}\) , so that \([w_{ij}, w'_{pq}] = \delta_{ip} \delta_{jq}\) . If \(w = \sum_p \sum_q \alpha_{pq} w_{pq}\) is an arbitrary bilinear form on \(\mathcal{U} \oplus \mathcal{V}\) , then \begin{align} w'_{ij}(w) &= [w, w'_{ij}]\\ &= \sum_p \sum_q \alpha_{pq} [w_{pq}, w'_{ij}] \\ &= \alpha_{ij}\\ &= w(x_i, y_j)\\ &= z_{ij}(w). \end{align} The conclusion follows from the fact that the vectors \(w'_{ij}\) do constitute a basis of \(\mathcal{U} \otimes \mathcal{V}\) . ◻
EXERCISES
Exercise 1. If \(x = (1, 1)\) and \(y = (1, 1, 1)\) are vectors in \(\mathbb{R}^2\) and \(\mathbb{R}^3\) respectively, find the coordinates of \(x \otimes y\) in \(\mathbb{R}^2 \otimes \mathbb{R}^3\) with respect to the product basis \(\{x_i \otimes y_j\}\) , where \(x_i = (\delta_{i1}, \delta_{i2})\) and \(y_j = (\delta_{1j}, \delta_{2j}, \delta_{3j})\) .
Exercise 2. Let \(\mathcal{P}_{n,m}\) be the space of all polynomials \(z\) with complex coefficients, in two variables \(s\) and \(t\) , such that either \(z = 0\) or else the degree of \(z(s, t)\) is \(\le m - 1\) for each fixed \(s\) and \(\le n - 1\) for each fixed \(t\) . Prove that there exists an isomorphism between \(\mathcal{P}_n \otimes \mathcal{P}_m\) and \(\mathcal{P}_{n,m}\) such that the element \(z\) of \(\mathcal{P}_{n,m}\) that corresponds to \(x \otimes y\) ( \(x\) in \(\mathcal{P}_n\) , \(y\) in \(\mathcal{P}_m\) ) is given by \(z(s, t) = x(s)y(t)\) .
Exercise 3. To what extent is the formation of tensor products commutative and associative? What about the distributive law \(\mathcal{U} \otimes (\mathcal{V} \oplus \mathcal{W}) = (\mathcal{U} \otimes \mathcal{V}) \oplus (\mathcal{U} \otimes \mathcal{W})\) ?
Exercise 4. If \(\mathcal{V}\) is a finite-dimensional vector space, and if \(x\) and \(y\) are in \(\mathcal{V}\) , is it true that \(x \otimes y = y \otimes x\) ?
Exercise 5.
- Suppose that \(\mathcal{V}\) is a finite-dimensional real vector space, and let \(\mathcal{U}\) be the set \(\mathbb{C}\) of all complex numbers regarded as a (two-dimensional) real vector space. Form the tensor product \(\mathcal{V}^+ = \mathcal{U} \otimes \mathcal{V}\) . Prove that there is a way of defining products of complex numbers with elements of \(\mathcal{V}^+\) so that \(\alpha(x \otimes y) = (\alpha x) \otimes y\) whenever \(\alpha\) and \(x\) are in \(\mathbb{C}\) and \(y\) is in \(\mathcal{V}\) .
- Prove that with respect to vector addition, and with respect to complex scalar multiplication as defined in (a), the space \(\mathcal{V}^+\) is a complex vector space.
- Find the dimension of the complex vector space \(\mathcal{V}^+\) in terms of the dimension of the real vector space \(\mathcal{V}\) .
- Prove that the vector space \(\mathcal{V}\) is isomorphic to a subspace in \(\mathcal{V}^+\) (when the latter is regarded as a real vector space).
The moral of this exercise is that not only can every complex vector space be regarded as a real vector space, but, in a certain sense, the converse is true. The complex vector space \(\mathcal{U}^+\) is called the complexification of \(\mathcal{V}\) .
Exercise 6. If \(\mathcal{U}\) and \(\mathcal{V}\) are finite-dimensional vector spaces, what is the dual space of \(\mathcal{U}' \otimes \mathcal{V}'\) ?