Tensor products

If and are vector spaces (over the same field), then their direct sum is another vector space; we propose to study certain functions on . (For present purposes the original definition of , via ordered pairs, is the convenient one.) The value of such a function, say , at an element of will be denoted by . The study of linear functions on is no longer of much interest to us; the principal facts concerning them were discussed in Section: Dual of a direct sum . The functions we want to consider now are the bilinear ones; they are, by definition, the scalar-valued functions on with the property that for each fixed value of either argument they depend linearly on the other argument. More precisely, a scalar-valued function on is a bilinear form (or bilinear functional ) if and identically in the vectors and scalars involved.

In one special situation we have already encountered bilinear functionals. If, namely, is the dual space of , , and if we write (see Section: Brackets ), then is a bilinear functional on . For an example in a more general situation, let and be arbitrary vector spaces (over the same field, as always), let and be elements of and respectively, and write for all in and in . An even more general example is obtained by selecting a finite number of elements in , say , selecting the same finite number of elements in , say , and writing . Which of the words, “functional” or “form,” is used depends somewhat on the context and, somewhat more, on the user’s whim. In this book we shall generally use “functional” with “linear” and “form” with “bilinear” (and its higher-dimensional generalizations).

If and are bilinear forms on , and if and are scalars, we write for the function on defined by It is easy to see that is a bilinear form; we denote it by . With this definition of the linear operations, the set of all bilinear forms on is a vector space. The chief purpose of the remainder of this section is to determine (in the finite-dimensional case) how the dimension of this space depends on the dimensions of and .

Theorem 1. If is an -dimensional vector space with basis , if is an -dimensional vector space with basis , and if is any set of scalars ( ; ), then there is one and only one bilinear form on such that for all and .

Proof. If , , and is a bilinear form on such that , then From this equation the uniqueness of is clear; the existence of a suitable is proved by reading the same equation from right to left, that is, defining by it. (Compare this result with Section: Dual bases , Theorem 1.) ◻

Theorem 2. If is an -dimensional vector space with basis , and if is an -dimensional vector space with basis , then there is a uniquely determined basis ( ; ) in the vector space of all bilinear forms on with the property that . Consequently the dimension of the space of bilinear forms on is the product of the dimensions of and .

Proof. Using Theorem 1, we determine (for each fixed and ) by the given condition . The bilinear forms so determined are linearly independent, since implies that If, moreover, is an arbitrary element of , and if , then . Indeed, if and , then and, consequently, It follows that the form a basis in the space of bilinear forms; this completes the proof of the theorem. (Compare this result with Section: Dual bases , Theorem 2.) ◻

EXERCISES

Exercise 1.

If is a bilinear form on , then there exist scalars , , such that if and , then . The scalars are uniquely determined by .
If is a linear functional on the space of all bilinear forms on , then there exist scalars such that (in the notation of (a)) for every . The scalars are uniquely determined by .

Exercise 2. A bilinear form on is degenerate if, as a function of one of its two arguments, it vanishes identically for some non-zero value of its other argument; otherwise it is non-degenerate .

Give an example of a degenerate bilinear form (not identically zero) on .
Give an example of a non-degenerate bilinear form on .

Exercise 3. If is a bilinear form on , if is in , and if a function is defined on by , then is a linear functional on . Is it true that if is non-degenerate, then every linear functional on can be obtained this way (by a suitable choice of )?

Exercise 4. Suppose that for each and in the function is defined by

In which of these cases is a bilinear form on ? In which cases is it non-degenerate?

Exercise 5. Does there exist a vector space and a bilinear form on such that is not identically zero but for every in ?

Exercise 6.

A bilinear form on is symmetric if for all and . A quadratic form on is a function on obtained from a bilinear form by writing . Prove that if the characteristic of the underlying scalar field is different from , then every symmetric bilinear form is uniquely determined by the corresponding quadratic form. What happens if the characteristic is ?
Can a non-symmetric bilinear form define the same quadratic form as a symmetric one?

EXERCISES

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