Tensor products

If $𝒰$ and $𝒱$ are vector spaces (over the same field), then their direct sum $𝒲 = 𝒰 \oplus 𝒱$ is another vector space; we propose to study certain functions on $𝒲$ . (For present purposes the original definition of $𝒰 \oplus 𝒱$ , via ordered pairs, is the convenient one.) The value of such a function, say $w$ , at an element $⟨ x, y ⟩$ of $𝒲$ will be denoted by $w (x, y)$ . The study of linear functions on $𝒲$ 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 $𝒲$ with the property that for each fixed value of either argument they depend linearly on the other argument. More precisely, a scalar-valued function $w$ on $𝒲$ is a bilinear form (or bilinear functional ) if $w (α_{1} x_{1} + α_{2} x_{2}, y) = α_{1} w (x_{1}, y) + α_{2} w (x_{2}, y)$ and $w (x, α_{1} y_{1} + α_{2} y_{2}) = α_{1} w (x, y_{1}) + α_{2} w (x, y_{2}),$ identically in the vectors and scalars involved.

In one special situation we have already encountered bilinear functionals. If, namely, $𝒱$ is the dual space of $𝒰$ , \mathcal{V} = \mathcal{U}' , and if we write $w (x, y) = [x, y]$ (see Section: Brackets ), then $w$ is a bilinear functional on \mathcal{U} \oplus \mathcal{U}' . For an example in a more general situation, let $𝒰$ and $𝒱$ be arbitrary vector spaces (over the same field, as always), let $u$ and $v$ be elements of \mathcal{U}' and \mathcal{V}' respectively, and write $w (x, y) = u (x) v (y)$ for all $x$ in $𝒰$ and $y$ in $𝒱$ . An even more general example is obtained by selecting a finite number of elements in \mathcal{U}' , say $u_{1}, \dots, u_{k}$ , selecting the same finite number of elements in \mathcal{V}' , say $v_{1}, \dots, v_{k}$ , and writing $w (x, y) = u_{1} (x) v_{1} (y) + \dots + u_{k} (x) v_{k} (y)$ . 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 $w_{1}$ and $w_{2}$ are bilinear forms on $𝒲$ , and if $α_{1}$ and $α_{2}$ are scalars, we write $w$ for the function on $𝒲$ defined by $w (x, y) = α_{1} w_{1} (x, y) + α_{2} w_{2} (x, y) .$ It is easy to see that $w$ is a bilinear form; we denote it by $α_{1} w_{1} + α_{2} w_{2}$ . With this definition of the linear operations, the set of all bilinear forms on $𝒲$ 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 $𝒰$ and $𝒱$ .

Theorem 1. If $𝒰$ is an $n$ -dimensional vector space with basis ${x_{1}, \dots, x_{n}}$ , if $𝒱$ is an $m$ -dimensional vector space with basis ${y_{1}, \dots, y_{m}}$ , and if ${α_{i j}}$ is any set of $n m$ scalars ( $i = 1, \dots, n$ ; $j = 1, \dots, m$ ), then there is one and only one bilinear form $w$ on $𝒰 \oplus 𝒱$ such that $w (x_{i}, y_{j}) = α_{i j}$ for all $i$ and $j$ .

Proof. If $x = \sum_{i} ξ_{i} x_{i}$ , $y = \sum_{j} η_{j} y_{j}$ , and $w$ is a bilinear form on $𝒰 \oplus 𝒱$ such that $w (x_{i}, y_{j}) = α_{i j}$ , then $w (x, y) = \sum_{i} \sum_{j} ξ_{i} η_{j} w (x_{i}, y_{j}) = \sum_{i} \sum_{j} ξ_{i} η_{j} α_{i j} .$ From this equation the uniqueness of $w$ is clear; the existence of a suitable $w$ is proved by reading the same equation from right to left, that is, defining $w$ by it. (Compare this result with Section: Dual bases , Theorem 1.) ◻

Theorem 2. If $𝒰$ is an $n$ -dimensional vector space with basis ${x_{1}, \dots, x_{n}}$ , and if $𝒱$ is an $m$ -dimensional vector space with basis ${y_{1}, \dots, y_{m}}$ , then there is a uniquely determined basis ${w_{p q}}$ ( $p = 1, \dots, n$ ; $q = 1, \dots, m$ ) in the vector space of all bilinear forms on $𝒰 \oplus 𝒱$ with the property that $w_{p q} (x_{i}, y_{j}) = δ_{i p} δ_{j q}$ . Consequently the dimension of the space of bilinear forms on $𝒰 \oplus 𝒱$ is the product of the dimensions of $𝒰$ and $𝒱$ .

Proof. Using Theorem 1, we determine $w_{p q}$ (for each fixed $p$ and $q$ ) by the given condition $w_{p q} (x_{i}, y_{j}) = δ_{i p} δ_{j q}$ . The bilinear forms so determined are linearly independent, since $\sum_{p} \sum_{q} α_{p q} w_{p q} = 0$ implies that $0 = \sum_{p} \sum_{q} α_{p q} δ_{i p} δ_{j q} = α_{i j} .$ If, moreover, $w$ is an arbitrary element of $𝒲$ , and if $w (x_{i}, y_{j}) = α_{i j}$ , then $w = \sum_{p} \sum_{q} α_{p q} w_{p q}$ . Indeed, if $x = \sum_{i} ξ_{i} x_{i}$ and $y = \sum_{j} η_{j} y_{j}$ , then $w_{p q} (x, y) = \sum_{i} \sum_{j} ξ_{i} η_{j} δ_{i p} δ_{j q} = ξ_{p} η_{q},$ and, consequently, $w (x, y) = \sum_{i} \sum_{j} ξ_{i} η_{j} α_{i j} = \sum_{p} \sum_{q} α_{p q} w_{p q} (x, y) .$ It follows that the $w_{p q}$ 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 $w$ is a bilinear form on $ℝ^{n} \oplus ℝ^{n}$ , then there exist scalars $α_{i j}$ , $i, j = 1, \dots, n$ , such that if $x = (ξ_{1}, \dots, ξ_{n})$ and $y = (η_{1}, \dots, η_{n})$ , then $w (x, y) = \sum_{i} \sum_{j} α_{i j} ξ_{i} η_{j}$ . The scalars $α_{i j}$ are uniquely determined by $w$ .
If $z$ is a linear functional on the space of all bilinear forms on $ℝ^{n} \oplus ℝ^{n}$ , then there exist scalars $β_{i j}$ such that (in the notation of (a)) $z (w) = \sum_{i} \sum_{j} α_{i j} β_{i j}$ for every $w$ . The scalars $β_{i j}$ are uniquely determined by $z$ .

Exercise 2. A bilinear form $w$ on $𝒰 \oplus 𝒱$ 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 $ℂ^{2} \oplus ℂ^{2}$ .
Give an example of a non-degenerate bilinear form on $ℂ^{2} \oplus ℂ^{2}$ .

Exercise 3. If $w$ is a bilinear form on $𝒰 \oplus 𝒱$ , if $y_{0}$ is in $𝒱$ , and if a function $y$ is defined on $𝒰$ by $y (x) = w (x, y_{0})$ , then $y$ is a linear functional on $𝒰$ . Is it true that if $w$ is non-degenerate, then every linear functional on $𝒰$ can be obtained this way (by a suitable choice of $y_{0}$ )?

Exercise 4. Suppose that for each $x$ and $y$ in $𝒫_{n}$ the function $w$ is defined by

$w (x, y) = \int_{0}^{1} x (t) y (t) d t$ ,
$w (x, y) = x (1) + y (1)$ ,
$w (x, y) = x (1) \cdot y (1)$ ,
$w (x, y) = x (1) {(\frac{d y}{d t})}_{t = 1}$ .

In which of these cases is $w$ a bilinear form on $𝒫_{n} \oplus 𝒫_{n}$ ? In which cases is it non-degenerate?

Exercise 5. Does there exist a vector space $𝒱$ and a bilinear form $w$ on $𝒱 \oplus 𝒱$ such that $w$ is not identically zero but $w (x, x) = 0$ for every $x$ in $𝒱$ ?

Exercise 6.

A bilinear form $w$ on $𝒱 \oplus 𝒱$ is symmetric if $w (x, y) = w (y, x)$ for all $x$ and $y$ . A quadratic form on $𝒱$ is a function $q$ on $𝒱$ obtained from a bilinear form $w$ by writing $q (x) = w (x, x)$ . Prove that if the characteristic of the underlying scalar field is different from $2$ , then every symmetric bilinear form is uniquely determined by the corresponding quadratic form. What happens if the characteristic is $2$ ?
Can a non-symmetric bilinear form define the same quadratic form as a symmetric one?

EXERCISES

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