We shall study several important general methods of making new vector spaces out of old ones; in this section we begin by studying the easiest one.

Definition 1. If \(\mathcal{U}\) and \(\mathcal{V}\) are vector spaces (over the same field), their direct sum is the vector space \(\mathcal{W}\) (denoted by \(\mathcal{U} \oplus \mathcal{V}\) ) whose elements are all the ordered pairs \(\langle x, y \rangle\) with \(x\) in \(\mathcal{U}\) and \(y\) in \(\mathcal{V}\) , with the linear operations defined by \[\alpha_1 \langle x_1, y_1 \rangle + \alpha_2 \langle x_2, y_2 \rangle = \langle \alpha_1 x_1 + \alpha_2 x_2, \alpha_1 y_1 + \alpha_2 y_2 \rangle.\] 

We observe that the formation of the direct sum is analogous to the way in which the plane is constructed from its two coordinate axes.

We proceed to investigate the relation of this notion to some of our earlier ones.

The set of all vectors (in \(\mathcal{W}\) ) of the form \(\langle x, 0 \rangle\) is a subspace of \(\mathcal{W}\) ; the correspondence \(\langle x, 0 \rangle \rightleftarrows x\) shows that this subspace is isomorphic to \(\mathcal{U}\) . It is convenient, once more, to indulge in a logical inaccuracy and, identifying \(x\) and \(\langle x, 0 \rangle\) , to speak of \(\mathcal{U}\) as a subspace of \(\mathcal{W}\) . Similarly, of course, the vectors \(y\) of \(\mathcal{V}\) may be identified with the vectors of the form \(\langle 0, y \rangle\) in \(\mathcal{W}\) , and we may consider \(\mathcal{V}\) as a subspace of \(\mathcal{W}\) . This terminology is, to be sure, not quite exact, but the logical difficulty is much easier to get around here than it was in the case of the second dual space. We could have defined the direct sum of \(\mathcal{U}\) and \(\mathcal{V}\) (at least in the case in which \(\mathcal{U}\) and \(\mathcal{V}\) have no non-zero vectors in common) as the set consisting of all \(x\) ’s in \(\mathcal{U}\) , all \(y\) ’s in \(\mathcal{V}\) , and all those pairs \(\langle x, y \rangle\) for which \(x \neq 0\) and \(y \neq 0\) . This definition yields a theory analogous in every detail to the one we shall develop, but it makes it a nuisance to prove theorems because of the case distinctions it necessitates. It is clear, however, that from the point of view of this definition \(\mathcal{U}\) is actually a subset of \(\mathcal{U} \oplus \mathcal{V}\) . In this sense then, or in the isomorphism sense of the definition we did adopt, we raise the question: what is the relation between \(\mathcal{U}\) and \(\mathcal{V}\) when we consider these spaces as subspaces of the big space \(\mathcal{W}\) ?

Theorem 1. If \(\mathcal{U}\) and \(\mathcal{V}\) are subspaces of a vector space \(\mathcal{W}\) , then the following three conditions are equivalent.

  1. \(\mathcal{W} = \mathcal{U} \oplus \mathcal{V}\) .
  2. \(\mathcal{U} \cap \mathcal{V} = \mathcal{O}\) and \(\mathcal{U} + \mathcal{V} = \mathcal{W}\) (i.e., \(\mathcal{U}\) and \(\mathcal{V}\) are complements of each other).
  3. Every vector \(z\) in \(\mathcal{W}\) may be written in the form \(z = x + y\) , with \(x\) in \(\mathcal{U}\) and \(y\) in \(\mathcal{V}\) , in one and only one way.

Proof. We shall prove the implications (1) \(\implies\) (2) \(\implies\) (3) \(\implies\) (1).

(1) \(\implies\) (2). We assume that \(\mathcal{W} = \mathcal{U} \oplus \mathcal{V}\) . If \(z = \langle x, y \rangle\) lies in both \(\mathcal{U}\) and \(\mathcal{V}\) , then \(x = y = 0\) , so that \(z = 0\) ; this proves that \(\mathcal{U} \cap \mathcal{V} = \mathcal{O}\) . Since the representation \(z = \langle x, 0 \rangle + \langle 0, y \rangle\) is valid for every \(z\) , it follows also that \(\mathcal{U} + \mathcal{V} = \mathcal{W}\) .

(2) \(\implies\) (3). If we assume (2), so that, in particular, \(\mathcal{U} + \mathcal{V} = \mathcal{W}\) , then it is clear that every \(z\) in \(\mathcal{W}\) has the desired representation, \(z = x + y\) . To prove uniqueness, we assume that \(z = x_1 + y_1\) and \(z = x_2 + y_2\) , with \(x_1\) and \(x_2\) in \(\mathcal{U}\) and \(y_1\) and \(y_2\) in \(\mathcal{V}\) . Since \(x_1 + y_1 = x_2 + y_2\) , it follows that \(x_1 - x_2 = y_2 - y_1\) . Since the left member of this last equation is in \(\mathcal{U}\) and the right member is in \(\mathcal{V}\) , the disjointness of \(\mathcal{U}\) and \(\mathcal{V}\) implies that \(x_1 = x_2\) and \(y_1 = y_2\) .

(3) \(\implies\) (1). This implication is practically indistinguishable from the definition of direct sum. If we form the direct sum \(\mathcal{U} \oplus \mathcal{V}\) , and then identify \(\langle x, 0 \rangle\) and \(\langle 0, y \rangle\) with \(x\) and \(y\) respectively, we are committed to identifying the sum \(\langle x, y \rangle = \langle x, 0 \rangle + \langle 0, y \rangle\) with what we are assuming to be the general element \(z = x + y\) of \(\mathcal{W}\) ; from the hypothesis that the representation of \(z\) in the form \(x + y\) is unique we conclude that the correspondence between \(\langle x, 0 \rangle\) and \(x\) (and also between \(\langle 0, y \rangle\) and \(y\) ) is one-to-one. ◻

If two subspaces \(\mathcal{U}\) and \(\mathcal{V}\) in a vector space \(\mathcal{W}\) are disjoint and span \(\mathcal{W}\) (that is, if they satisfy (2)), it is usual to say that \(\mathcal{W}\) is the internal direct sum of \(\mathcal{U}\) and \(\mathcal{V}\) ; symbolically, as before, \(\mathcal{W} = \mathcal{U} \oplus \mathcal{V}\) . If we want to emphasize the distinction between this concept and the one defined before, we describe the earlier one by saying that \(\mathcal{W}\) is the external direct sum of \(\mathcal{U}\) and \(\mathcal{V}\) . In view of the natural isomorphisms discussed above, and, especially, in view of the preceding theorem, the distinction is more pedantic than conceptual. In accordance with our identification convention, we shall usually ignore it.