Direct sums

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 and are vector spaces (over the same field), their direct sum is the vector space (denoted by ) whose elements are all the ordered pairs with in and in , with the linear operations defined by

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 ) of the form is a subspace of ; the correspondence shows that this subspace is isomorphic to . It is convenient, once more, to indulge in a logical inaccuracy and, identifying and , to speak of as a subspace of . Similarly, of course, the vectors of may be identified with the vectors of the form in , and we may consider as a subspace of . 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 and (at least in the case in which and have no non-zero vectors in common) as the set consisting of all ’s in , all ’s in , and all those pairs for which and . 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 is actually a subset of . In this sense then, or in the isomorphism sense of the definition we did adopt, we raise the question: what is the relation between and when we consider these spaces as subspaces of the big space ?

Theorem 1. If and are subspaces of a vector space , then the following three conditions are equivalent.

.
and (i.e., and are complements of each other).
Every vector in may be written in the form , with in and in , in one and only one way.

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

(1) (2). We assume that . If lies in both and , then , so that ; this proves that . Since the representation is valid for every , it follows also that .

(2) (3). If we assume (2), so that, in particular, , then it is clear that every in has the desired representation, . To prove uniqueness, we assume that and , with and in and and in . Since , it follows that . Since the left member of this last equation is in and the right member is in , the disjointness of and implies that and .

(3) (1). This implication is practically indistinguishable from the definition of direct sum. If we form the direct sum , and then identify and with and respectively, we are committed to identifying the sum with what we are assuming to be the general element of ; from the hypothesis that the representation of in the form is unique we conclude that the correspondence between and (and also between and ) is one-to-one. ◻

If two subspaces and in a vector space are disjoint and span (that is, if they satisfy (2)), it is usual to say that is the internal direct sum of and ; symbolically, as before, . If we want to emphasize the distinction between this concept and the one defined before, we describe the earlier one by saying that is the external direct sum of and . 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.

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