EXTERNAL DIRECT SUM AND INTERNAL DIRECT SUM OF VECTOR SPACES
1. Direct Sum of Vector Spaces Let V and W be vector spaces over a field F. On the cartesian product
V × W = {(v, w) : v ∈ V, w ∈ W }
of V and W, we define the addition and the scalar multiplication of elements as follows. Let (v, w) and (v1, w1) and (v2, w2) be elements of V × W and a ∈ F. We define
(v1, w1) + (v2, w2) = (v1+ v2, w1+ w2), a · (v, w) = (av, aw).
Lemma 1.1. (V × W, +, ·) forms a vector space over F and is denoted by V ⊕eW.
Proof. This is left to the reader as an exercise.
Definition 1.1. The vector space V ⊕eW over F defined above is called the external direct sum of V and W.
Let Z be a vector space over F and X and Y be vector subspaces of Z. Suppose that X and Y satisfy the following properties:
(1) for each z ∈ Z, there exist x ∈ X and y ∈ Y such that z = x + y;
(2) X ∩ Y = {0}.
In this case, we write Z = X ⊕iY and say that Z is the internal direct sum of vector subspaces X and Y.
Theorem 1.1. Let X and Y be vector subspaces of a vector space Z over F such that Z is the internal direct sum of X and Y, i.e. Z = X ⊕iY. Then there is a linear isomorphism from Z onto X ⊕eY, i.e. X ⊕iY is isomorphic to X ⊕eY.
Proof. Define f : X ⊕eY → Z by f (x, y) = x + y. Then f is a linear map. (Readers need to check).
Since Z is the internal direct sum of X and Y, for any z ∈ Z, there exist x ∈ X and y ∈ Y such that z = x + y. Hence f (x, y) = z. This proves that f is surjective. To show that f is injective, we check that ker f = {(0, 0)}. Let (x, y) ∈ ker f. Then f (x, y) = x + y = 0. We see that x = −y in Z.
Therefore x = −y ∈ X ∩ Y = {0} (Z is the internal direct sum of X and Y.) We find x = y = 0.
Hence (x, y) = (0, 0). We conclude that f : X ⊕eY → Z is a linear isomorphism. Since X ⊕iY is isomorphic to X ⊕eY, if X ∩ Y = {0}, we do not distinguish X ⊕iY and X ⊕eY when X ∩ Y = {0} and X, Y are vector subspaces of Z. We use the notation X ⊕ Y for both of them when X ∩ Y = {0}. We call X ⊕ Y the direct sum of X and Y for simplicity.
1
