Linear Algebra

132 Chapter Two. Vector SpacesFinally, for (3) =⇒ (1), assume that, among nonzero vectors from differentW i ’s, any linear relationship is trivial. Consider two decompositions of a vector⃗v = ⃗w 1 + · · · + ⃗w k and ⃗v = ⃗u 1 + · · · + ⃗u k in order to show that the two are thesame. We have⃗0 = (⃗w 1 + · · · + ⃗w k ) − (⃗u 1 + · · · + ⃗u k ) = (⃗w 1 − ⃗u 1 ) + · · · + (⃗w k − ⃗u k )which violates the assumption unless each ⃗w i − ⃗u i is the zero vector. Hence,decompositions are unique.QED4.9 Definition A collection of subspaces {W 1 , . . . , W k } is independent if nononzero vector from any W i is a linear combination of vectors from the othersubspaces W 1 , . . . , W i−1 , W i+1 , . . . , W k .4.10 Definition A vector space V is the direct sum (or internal direct sum)of its subspaces W 1 , . . . , W k if V = W 1 + W 2 + · · · + W k and the collection{W 1 , . . . , W k } is independent. We write V = W 1 ⊕ W 2 ⊕ · · · ⊕ W k .4.11 Example Our prototype works: R 3 = x-axis ⊕ y-axis ⊕ z-axis.4.12 Example The space of 2×2 matrices is this direct sum.( ) ( ) ( )a 0 ∣∣ 0 b ∣∣ 0 0 ∣∣{ a, d ∈ R} ⊕ { b ∈ R} ⊕ { c ∈ R}0 d0 0c 0It is the direct sum of subspaces in many other ways as well; direct sumdecompositions are not unique.4.13 Corollary The dimension of a direct sum is the sum of the dimensions of itssummands.Proof In Lemma 4.8, the number of basis vectors in the concatenation equals thesum of the number of vectors in the sub-bases that make up the concatenation.QEDThe special case of two subspaces is worth mentioning separately.4.14 Definition When a vector space is the direct sum of two of its subspacesthen they are complements.4.15 Lemma A vector space V is the direct sum of two of its subspaces W 1 andW 2 if and only if it is the sum of the two V = W 1 + W 2 and their intersectionis trivial W 1 ∩ W 2 = {⃗0 }.Proof Suppose first that V = W 1 ⊕ W 2 . By definition, V is the sum of the two.To show that they have a trivial intersection, let ⃗v be a vector from W 1 ∩ W 2and consider the equation ⃗v = ⃗v. On the left side of that equation is a member