Linear Algebra

Section III. Basis and Dimension 1354.9 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 The benchmark model fits: 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 sum decompositionsare not unique.4.13 Corollary The dimension of a direct sum is the sum of the dimensionsof its summands.Proof. In Lemma 4.8, the number of basis vectors in the concatenation equalsthe sum of the number of vectors in the subbases 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 subspaces,then they are said to be complements.4.15 Lemma A vector space V is the direct sum of two of its subspaces W 1and W 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 thetwo. To show that the two have a trivial intersection, let ⃗v be a vector fromW 1 ∩ W 2 and consider the equation ⃗v = ⃗v. On the left side of that equationis a member of W 1 , and on the right side is a linear combination of members(actually, of only one member) of W 2 . But the independence of the spaces thenimplies that ⃗v = ⃗0, as desired.For the other direction, suppose that V is the sum of two spaces with a trivialintersection. To show that V is a direct sum of the two, we need only showthat the spaces are independent — no nonzero member of the first is expressibleas a linear combination of members of the second, and vice versa. This istrue because any relationship ⃗w 1 = c 1 ⃗w 2,1 + · · · + d k ⃗w 2,k (with ⃗w 1 ∈ W 1 and⃗w 2,j ∈ W 2 for all j) shows that the vector on the left is also in W 2 , since theright side is a combination of members of W 2 . The intersection of these twospaces is trivial, so ⃗w 1 = ⃗0. The same argument works for any ⃗w 2 . QED