Linear Algebra

134 Chapter Two. Vector SpacesProof. We will show that (1) =⇒ (2), that (2) =⇒ (3), and finally that(3) =⇒ (1). For these arguments, observe that we can pass from a combinationof ⃗w’s to a combination of ⃗ β’sd 1 ⃗w 1 + · · · + d k ⃗w k= d 1 (c 1,1⃗ β1,1 + · · · + c 1,n1⃗ β1,n1 ) + · · · + d k (c k,1⃗ βk,1 + · · · + c k,nk⃗ βk,nk )= d 1 c 1,1 · ⃗β 1,1 + · · · + d k c k,nk · ⃗β k,nk (∗)and vice versa.For (1) =⇒ (2), assume that all decompositions are unique. We will showthat B 1⌢· · ·⌢Bk spans the space and is linearly independent. It spans thespace because the assumption that V = W 1 + · · · + W k means that every ⃗vcan be expressed as ⃗v = ⃗w 1 + · · · + ⃗w k , which translates by equation (∗) to anexpression of ⃗v as a linear combination of the ⃗ β’s from the concatenation. Forlinear independence, consider this linear relationship.⃗0 = c 1,1⃗ β1,1 + · · · + c k,nk⃗ βk,nkRegroup as in (∗) (that is, take d 1 , . . . , d k to be 1 and move from bottom totop) to get the decomposition ⃗0 = ⃗w 1 + · · · + ⃗w k . Because of the assumptionthat decompositions are unique, and because the zero vector obviously has thedecomposition ⃗0 = ⃗0+· · ·+⃗0, we now have that each ⃗w i is the zero vector. Thismeans that c i,1βi,1 ⃗ + · · · + c i,niβi,ni⃗ = ⃗0. Thus, since each B i is a basis, we havethe desired conclusion that all of the c’s are zero.For (2) =⇒ (3), assume that B 1⌢· · ·⌢Bk is a basis for the space. Considera linear relationship among nonzero vectors from different W i ’s,⃗0 = · · · + d i ⃗w i + · · ·in order to show that it is trivial. (The relationship is written in this waybecause we are considering a combination of nonzero vectors from only some ofthe W i ’s; for instance, there might not be a ⃗w 1 in this combination.) As in (∗),⃗0 = · · ·+d i (c i,1βi,1 ⃗ +· · ·+c i,niβi,ni ⃗ )+· · · = · · ·+d i c i,1·⃗β i,1 +· · ·+d i c i,ni·⃗β i,ni +· · ·and the linear independence of B 1⌢· · ·⌢Bk gives that each coefficient d i c i,j iszero. Now, ⃗w i is a nonzero vector, so at least one of the c i,j ’s is not zero, andthus d i is zero. This holds for each d i , and therefore the linear relationship istrivial.Finally, 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.QED