Linear Algebra

Section III. Basis and Dimension 133is a reflection of the fact that E 3 spans the space — this equation⎛⎝ x ⎞ ⎛y⎠ = c 1⎝ 1 ⎞ ⎛0⎠ + c 2⎝ 0 ⎞ ⎛1⎠ + c 3⎝ 0 ⎞0⎠z 0 0 1has a solution for any x, y, z ∈ R. And, the fact that each such expression isunique reflects that fact that E 3 is linearly independent — any equation like theone above has a unique solution.4.6 Example We don’t have to take the basis vectors one at a time, the sameidea works if we conglomerate them into larger sequences. Consider again thespace R 3 and the vectors from the standard basis E 3 . The subspace with thebasis B 1 = 〈⃗e 1 , ⃗e 3 〉 is the xz-plane. The subspace with the basis B 2 = 〈⃗e 2 〉 isthe y-axis. As in the prior example, the fact that any member of the space is asum of members of the two subspaces in one and only one way⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎝ x y⎠ =z⎝ x 0z⎠ +⎝ 0 y⎠0is a reflection of the fact that these vectors form a basis — this system⎛⎝ x ⎞ ⎛y⎠ = (c 1⎝ 1 ⎞ ⎛0⎠ + c 3⎝ 0 ⎞ ⎛0⎠) + c 2⎝ 0 ⎞1⎠z 0 1 0has one and only one solution for any x, y, z ∈ R.These examples illustrate a natural way to decompose a space into a sumof subspaces in such a way that each vector decomposes uniquely into a sum ofvectors from the parts. The next result says that this way is the only way.4.7 Definition The concatenation of the sequences B 1 = 〈 ⃗ β 1,1 , . . . , ⃗ β 1,n1 〉,. . . , B k = 〈 ⃗ β k,1 , . . . , ⃗ β k,nk 〉 is their adjoinment.B 1⌢B2⌢· · ·⌢Bk = 〈 ⃗ β 1,1 , . . . , ⃗ β 1,n1 , ⃗ β 2,1 , . . . , ⃗ β k,nk 〉4.8 Lemma Let V be a vector space that is the sum of some of its subspacesV = W 1 + · · · + W k . Let B 1 , . . . , B k be any bases for these subspaces. Thenthe following are equivalent.(1) For every ⃗v ∈ V , the expression ⃗v = ⃗w 1 + · · · + ⃗w k (with ⃗w i ∈ W i ) isunique.(2) The concatenation B 1⌢· · ·⌢Bk is a basis for V .(3) The nonzero members of { ⃗w 1 , . . . , ⃗w k } (with ⃗w i ∈ W i ) form a linearlyindependent set — among nonzero vectors from different W i ’s, every linearrelationship is trivial.