Linear Algebra, 2020a

146 Chapter Two. Vector Spaces
has a solution for any x, y, z ∈ R. And the fact that each such expression is
unique reflects that fact that E 3 is linearly independent, so any equation like
the one above has a unique solution.
4.6 Example We don’t have to take the basis vectors one at a time, we can
conglomerate them into larger sequences. Consider again the space R 3 and the
vectors from the standard basis E 3 . The subspace with the basis B 1 = 〈⃗e 1 ,⃗e 3 〉
is the xz-plane. The subspace with the basis B 2 = 〈⃗e 2 〉 is the y-axis. As in the
prior example, the fact that any member of the space is a sum of members of
the two subspaces in one and only one way
⎛ ⎞ ⎛
x
⎜ ⎟ ⎜
x
⎞ ⎛ ⎞
0
⎟ ⎜ ⎟
⎝y⎠ = ⎝0⎠ + ⎝y⎠
z z 0
is a reflection of the fact that these vectors form a basis — this equation
⎛ ⎞ ⎛
x
⎜ ⎟ ⎜
1
⎞ ⎛
⎟ ⎜
0
⎞ ⎛
⎟ ⎜
0
⎞
⎟
⎝y⎠ =(c 1 ⎝0⎠ + c 3 ⎝0⎠)+c 2 ⎝1⎠
z 0 1 0
has one and only one solution for any x, y, z ∈ R.
4.7 Definition The concatenation of the sequences B 1 = 〈⃗β 1,1 ,...,⃗β 1,n1 〉, ...,
B k = 〈⃗β k,1 ,...,⃗β k,nk 〉 adjoins them into a single sequence.
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 subspaces
V = W 1 + ···+ W k .LetB 1 , ..., B k be bases for these subspaces. The following
are equivalent.
(1) The expression of any ⃗v ∈ V as a combination ⃗v = ⃗w 1 + ···+ ⃗w k with
⃗w i ∈ W i is unique.
(2) The concatenation B 1
⌢···⌢
Bk is a basis for V.
(3) Among nonzero vectors from different W i ’s every linear relationship is
trivial.
Proof We will show that (1) =⇒ (2), that (2) =⇒ (3), and finally that
(3) =⇒ (1). For these arguments, observe that we can pass from a combination
of ⃗w’s to a combination of ⃗β’s
d 1 ⃗w 1 + ···+ d k ⃗w k = d 1 (c 1,1
⃗β 1,1 + ···+ c 1,n1
⃗β 1,n1 )
+ ···+ d k (c k,1
⃗β k,1 + ···+ c k,nk
⃗β k,nk )
= d 1 c 1,1 · ⃗β 1,1 + ···+ d k c k,nk · ⃗β k,nk (∗)