Linear Algebra, 2020a

130 Chapter Two. Vector Spaces
2.2 Example Here is a basis for R 3 and a vector given as a linear combination of
members of that basis.
⎛
⎜
1
⎞ ⎛
⎟ ⎜
1
⎞ ⎛
⎟ ⎜
0
⎞ ⎛
⎟ ⎜
1
⎞ ⎛
⎟ ⎜
1
⎞ ⎛
⎟ ⎜
1
⎞ ⎛
⎟ ⎜
0
⎞
⎟
B = 〈 ⎝0⎠ , ⎝1⎠ , ⎝0⎠〉
⎝2⎠ =(−1) · ⎝0⎠ + 2 ⎝1⎠ + 0 · ⎝0⎠
0 0 2 0
0 0 2
Two of the basis vectors have non-zero coefficients. Pick one, for instance the
first. Replace it with the vector that we’ve expressed as the combination
⎛
⎜
1
⎞ ⎛
⎟ ⎜
1
⎞ ⎛
⎟ ⎜
0
⎞
⎟
ˆB = 〈 ⎝2⎠ , ⎝1⎠ , ⎝0⎠〉
0 0 2
and the result is another basis for R 3 .
2.3 Lemma (Exchange Lemma) Assume that B = 〈⃗β 1 ,...,⃗β n 〉 is a basis for a
vector space, and that for the vector ⃗v the relationship ⃗v = c 1
⃗β 1 + c 2
⃗β 2 + ···+
c n
⃗β n has c i ≠ 0. Then exchanging ⃗β i for ⃗v yields another basis for the space.
Proof Call the outcome of the exchange ˆB = 〈⃗β 1 ,...,⃗β i−1 ,⃗v, ⃗β i+1 ,...,⃗β n 〉.
We first show that ˆB is linearly independent. Any relationship d 1
⃗β 1 + ···+
d i ⃗v + ···+ d n
⃗β n = ⃗0 among the members of ˆB, after substitution for ⃗v,
d 1
⃗β 1 + ···+ d i · (c 1
⃗β 1 + ···+ c i
⃗β i + ···+ c n
⃗β n )+···+ d n
⃗β n = ⃗0
(∗)
gives a linear relationship among the members of B. The basis B is linearly
independent so the coefficient d i c i of ⃗β i is zero. Because we assumed that c i is
nonzero, d i = 0. Using this in equation (∗) gives that all of the other d’s are
also zero. Therefore ˆB is linearly independent.
We finish by showing that ˆB has the same span as B. Half of this argument,
that [ˆB] ⊆ [B], is easy; we can write any member d 1
⃗β 1 +···+d i ⃗v+···+d n
⃗β n of [ˆB]
as d 1
⃗β 1 +···+d i ·(c 1
⃗β 1 +···+c n
⃗β n )+···+d n
⃗β n , which is a linear combination
of linear combinations of members of B, and hence is in [B]. For the [B] ⊆ [ˆB]
half of the argument, recall that if ⃗v = c 1
⃗β 1 +···+c n
⃗β n with c i ≠ 0 then we can
rearrange the equation to ⃗β i =(−c 1 /c i )⃗β 1 + ···+(1/c i )⃗v + ···+(−c n /c i )⃗β n .
Now, consider any member d 1
⃗β 1 + ···+ d i
⃗β i + ···+ d n
⃗β n of [B], substitute for
⃗β i its expression as a linear combination of the members of ˆB, and recognize,
as in the first half of this argument, that the result is a linear combination of
linear combinations of members of ˆB, and hence is in [ˆB].
QED