A First Course in Linear Algebra, 2017a

9.4. Subspaces and Basis 485
Then {⃗v 1 ,⃗v 2 } is a basis for V and we are done. If span{⃗v 1 ,⃗v 2 } ≠ V , then there exists ⃗v 3 /∈ span{⃗v 1 ,⃗v 2 }
and {⃗v 1 ,⃗v 2 ,⃗v 3 } is a larger linearly independent set of vectors. Continuing this way, the process must stop
before n+1 steps because if not, it would be possible to obtain n+1 linearly independent vectors contrary
to the exchange theorem, Theorem 9.35.
♠
If in fact W has n vectors, then it follows that W = V .
Theorem 9.41: Subspace of Same Dimension
Let V be a vector space of dimension n and let W be a subspace. Then W = V if and only if the
dimension of W is also n.
Proof. First suppose W = V. Then obviously the dimension of W = n.
Now suppose that the dimension of W is n. Let a basis for W be {⃗w 1 ,···,⃗w n }.IfW is not equal to
V ,thenlet⃗v be a vector of V which is not contained in W. Thus ⃗v is not in span{⃗w 1 ,···,⃗w n } and by
Lemma 9.74, {⃗w 1 ,···,⃗w n ,⃗v} is linearly independent which contradicts Theorem 9.35 because it would be
an independent set of n + 1 vectors even though each of these vectors is in a spanning set of n vectors, a
basis of V .
♠
Consider the following example.
Example 9.42: Basis of a Subspace
{ ∣ [ ]
∣∣∣ 1 0
Let U = A ∈ M 22 A =
1 −1
U, and hence dim(U).
[
1 1
0 −1
[ ]
a b
Solution. Let A = ∈ M
c d 22 .Then
[ ] [
1 0 a b
A =
1 −1 c d
] }
A .ThenU is a subspace of M 22 Find a basis of
][
1 0
1 −1
] [ ]
a + b −b
=
c + d −d
and [ ] [ ][ ] [ ]
1 1 1 1 a b a + c b+ d
A =
=
.
0 −1 0 −1 c d −c −d
[ ] [ ]
a + b −b a + c b+ d
If A ∈ U,then
=
.
c + d −d −c −d
Equating entries leads to a system of four equations in the four variables a,b,c and d.
a + b = a + c
−b = b + d
c + d = −c
−d = −d
or
b − c = 0
−2b − d = 0
2c + d = 0
The solution to this system is a = s, b = − 1 2 t, c = − 1 2t, d = t for any s,t ∈ R, and thus
[ ] [ ] [
s
t
A = 2 1 0 0 − 1 ]
−
2
t = s +t
2
t 0 0 − 1 .
2
1
.