A First Course in Linear Algebra, 2017a

9.4. Subspaces and Basis 479
1. U ⊆ W
Notice that 2p(x)−q(x) and p(x)+3q(x) are both in W = span{p(x),q(x)}. Then by Theorem 9.28
W must contain the span of these polynomials and so U ⊆ W .
2. W ⊆ U
Notice that
p(x) = 3 7 (2p(x) − q(x)) + 1 7 (p(x)+3q(x))
q(x) = − 1 7 (2p(x) − q(x)) + 2 7 (p(x)+3q(x))
Hence p(x),q(x) are in span{2p(x) − q(x), p(x)+3q(x)}. By Theorem 9.28 U must contain the
span of these polynomials and so W ⊆ U.
To prove that a set is a vector space, one must verify each of the axioms given in Definition 9.2 and
9.3. This is a cumbersome task, and therefore a shorter procedure is used to verify a subspace.
Procedure 9.30: Subspace Test
Suppose W is a subset of a vector space V. TodetermineifW is a subspace of V , it is sufficient to
determine if the following three conditions hold, using the operations of V:
1. The additive identity⃗0 of V is contained in W.
2. For any vectors ⃗w 1 ,⃗w 2 in W , ⃗w 1 + ⃗w 2 is also in W .
3. For any vector ⃗w 1 in W and scalar a, the product a⃗w 1 is also in W.
♠
Therefore it suffices to prove these three steps to show that a set is a subspace.
Consider the following example.
Example 9.31: Improper Subspaces
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Let V be an arbitrary vector space. Then V is a subspace of itself. Similarly, the set ⃗ 0 containing
only the zero vector is also a subspace.
{ }
Solution. Using the subspace test in Procedure 9.30 we can show that V and ⃗ 0 are subspaces of V.
Since V satisfies the vector space axioms it also satisfies the three steps of the subspace test. Therefore
V is a subspace.
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Let’s consider the set ⃗ 0 .
{ }
1. The vector⃗0 is clearly contained in ⃗ 0 , so the first condition is satisfied.