A First Course in Linear Algebra, 2017a

480 Vector Spaces
{ }
2. Let ⃗w 1 ,⃗w 2 be in ⃗ 0 .Then⃗w 1 =⃗0 and⃗w 2 =⃗0 andso
⃗w 1 + ⃗w 2 =⃗0 +⃗0 =⃗0
{ }
It follows that the sum is contained in ⃗ 0 and the second condition is satisfied.
{ }
3. Let ⃗w 1 be in ⃗ 0 and let a be an arbitrary scalar. Then
a⃗w 1 = a⃗0 =⃗0
{ }
Hence the product is contained in ⃗ 0 and the third condition is satisfied.
{ }
It follows that ⃗ 0 is a subspace of V .
♠
The two subspaces described { } above are called improper subspaces. Any subspace of a vector space
V which is not equal to V or ⃗ 0 is called a proper subspace.
Consider another example.
Example 9.32: Subspace of Polynomials
Let P 2 be the vector space of polynomials of degree two or less. Let W ⊆ P 2 be all polynomials of
degree two or less which have 1 as a root. Show that W is a subspace of P 2 .
Solution. First, express W as follows:
W = { p(x)=ax 2 + bx + c,a,b,c,∈ R|p(1)=0 }
We need to show that W satisfies the three conditions of Procedure 9.30.
1. The zero polynomial of P 2 is given by 0(x)=0x 2 +0x+0 = 0. Clearly 0(1)=0so0(x) is contained
in W .
2. Let p(x),q(x) be polynomials in W. It follows that p(1)=0andq(1)=0. Now consider p(x)+q(x).
Let r(x) represent this sum.
r(1) = p(1)+q(1)
= 0 + 0
= 0
Therefore the sum is also in W and the second condition is satisfied.
3. Let p(x) be a polynomial in W and let a be a scalar. It follows that p(1)=0. Consider the product
ap(x).
ap(1) = a(0)
= 0
Therefore the product is in W and the third condition is satisfied.