Page 196 - 35Linear Algebra

196 Subspaces and Spanning Sets

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Example 109 Consider a plane P in R through the origin:
ax + by + cz = 0.

 
x

y
This equation can be expressed as the homogeneous system a b c   = 0, or
z

MX = 0 with M the matrix a b c . If X 1 and X 2 are both solutions to MX = 0,
then, by linearity of matrix multiplication, so is µX 1 + νX 2 :
M(µX 1 + νX 2 ) = µMX 1 + νMX 2 = 0.
So P is closed under addition and scalar multiplication. Additionally, P contains the
origin (which can be derived from the above by setting µ = ν = 0). All other vector
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space requirements hold for P because they hold for all vectors in R .
Theorem 9.1.1 (Subspace Theorem). Let U be a non-empty subset of a
vector space V . Then U is a subspace if and only if µu 1 + νu 2 ∈ U for
arbitrary u 1 , u 2 in U, and arbitrary constants µ, ν.

Proof. One direction of this proof is easy: if U is a subspace, then it is a vector
space, and so by the additive closure and multiplicative closure properties of
vector spaces, it has to be true that µu 1 + νu 2 ∈ U for all u 1 , u 2 in U and all
constants constants µ, ν.
The other direction is almost as easy: we need to show that if µu 1 +νu 2 ∈
U for all u 1 , u 2 in U and all constants µ, ν, then U is a vector space. That
is, we need to show that the ten properties of vector spaces are satisfied. We
already know that the additive closure and multiplicative closure properties
are satisfied. Further, U has all of the other eight properties because V has
them.

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