Linear Algebra in Physics - Subspaces

Let V be a vector space over a field F and let W ⊂ V be some subset. IfW is itself
a vector space over F, considered using the addition and scalar multiplication in V,
then we say thatWis a subspace of V. Analogously, a subset H of a group G, which
is itself a group using the multiplication operation from G, is called a subgroup of
G. Subfields are similarly defined.
Theorem 2. Let W ⊂ V be a subset of a vector space over the field F. Then
W is a subspace of V ⇔ a · v + b · w ∈W,
for all v,w ∈W and a, b ∈ F.
Proof. The direction ‘⇒’ is trivial.
For ‘⇐’, begin by observing that 1 · v + 1 · w = v + w ∈W, and a · v + 0 · w =
a · v ∈ W, for all v,w ∈ W and a ∈ F. Thus W is closed under vector addition
and scalar multiplication.
Is W a group with respect to vector addition? We have 0 · v = 0 ∈ W, for
v ∈W; therefore the neutral element 0 is contained in W. For an arbitrary v ∈W
we have
v + (−1) · v = 1 · v + (−1) · v
= (1 + (−1)) · v
= 0 · v
= 0.
Therefore (−1) · v is the inverse element to v under addition, and so we can simply
write (−1) · v = −v.
The other axioms for a vector space can be easily checked.
The method of this proof also shows that we have similar conditions for subsets
of groups or fields to be subgroups, or subfields, respectively.
Theorem 3. Let H ⊂ G be a (non-empty) subset of the group G. Then H is a
subgroup of G ⇔ ab−1 ∈ H, for all a, b ∈ H.
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Proof. The direction ‘⇒’ is trivial. As for ‘⇐’, let a ∈ H. Then aa−1 = e ∈ H. Thus
the neutral element of the group multiplication is contained in H. Also ea−1 = a−1 ∈ H. Furthermore, for all a, b ∈ H, we have a(b−1)−1 = ab ∈ H. Thus H is closed
under multiplication. The fact that the multiplication is associative (a(bc) = (ab)c,
for all a, b and c ∈ H) follows since G itself is a group; thus the multiplication
throughout G is associative.
Theorem 4. Let U,W ⊂ V be subspaces of the vector space V over the field F.
Then U ∩W is also a subspace.
Proof. Let v,w ∈ U ∩W be arbitrary vectors in the intersection, and let a, b ∈ F
be arbitrary elements of the field F. Then, since U is a subspace of V, we have
a · v + b ·w ∈ U. This follows from theorem 2. Similarly a · v + b · w ∈W. Thus it
is in the intersection, and so theorem 2 shows that U ∩W is a subspace.


http://www.math.uni-bielefeld.de/~hemion/Linear_Algebra_in_Physics/LinearAlg.Physics.pdf