Tuesday, June 3, 2014

Linear Algebra: #3 Subspaces

  • Linear Algebra: #3 Subspaces

Let V be a vector space over a field F and let WV be some subset. If W is itself a vector space over F, considered using the addition and scalar multiplication in V, then we say that W is 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 WV 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, wW and a, b ∈ F.

Proof. 
The direction ‘⇒’ is trivial.

For ‘⇐’, begin by observing that 1 · v + 1 · w = v + wW, and a · v + 0 · w = a · vW, for all v, wW 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 = 0W, for vW; therefore the neutral element 0 is contained in W. For an arbitrary vW we have

Linear Algebra: #3 Subspaces equation pic 1


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.

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, WV be subspaces of the vector space V over the field F. Then UW is also a subspace.

Proof 
Let v, wUW 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 · wU. This follows from theorem 2. Similarly a · v + b · wW. Thus it is in the intersection, and so theorem 2 shows that UW is a subspace.

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