Linear Subspace - Definition and Useful Characterization of Subspace

Definition and Useful Characterization of Subspace

Let K be a field (such as the field of real numbers), and let V be a vector space over K. As usual, we call elements of V vectors and call elements of K scalars. Suppose that W is a subset of V. If W is a vector space itself, with the same vector space operations as V has, then it is a subspace of V.

To use this definition, we don't have to prove that all the properties of a vector space hold for W. Instead, we can prove a theorem that gives us an easier way to show that a subset of a vector space is a subspace.

Theorem: Let V be a vector space over the field K, and let W be a subset of V. Then W is a subspace if and only if it satisfies the following three conditions:

1. The zero vector, 0, is in W.
2. If u and v are elements of W, then the sum u + v is an element of W;
3. If u is an element of W and c is a scalar from K, then the scalar product cu is an element of W;

Proof: Firstly, property 1 ensures W is nonempty. Looking at the definition of a vector space, we see that properties 2 and 3 above assure closure of W under addition and scalar multiplication, so the vector space operations are well defined. Since elements of W are necessarily elements of V, axioms 1, 2 and 5-8 of a vector space are satisfied. By the closure of W under scalar multiplication

Conversely, if W is subspace of V, then W is itself a vector space under the operations induced by V, so properties 2 and 3 are satisfied. By property 3, -w is in W whenever w is, and it follows that W is closed under subtraction as well. Since W is nonempty, there is an element x in W, and is in W, so property 1 is satisfied. One can also argue that since W is nonempty, there is an element x in W, and 0 is in the field K so and therefore property 1 is satisfied.

Example I: Let the field K be the set R of real numbers, and let the vector space V be the Euclidean space R3. Take W to be the set of all vectors in V whose last component is 0. Then W is a subspace of V.

Proof:

1. Given u and v in W, then they can be expressed as u = (u₁,u₂,0) and v = (v₁,v₂,0). Then u + v = (u₁+v₁,u₂+v₂,0+0) = (u₁+v₁,u₂+v₂,0). Thus, u + v is an element of W, too.
2. Given u in W and a scalar c in R, if u = (u₁,u₂,0) again, then cu = (cu₁, cu₂, c0) = (cu₁,cu₂,0). Thus, cu is an element of W too.

Example II: Let the field be R again, but now let the vector space be the Euclidean geometry R2. Take W to be the set of points (x,y) of R2 such that x = y. Then W is a subspace of R2.

Proof:

1. Let p = (p₁,p₂) and q = (q₁,q₂) be elements of W, that is, points in the plane such that p₁ = p₂ and q₁ = q₂. Then p + q = (p₁+q₁,p₂+q₂); since p₁ = p₂ and q₁ = q₂, then p₁ + q₁ = p₂ + q₂, so p + q is an element of W.
2. Let p = (p₁,p₂) be an element of W, that is, a point in the plane such that p₁ = p₂, and let c be a scalar in R. Then cp = (cp₁,cp₂); since p₁ = p₂, then cp₁ = cp₂, so cp is an element of W.

In general, any subset of a Euclidean space Rn that is defined by a system of homogeneous linear equations will yield a subspace. (The equation in example I was z = 0, and the equation in example II was x = y.) Geometrically, these subspaces are points, lines, planes, and so on, that pass through the point 0.