Linear subspace

The concept of a linear subspace (or vector subspace) is important in linear algebra and related fields of mathematics. A linear subspace is usually called simply a subspace when the context serves to distinguish it from other kinds of subspaces.

Definition and useful characterization

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 3 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 a fortiori. By the closure of W under scalar multiplication (specifically by 0 and -1), axioms 3 and 4 of a vector space are satisfied.

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 Failed to parse (Missing texvc executable; please see math/README to configure.): x - x = {\bold 0}

is in W, so property 1 is satisfied.

Examples

Examples related to analytic geometry

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 an 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.

Examples related to calculus

Example III: Again take the field to be R, but now let the vector space V be the set R^R of all functions from R to R. Let C(R) be the subset consisting of continuous functions. Then C(R) is a subspace of R^R.

Proof:

1. We know from calculus the sum of continuous functions is continuous.
2. Again, we know from calculus that the product of a continuous function and a number is continuous.

Example IV: Keep the same field and vector space as before, but now consider the set Diff(R) of all differentiable functions. The same sort of argument as before shows that this is a subspace too.

Examples that extend these themes are common in functional analysis.

Properties of subspaces

A way to characterise subspaces is that they are closed under linear combinations. That is, W is a subspace if and only if every linear combination of (finitely many) elements of W also belongs to W. Conditions 1 and 2 for a subspace are simply the most basic kinds of linear combinations.

Operations on subspaces

Given subspaces U and W of a vector space V, then their intersection U ∩ W := {v ∈ V : v is an element of both U and W} is also a subspace of V.

Proof:

1. Let v and w be elements of U ∩ W. Then v and w belong to both U and W. Because U is a subspace, then v + w belongs to U. Similarly, since W is a subspace, then v + w belongs to W. Thus, v + w belongs to U ∩ W.
2. Let v belong to U ∩ W, and let c be a scalar. Then v belongs to both U and W. Since U and W are subspaces, cv belongs to both U and W.

Furthermore, the sum

Failed to parse (Missing texvc executable; please see math/README to configure.): U+W = \{ \mathbf{u} + \mathbf{w} : \mathbf{u} \in U \mbox{ and } \mathbf{w} \in W \}

is also a subspace of V. The dimensions of U ∩ W and U + W satisfy

Failed to parse (Missing texvc executable; please see math/README to configure.): \dim (U+W) = \dim U + \dim W - \dim (U \cap W)

For every vector space V, the set {0} and V itself are subspaces of V.

If V is an inner product space, then the orthogonal complement of any subspace of V is again a subspace.

External links

• MIT Linear Algebra Lecture on the Four Fundamental Subspaces at Google Video, from MIT OpenCourseWare
• Vector subspace on PlanetMath.es:Subespacio vectorial

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