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 2 conditions:

1. If u and v are elements of W, then the sum u + v is an element of W;
2. 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: Looking at the definition of a vector space, we see that properties 1 and 2 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, the other properties of a vector space are satisfied a fortiori.

Vector subspaces are subgroups of vector spaces under vector addition.

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 = (c'u₁,c'u₂,c0) = (c'u₁,c'u₂,0). 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 = (c'p₁,c'p₂); since p₁ = p₂, then c'p₁ = c'p₂, 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 iff 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

$U+W=\{\mathbf {u} +\mathbf {w} :\mathbf {u} \in U{\mbox{ and }}\mathbf {w} \in W\}$

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

$\dim(U\cap W)+\dim(U+W)=\dim U+\dim W.$

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

• "Vector subspace". PlanetMath..