Linear subspace : Linear algebra/Subspace 

Definition and useful characterisation

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. If u and v are elements of W, then the sum u + v of u and 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 cv is an element of W;
3. W isn't empty.

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. By property 3, there is some element w of W. Multiplying this by the scalar 0, we get an additive identity, 0w = 0 in W, and we get the additive inverse of any vector in W by multiplying it by the scalar −1. Since elements of W are necessarily elements of V, the other properties of a vector space are satisfied a fortiori.

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.
3. 0 = (0,0,0) is immediately an element of W.

Example II: Let the field be R again, but now let the vector space be the Euclidean plane 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.
3. Again, the point 0 = (0,0) belongs to W, since 0 = 0.

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.
3. Consider the function 0 from R to R defined by 0(x) = 0 for all x in R. This zero function is continuous from R into R.

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

In the examples above, we always found that the subspace was nonempty because the zero vector belonged to it. This is no coincidence; in fact, 0 always belongs to every subspace, and it's usually the easiest and most natural was to check condition 3.

Another 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, 2, and 3 for a subspace are simply the most basic kinds of linear combinations (where for condition 3, we remember that a linear combination of no vectors at all yields the zero vector).

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.
3. Since U and W are subspaces of V, 0 belongs to both U and W by a property in the previous section. Thus, U ∩ W is nonempty.

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

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