Linear algebra/Subspace

Before we define a subspace, in the interests of simplicity we will revise some of the cumbersome notation used in the definition of a vector space.

1. If u, v are vectors of an arbitrary vector space, then we will write u+(-v) as u-v.
2. We shall use 0 to denote the scalar? (i.e. real or complex number) zero and use 0 or 0 to denote the vector additive identity.
3. The notation u+v+w+z will represent ((u+v)+w)+z. This simplifies notation for addition and subtraction of multiple vectors while realizing that addition is still a binary operation.

Definition: Let V be a vector space over a field K and let W be a subset of V. Then if W is a vector space, itself, it is a subspace of V over K.

To use this definition means that given a subset W of V, one would have to prove that all 10 properties of a vector space hold for W. We can prove a theorem that gives us an easier way to prove that a subset of a vector space, is a vector space itself.

Theorem: Let V be a vector space and let W be a subset of V. Assume that W satisfies the following 3 conditions:

1. If u and v are elements of W, then the sum of u and v, u+v is an element of W.
2. If u is an element of W, and c is a member of K, the c*v is an element of W.
3. The element 0 of V is also 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 in W. Given that we have an additive identity, 0 in W, by property 3, since elements of W are necessarily elements of V, the other 8 properties of a vector space are satisfied a fortiori.

Note: For every vector space V, the empty vector space { } and V, itself, are subsets of V.

Example I:

Let V=Rⁿ and let W be the set of all vectors in V whose last component is 0. Then W is a subspace of V.

Proof:

1. Property 1: Given u,v in W, then u=(u1,u2,…,0) and v=(v1,v2,…,0). Then u+v=(u1+v1,u2+v2,...,0+0)=(u1+v1,u2+v2,...,0). Then u+v is an element in W.
2. Property 2: Given u in W and c in R, then c*v=c*(u1,u2,...0)=(c*u1, c*u2,…,c*0)=(c*u1,c*u2,...,0). Then c*u is an element of W.
3. Property 3: Since (0,0,...,0) is an element of W, by definition, 0 is an element of W.

Example II: Let us remember that the set of points in a line are points in a plane. Then for the set of points (x,y) of R² such that x=y is a subspace of Rⁿ.

Proof:

1. Property 1: Let P=(p₁,p₂) and Q=(q₁,q₂) be elements of W, that is, points such that p₁=p₂ and p₁=q₂. Then P+Q=p₁+q₁,p₂+q₂). Then since p₁=p₂ and q₁=q₂, P+Q=(p₁+q₁,p₁+q₁)= (2p₁,2q₁). Then P+Q is an element of W.
2. Property 2: Let P=(p,q) be an element of W, that is, a point such that p=q, and let c be an element of R. Then c*P=(c*p,c*q). Then since p=q, c*p=C*q and cP is an element of W.
3. Property 3: Consider the point 0= (0,0) in Rⁿ. Then, since 0=0, 0 is an element of W.

Example III: (Calculus required)

First let us define a vector space V. Let V be the vector space over R of all functions from R into R. (Try and prove this is a vector space yourself.)

Let W be the subset of V of continuous functions. Then W is a subspace of V.

Proof:

1. Property 1: Let f and g be elements of W, that is, continuous functions from R into R. Then we know from Calculus (specifically the linearity of sums of functions), that f+g is a continuous function from R into R.
   
2. Property 2: Let f be a element of W, that is, a continuous function from R into R, and let c be an element of R. Then we know from Calculus, that c*f is a continuous function from R into R.
   
3. Property 3: Consider the function 0 from R into R, defined by 0(x)=0 for all x in R. This is called the zero function and we know from Calculus that it is continuous from R into R. Then 0 is an element in W.
   

Note: We can define U as the subset of W of differentiable? functions. Then, basically, since every differentiable function is continuous, U is a subspace of W, which, in turn, is a subspace of V.

Example IV: Given U and W, subspaces of a vector space V over a filed K. Then if U^W={all v: v is an element of U and v is an element of W}, U^W is a subspace of V.

Proof:

1. Property 1: Let v1 and v2 be elements of U^W. Let v1 and v2 belong to U^W. Then vi and v2 belong to both U and W. Consider v1+v2. Since v1 and v2 belong to U, then v1+v2 belongs to U. Similarly since v1 and v2 belong to W, v1+v2 belongs to W. Then, v1+v2 belongs to U^W.
2. Property 2: Let v belong to U^W. Then v belongs to U and v belongs to W. Given c an element of K, consider c*v. Since U and W are vector spaces, cv belongs to both U and W.
3. Property 3: Since U and W are subspaces of V, 0 belongs to both U and W.