Wikipedia: History of Linear algebra/Subspace

Difference (from prior major revision) (no other diffs)

Changed: 1c1

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.

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.

Changed: 3c3

#We shall use 0 to denote the number zero and use 0 to denote the additive identity, which is a vector not a number.

#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.

Changed: 6c6

Definition : Let V be a vector space over a field K and let W be a subset of V.

Definition: Let V be a vector space over a field K and let W be a subset of V.

Changed: 9c9

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.

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.

Changed: 19c19

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.

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.

Changed: 27,33c27,30

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 a
an element of W.

Proof:
#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.
#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.
#Property 3: Since (0,0,...,0) is an element of W, by definition, 0 is an element of W.

Changed: 37c34,37

Proof:

Proof:
#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.
#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.
#Property 3: Consider the point 0= (0,0) in Rⁿ. Then, since 0=0, 0 is an element of W.

Changed: 39,54c39

1. Property I: 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: (Calulus required)

Example III: (Calculus required)

Changed: 61,73c46,49

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, 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.

Proof:
#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.

#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.

#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.

Changed: 79,89c55,58

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.

Proof:
#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.
#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.
#Property 3: Since U and W are subspaces of V, 0 belongs to both U and W.

	Revision 2 . . November 11, 2001 12:03 am by Damian Yerrick [Added 'scalar' because vectors are numbers too; converted examples' preformatting into ordered lists]
	Revision 1 . . November 7, 2001 12:13 pm by Damian Yerrick [moved from Linear Algebra/Subspace]