## Thursday, May 03, 2007

### Vector spaces

Today's blog is taken straight from Matrices and Linear Algebra by Hans Schneider and George Philip Barker.

Definition 1: Vector

A vector is any matrix of the form 1 x n or m x 1. [See Definition 1, here for review of matrices if needed]

Definition 2: Vector Space

A vector space V is a set of vectors such that the following 10 properties are met:

(1) V is nonempty.

(2) V is closed on addition:

If x,y ∈ V, then x + y ∈ V.

(3) V is closed on scalar multiplication:

For any number α, if x ∈ V, then αx ∈ V.

For x,y,z ∈ V, (x + y) + z = x + (y + z)

(5) Existence of 0:

0 ∈ V such that for all x ∈ V, x + 0 = x = 0 + x.

(6) Existence of negative:

For each vector x ∈ V, there is a vector y ∈ V such that:

x + y = 0 = y + x

For all vectors x,y ∈ V, x + y = y + x.

(8) All vectors in V are distributive:

For all numbers α and for all x,y ∈ V:
α(x + y) = αx + αy

For all numbers α, β and for all x ∈ V:
(α + β)x = αx + βx

(9) Scalar multiplication is associative:

For all numbers α, β and for all x ∈ V:
(αβ)x = α(βx)

(10) Existence of 1:

For all x ∈ V, 1x = x.

This definition is very important. For example, it is used to establish Cramer's Rule regarding matrices.

Definition 3: subspace

A vector space W is a subspace if and only if there exists a vector space V such that W is a nonempty subset of V.

We can use the following to determine if a nonepty subset of a vector space is a subspace:

Theorem 1: Criteria for a subspace

Let V be a vector space and let W be a nonempty subset of V. Then W is a subspace of V if and only if W is closed under the addition and under scalar multiplication.

Proof:

(1) Assume that W is a subspace.

(2) Then by Definition 3 above, W is a vector space.

(3) Then, W is closed under addition and under scalar multiplication. [See Definition 2, #2 and #3].

(4) Assume that W is closed under addition and under scalar multiplication.

(5) Them W is a vector space by Definition 2 above since:

(a) W is nonempty [from the given]

(b) W is closed under addition. [from the assumption in step #4]

(c) W is closed under scalar multiplication. [from the assumption in step #4]

(d) Addition in W is associative since:

Assume x,y,z ∈ W

Then x,y,z ∈ V [since W ⊆ V from the given]

Then (x + y) + z = x + (y + z) [since V is vector space, from definition 2, #4]

(e) Existence of 0 since:

Assume that x ∈ W

0x ∈ W since W is closed under scalar multiplication and 0 is a number which is scalar.

It is therefore clear that 0 exists since x + 0x = x = 0x + x.

(f) Existence of negative since:

Assume that x ∈ W

Then (-1)*x = -x ∈ W since W is closed under scalar multiplication and (-1) is a number.

And x + (-1)*x = 0*x = (-1)*x + x.

Assume that x,y ∈ W

Then x,y ∈ V [since V ⊆ W from the given]

Then x + y = y + x [since V is a vector space, from definition 2, #7]

(h) W is distributive since:

Assume that x,y ∈ W

Then (x + y) ∈ W since W is closed under addition.

Then α(x + y) ∈ W since W is closed under scalar multiplication.

If α, β are numbers, then (α + β) is a number since numbers are closed on addition.

If x ∈ W, then it follows that (α + β)x ∈ W since W is closed under scalar multiplication.

(i) Scalar multiplication on W is associative since:

Assume x ∈ W

Then x ∈ V [since W ⊆ V from the given]

Then for any numbers α, β, (αβ)x = α(βx) [since V is a vector space, from Definition 2, #9]

(j) Existence of 1 since:

Assume x ∈ W

Then x ∈ V [since W ⊆ V from the given]

Then 1*x = x [from Definition 2, #10]

QED

Definition 4: family

A family is an ordered sequence (v1, v2, v3, ..., vi) of vectors from a given vector space such that if a vector u ≠ vector v, then (u,u,v) ≠ (u,v,u) ≠ (v,u,u) ≠ (u,v).

And here is the last definition:

Definition 5: linear combination

Let (x1, ..., xt) be a family of vectors in a vector space V. We call u in V a linear combination of (x1, ..., xt) if and only if there exists scalars α1, ..., αt such that:

u = α1x1 + ... + αtxt = ∑ (i=1,t) αixi.

Theorem 2: Let V be a vector space and let x1, ..., xt be in V. Then the set of all linear combinations of (x1, ..., xt) is a subspace of V.

Proof:

(1) Let W be the set of all linear combinations of (x1, ..., xt). [See Definition 5 above]

(2) For all x ∈ W, it is clear that x ∈ V. [from Definition 5 above]

(3) Thus, W ⊆ V

(4) W is nonempty since if if αi = 1,

then x1 + ... + xt ∈ W [since W includes all possible linear combinations of (x1, ..., xt).]

(5) W is closed on addition since:

(a) Assume u,v ∈ W

(b) Then, there exists αi such that:

u = α1x1 + ... + αtxt = ∑ (i=1,t) αixi. [since W is the set of all linear combinations of (x1, ..., xt).]

(c) Then, there exists βi such that:

v = β1x1 + ... + βtxt = ∑ (i=1,t) βixi. [since W is the set of all linear combinations of (x1, ..., xt).]

(d) And then (u+v) ∈ W since:

(u + v) = (α1 + β1)x1 + ... + (αt + βt)xt = ∑ (i=1,t) (αi + βi)xi. [since W is the set of all linear combinations of (x1, ..., xt).]

(6) Likewise, W is closed under scalar multiplication since:

(a) Assume that u ∈ W.

(b) Then, there exists αi such that:

u = α1x1 + ... + αtxt = ∑ (i=1,t) αixi. [since W is the set of all linear combinations of (x1, ..., xt).]

(c) But then for any scalar γ, γu ∈ W since:

γu = (γα1)x1 + ... + (γαt)xt = ∑ (i=1,t) (γαi)xi. [since W is the set of all linear combinations of (x1, ..., xt).]

(7) We can now use Theorem 1 above to establish that W is a subspace of V since:

(a) V is a vector space [given]

(b) W is a nonempty subset of V. [see step #3 and step #4]

(c) W is closed under addition [see step #5] and W is closed under scalar multiplication [see step #6]

QED

References
• Hans Schneider, George Philip Barker, , 1989.