Difference between revisions of "Subspace MA265F12Alvarado" - Rhea

Latest revision as of 12:12, 3 December 2012

VECTOR SPACE

A vector space is a set of vectors that defines addition V x V --> V and scalar multiplcation cV --> V that satisfy the following properties:

1. Communative Property: u + v = v + u

2. Associative Property:

· Of addition: (u + v) + w = u = (v + w)

· Of multiplication: (ab)v = a(bv)

3. Zero Property: There exist some 0∈V such that 0 + v = v

4. Inverse Property: For every v∈V there is some -v where v + -v = 0

5. Identity Property: 1v=v

6. Distributive Property: a(u + v) = au + av & (a + b)u = au + bu & c(du) = (cd)u

Example

Prove that vector addition and scalar multiplication define R^2 as a vector space.

$ \left[\begin{array}{cc}x1\\x2\end{array}\right] + \left[\begin{array}{cc}y1\\y2\end{array}\right] = \left[\begin{array}{cc}x1+y1\\x2+y2\end{array}\right] $

$ r*\left[\begin{array}{cc}x1\\x2\end{array}\right] = \left[\begin{array}{cc}rx1\\rx2\end{array}\right] $

First, define v1, v2, w1, w2, u1, u2 as elements in R^2 and a,b,c,d as scalars.

Second, check each of the properties in the defintion of a vector space. If a single property fails the entire proof fails.

1. Communative Property:

$ \left[\begin{array}{cc}v1\\v2\end{array}\right] + \left[\begin{array}{cc}w1\\w2\end{array}\right] = \left[\begin{array}{cc}v1+w1\\v2+w2\end{array}\right] $

2. Associative Property:

· Of addition:

$ \left[\begin{array}{cc}u1+v1\\u2+v2\end{array}\right] + \left[\begin{array}{cc}w1\\w2\end{array}\right] =\left[\begin{array}{cc}u1\\u2\end{array}\right] + \left[\begin{array}{cc}v1+w1\\v2+w2\end{array}\right] $

· Of multiplication:

$ (ab)*\left[\begin{array}{cc} v1\\v2\end{array}\right] = \left[\begin{array}{cc}(ab)v1\\(ab)v2\end{array}\right] = \left[\begin{array}{cc}a(bv1)\\a(bv1)\end{array}\right] = a*(b*\left[\begin{array}{cc}v1\\v2\end{array}\right] $

3. Zero Property:

$ \left[\begin{array}{cc}v1\\v2\end{array}\right] + \left[\begin{array}{cc}0\\0\end{array}\right] = \left[\begin{array}{cc}v1\\v2\end{array}\right] $

4. Inverse Property:

$ \left[\begin{array}{cc}v1\\v2\end{array}\right] + \left[\begin{array}{cc}-v1\\-v2\end{array}\right] = \left[\begin{array}{cc}0\\0\end{array}\right] $

5. Identity Property:

$ 1*\left[\begin{array}{cc}v1\\v2\end{array}\right] = \left[\begin{array}{cc}v1\\v2\end{array}\right] $

6. Distributive Property:

$ a*\left[\begin{array}{cc}u1+v1\\u2+v2\end{array}\right] = \left[\begin{array}{cc}au1\\au2\end{array}\right] + \left[\begin{array}{cc}av1\\av2\end{array}\right] $

&

$ (a + b)*\left[\begin{array}{cc}u1\\u2\end{array}\right] = a*\left[\begin{array}{cc}u1\\u2\end{array}\right] + b*\left[\begin{array}{cc}u1\\u2\end{array}\right] $

&

$ c*\left[\begin{array}{cc}du1\\du2\end{array}\right] = (cd)*\left[\begin{array}{cc}u1\\u2\end{array}\right] $

Because all of these are true you can conclude that vector addition and scalar multiplication indeed define R^2 as a vector space.

SUBSPACE

A subspace is a subset of a vector space. To be a subspace of vectors the following must be true:

1. One set must be a subset of another set

2. The set must be closed under scalar multiplication

3. The set must be closed under vector addition

Proving one set is a subset of another set

Given sets A and B we say that is is a subset of B if every element of A is also an element of B, that is,

x∈A implies x∈B

Basic Outline of the Proof that A is a subset of B:

· Suppose x ∈ A

1. Say what it means for x to be in A

2. Mathematical details

3. Conclude that x satisfies what it means to be in B

· Conclude x∈B

Example

Let A be the set of scalars divisible by 6 and let B be the even numbers. Prove that A is a subset of B.

· Suppose x ∈ A:

1. What it means for x to be in A: x = 6k for any scalar k

2. x = 2 × (3k)

  3k = C

3. What it means for x to be in B: x = 2C

· Conclude x∈B

Closed Under Scalar Multiplication

A set of vectors is closed under scalar multiplication if for every v∈V and every c∈\mathbb{R} we have cv∈V

Basic Outline of the Proof V is Closed Under Scalar Multiplication:

· Suppose v∈V and c∈\mathbb{R}

1. Say what it means for v to be in V

2. Mathematical details

3. Conclude that cv satisfies what it means to be in V

· Conclude cv∈V

Closed Under Vector Addition

A set of vectors is closed under vector addition if for every v and w ∈ V we have v + w ∈ V

Basic Outline of the Proof V is Closed Under Vector Addition:

· Suppose v and w ∈ V

1. Say what it means for v and w to be in V

2. Mathematical details

3. Conclude that v+ w satisfies what it means to be in V

· Conclude v + w ∈ V

Example

Let V be the set of points in R^2 such that x=y

· Suppose v and w ∈ V

1. What it means for v and w to be in V :

v = (v1, v2) and v1 = v2

w = (w1, w2) and w1 = w2

2. z = v + w = (v1+w1, v2+w2) = (v1+w1, v1+w1)

3. What it means for z to be in V: v1+w1 = v2+w2

· Conclude z = v + w ∈ V

Explanation of how to determine a subspace. Information referenced from Wabash College MA 223, Spring 2011