(New page: ===SUBSPACE=== 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. T...)
 
 
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=== VECTOR SPACE ===
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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:
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1. '''Communative Property''': u + v = v + u
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2. '''Associative Property''':
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'''·''' Of addition: (u + v) + w = u = (v + w)
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'''·''' Of multiplication: (ab)v = a(bv)
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3. '''Zero Property''': There exist some '''0'''∈V such that '''0''' + v = v
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4. '''Inverse Property''': For every v∈V there is some -v where v + -v = 0
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5. '''Identity Property''': 1v=v
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6. '''Distributive Property''': a(u + v) = au + av & (a + b)u = au + bu & c(du) = (cd)u
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'''Example'''
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Prove that vector addition and scalar multiplication define R^2 as a vector space.
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<math>\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]</math>
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<math>r*\left[\begin{array}{cc}x1\\x2\end{array}\right] = \left[\begin{array}{cc}rx1\\rx2\end{array}\right]</math>
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First, define v1, v2, w1, w2, u1, u2 as elements in R^2 and a,b,c,d as scalars.
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Second, check each of the properties in the defintion of a vector space. If a single property fails the entire proof fails.
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1. '''Communative Property''':
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<math>\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]</math>
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2. '''Associative Property''':
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'''&middot;''' Of addition:
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<math>\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]</math>
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'''&middot;''' Of multiplication:
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<math>(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]</math>
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3. '''Zero Property''':
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<math>\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]</math>
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4. '''Inverse Property''':
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<math>\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]</math>
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5. '''Identity Property''':
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<math>1*\left[\begin{array}{cc}v1\\v2\end{array}\right] = \left[\begin{array}{cc}v1\\v2\end{array}\right]</math>
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6. '''Distributive Property''':
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<math>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]</math>
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&
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<math>(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]</math>
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&
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<math>c*\left[\begin{array}{cc}du1\\du2\end{array}\right] = (cd)*\left[\begin{array}{cc}u1\\u2\end{array}\right]</math>
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Because all of these are true you can conclude that vector addition and scalar multiplication indeed define R^2 as a vector space.
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===SUBSPACE===
 
===SUBSPACE===
  
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To be a subspace of vectors the following must be true:
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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
 
1. One set must be a '''subset''' of another set
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'''Example'''
 
'''Example'''
  
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Let V be the set of points above the graph of the absolute value function
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Let V be the set of points in R^2 such that x=y
  
  
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1. What it means for '''v''' and '''w''' to be in V :  
 
1. What it means for '''v''' and '''w''' to be in V :  
  
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'''v''' = (v1, v2) and v2 > |v1|
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'''v''' = (v1, v2) and v1 = v2
  
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'''w''' = (w1, w2) and w2 > |w1|
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'''w''' = (w1, w2) and w1 = w2
  
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2. '''z''' = (z1,z2) = '''v''' + '''w''' = (v1+w1, v2+w2)
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2. '''z''' = '''v''' + '''w''' = (v1+w1, v2+w2) = (v1+w1, v1+w1)
  
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So, |z1| = |v1 + w1| &le; |v1| + |w1| < v2 + w2 = z2
 
  
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3. What it means for '''z''' to be in V: |z1| < z2
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3. What it means for '''z''' to be in V: v1+w1 = v2+w2
  
  

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

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