Contents
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
