Vector Spaces

Definition of a Vector Space

We have seen that vectors in Rn enjoy a collection of properties such as commutative, associative, and distributive properties.  Other mathematical objects such as matrices and polynomials share the same properties.  Instead of proving theorems separately for each of these objects, it is convenient to give a single proof for anything that has these properties.  Below is a definition that collects some of the most common properties. 

Definition

A vector space V is a set with two operations + and * that satisfy the following properties.

 a.  If u and v are elements of V, then u + v is and element of V (closure under +)

  1. u + v  =  v + u
  2. u + (v + w)  =  (u + v) + w
  3. There is an element 0 in V such that
              u + 0  =  0 + u  =  u 
  4. For every u in V there is an element -u with
              u + (-u)  =  0 

b.  If u is in V and c is a real number then c*u is in V (closure under *)

  1. c * (u + v)  =  c * u + c * v 
  2. (c + d) * u  =  c * u + d * u 
  3. c * (d * u)  =  (cd) * u 
  4. 1 * u  =  u

 

You will recognize these properties as properties of vectors in Rn, however there is a large class of vector spaces that do not look at all like Rn.  


Examples of Vector Spaces

Example 1  P2 

Consider the set P2 of polynomials of degree less than or equal to 2.  Define + to be polynomial addition

        (a1t2 + b1t + c2)  +  (a2t2 + b2t + c2)  =  (a1 + a2)t2 + (b1 + b2)t + (c1 + c2)

and * is defined by

        k * (at2 + bt + c)  =  (ak)t2 + (kb)t + (kc)

This is a vector space.  Most of the properties clearly hold.  We will demonstrate a few of the properties.  For example the 0 vector is the zero polynomial (0).  We have

        (at2 + bt + c) + 0  =  0 + (at2 + bt + c)  =  at2 + bt + c

Property b2 holds since 

        (r + s) * (at2 + bt + c)  =  (r + s)at2 + (r + s)bt + (r + s) c

        =  (ra + sa)t2 + (rb + sb)t + (rc + sc)  =  (ra)t2 + (sa)t2 + (rb)t + (sb)t + rc + sc

        =  (ra)t2  + (rb)t + rc + (sa)t2+ (sb)t + sc  =  r(at2 + bt + c) + s(at2 + bt + c)


Example 2  Pn 

We can generalize Example 1 and let Pn be the set of all polynomials of degree less than n.  We define + to mean polynomial addition and * to be scalar multiplication as in Example 1.  This is a vector space as you can check.  


Example 3  M2x3 

Consider the set M2x3 of 2 x 3 matrices and let + be defined by matrix addition and * be defined by matrix scalar multiplication.  Then M2x3 is a vector space.  We have stated all of the required properties previously.


Example 4  Mmxn 

We can generalize Example 3 by letting Mmxn be the set of all m x n matrices with matrix addition and scalar multiplication as before.  


Example 5

Consider the set V of all differentiable functions f such that f '(1)  =  0.  Let + be defined as addition of functions and * be defined as regular scalar multiplication.  This is a vector space.  We will demonstrate a few of the properties.  Let f and g be elements of this set.  Then 

        f '(1)  =  g '(1)  =  0

To show additive closure, we have

        (f + g)'(1)  =  f '(1) + g'(1)  =  0 + 0  =  0

so that f + g is in V.  To show multiplicative closure we have

        (cf)'(1)  =  c(f '(1))  =  c(0)  =  0

The rest of the properties follow from the properties of function arithmetic and derivatives.  


Example 6

Let S be the set of ordered pairs in R2 with + defined by

        (x1, y1) + (x2, y2)  =  (x1 + 2x2, y1 + 2y2)

and * defined by

        c(x,y)  =  (cx,cy)

then S is not a vector space, since property a1 fails.  For example

        (2,3) + (4,5)  =  (10,13)

but 

        (4,5) + (2,3)  =  (8,11)


A Few Properties of Vector Spaces

Vector spaces enjoy several additional properties that we will later explore.  Below are some of the most basic ones.

Theorem

Let V be a vector space then

  1. 0u  =  0  for all u in V
  2. c0  =  0  for all scalars c
  3. If cu  =  0  then either c  =  0 or u  =  0
  4. (-1)u  =  -u  for all u in V

 

We will prove 4.  We have

        (-1)u + u  =  (-1)u + (1)u  

        =  (-1 + 1)u  =  0u  =  0

Since

        (-1)u + u  =  0 

We can conclude that (-1)u is the additive inverse of u.  

 



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