4 Vector Spaces
4.1 Vector Spaces and Subspaces
4.2 Null Spaces, Column Spaces, and Linear Transformations
4.3 Linearly Independent Sets; Bases
4.4 Coordinate systems
Definition
Let H be a subspace of a vector space V.
An indexed set of vectors   b1 ,, bp  in V is a basis for H if
i)  is a linearly independent set, and
ii) the subspace spanned by  coincides with H; i.e.
H  Spanb1 ,, bp 
REVIEW
The Spanning Set Theorem
Let S  v1 ,, v p  be a set in V, and let H  Spanv1 ,, v p .
a. If one of the vectors in S, say vk , is a linear combination
of the remaining vectors in S, then the set formed from S by
removing vk still spans H.
b. If H  0 , some subset of S is a basis for H.
REVIEW
Theorem
The pivot columns of a matrix A form a basis for Col A.
REVIEW
4.4 Coordinate Systems
Why is it useful to specify a basis for a
vector space?
 One reason is that it imposes a “coordinate system” on the vector
space.
 In this section we’ll see that if the basis contains n vectors, then
the coordinate system will make the vector space act like Rn.
Theorem: Unique Representation Theorem
Suppose   b1 ,, bp  is a basis for V and x is in V. Then
there exists a unique set of scalars c1 ,, c p such that
. x  c1b1   c pb p
Definition:
Suppose   b1 ,, bp  is a basis for V and x is in V. The
coordinates of x relative to the basis  (the  - coordinates of x )
are the weights c1 ,, c p such that
. x  c b   c b
1 1
If c1 ,, c p
p
p
n

R
x
are the
- coordinates of , then the vector in
x
 c1 
 
 
c p 
 
is the coordinate vector of x relative to  , or the  - coordinate
vector of x .
Example:
2
0
2
1. Consider a basis   b1 ,b2  for R , where b1   , b2   
 1
1
 2
2
Find an x in R such that x    .
 3
 4
2. For x    , find x where  is the standard basis for R 2.
1 
 4
x 
1 
on standard basis
 2
x   
 3
on 
http://webspace.ship.edu/msrenault/ggb/linear_transformations_points.html

The Coordinate Mapping
x
 
x

Theorem

Let   b1,L ,bn  be a basis for a vector space V.
Then the coordinate mapping x  [x] is a one-to-one and
onto linear transformation from V onto R n .
Example:

1
2 
 4
For x    and   b1    , b2     , find
  1
1  
1 

x .

For   b1,L ,bp  , let P  b1,L ,bp

.
Then x  c1b1   c pbp is equivalent to x  P x .
P : the change-of-coordinates

matrix from β to the standard
basis
Example:
 2
1
7
 
 
 
Let v1  3, v2   1, x   3 ,   {v1 , v2 },& H  Span{v1 , v2 }.
6
 0 
12
Determine if x is in H, and if it is, find the coordinate vector of
x relative to  .
Application to Discrete Math
Let  = {C(t,0), C(t,1), C(t,2)} be a basis for P2, so we can
write each of the standard basis elements as follows:
 C(t,0) = 1(1) + 0t + 0t2
 C(t,1) = 0(1) + 1t + 0t2
 C(t,2) = 0(1) – ½ t + ½ t2
This means that following matrix converts polynomials in the
“combinatorics basis” into polynomials in the standard basis:
1 0
0 


M   0 1 1/2

0 0 1/2 

Application to Discrete Math
Recall that  = {C(t,0), C(t,1), C(t,2)} is
a basis for P2.
The following matrix converts
polynomials in the “combinatorics basis”
to polynomials in the standard basis:
Therefore, the following matrix converts
polynomials in the the standard basis to

polynomials in “combinatorics basis”:
1 0
0 


M   0 1 1/2

0 0 1/2 

1 0 0


1
M   0 1 1

0 0 2

Application to Discrete Math
The polynomial p(t) = 3 + 5t – 7t2 has the
3
following coordinate vector in the
5
p
(
t
)

 
standard basis S = {1, t, t2}:
 7 S
We now want to find the coordinate vector of p(t)
in the “combinatorics basis”
 = {C(t,0), C(t,1), C(t,2)}: 1 0 0  3 
 3 
0 1 1   5     2 

  

0 0 2  7  14 
That is, p(t) = 3 C(t,0) – 2 C(t,1) – 14 C(t,2)
Application to Discrete Math
Find a formula in terms of n for the following sum
n
2
3

5k

7k


k 0
Solution. Using the form we found on the previous slide
n
n
 3  5k  7k   3C(k,0)  2C(k,1) 14C(k,2)
2
k 0

k 0
 3C(n 1,1)  2C(n 1,2) 14C(n 1,3)
Aside: Why are these true?
n
C(k,0)  C(n 1,1)
k 0
n
C(k,1)  C(n 1,2)
k 0
n
C(k,2)  C(n 1,3)
k 0

http://webspace.ship.edu/deensley/DiscreteMath/flash/ch5/sec5_3/hockey_stick.html

Application to Discrete Math
Find the matrix M that converts
polynomials in the “combinatorics
basis” into polynomials in the standard
basis for P3:
Use this matrix to find a version of the
following expression in terms of the
standard basis:
3C(n 1,1) 2C(n 1,2) 14C(n 1,3)
Application to Discrete Math
Find a formula in terms of n for the following sum
n
2
3

5k

7k


k 0
Solution. Continued…..
n
n
5k  7k   3C(k,0)  2C(k,1) 14C(k,2)
 3 
2
k 0
k 0
 3C(n 1,1)  2C(n 1,2) 14C(n 1,3)
 ___(n 1)  ___(n 1) 2  ___(n 1) 3
Final Steps…
 We can finish by multiplying by a matrix that converts
vectors written in {1,(t+1),(t+1)2,(t+1)3} coordinates to a
vector written in terms of the standard basis.
 Find such a matrix and multiply it by the answer on the
previous slide to get a final answer of the form
___(1)  ___( n)  ___( n 2 )  ___( n 3 )
