4.1 Vector Spaces and Subspaces4.2 Null Spaces, Column Spaces, and Linear Transformations4.3 Linearly Independent Sets; Bases4.4 Coordinate systems

4 Vector Spaces

Definition

Let H be a subspace of a vector space V. An indexed set of vectors in V is a basis for H if

i) is a linearly independent set, andii) the subspace spanned by coincides with H; i.e.

pbb ,,1

pbbH ,,Span 1

REVIEW

The Spanning Set Theorem

Let be a set in V, and let .

a. If one of the vectors in S, say , is a linear combination of the remaining vectors in S, then the set formed from S by removing still spans H.

b. If , some subset of S is a basis for H.

pvvS ,,1 pvvH ,,Span 1

kv

kv

0H

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 is a basis for V and is in V. Then there exists a unique set of scalars such that

.

pbb ,,1 pcc ,,1

ppcc bbx 11

x

Definition:

Suppose is a basis for V and is in V. Thecoordinates of relative to the basis (the - coordinates of )are the weights such that .

pbb ,,1

pcc ,,1

If are the - coordinates of , then the vector in

is the coordinate vector of relative to , or the - coordinatevector of .

pcc ,,1

pc

c

x 1

xx x

x nR

xx

ppcc bbx 11

Example:1. Consider a basis for , where

Find an x in such that .

2. For , find where is the standard basis for .

21,bb 2R

1

0,

1

221 bb

2R

3

2x

1

4x x 2R

on standard basis on

1

4x

3

2x

http://webspace.ship.edu/msrenault/ggb/linear_transformations_points.html

The Coordinate Mapping

Theorem

Let be a basis for a vector space V. Then the coordinate mapping is a one-to-one and onto linear transformation from V onto .

€

x[ ]βx

€

[ ]

€

b1,L ,bn{ }

][xx nR

1

4x

€

x[ ]β

1

2,

1

121 bb

Example:

For and , find .

For , let .

Then is equivalent to .

€

b1,L ,bp{ }

€

Pβ = b1,L ,bp[ ]

ppbcbcx 11

€

x = Pβ x[ ]β

: the change-of-coordinates matrix from β to the standard basis

P

Example:

Let

Determine if x is in H, and if it is, find the coordinate vector of x relative to .

},&,{,

12

3

7

,

0

1

1

,

6

3

2

2121 vvxvv

}.,Span{ 21 vvH

Application to Discrete MathLet = {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:

€

Mβ =

1 0 0

0 1 −1/2

0 0 1/2

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

Application to Discrete MathRecall 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”:

€

Mβ−1 =

1 0 0

0 1 1

0 0 2

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

€

Mβ =

1 0 0

0 1 −1/2

0 0 1/2

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

The polynomial p(t) = 3 + 5t – 7t2 has the following coordinate vector in the standard basis S = {1, t, t2}:

S

t

7

5

3

)(p

We now want to find the coordinate vector of p(t) in the “combinatorics basis” = {C(t,0), C(t,1), C(t,2)}:

14

2

3

7

5

3

200

110

001

That is, p(t) = 3 C(t,0) – 2 C(t,1) – 14 C(t,2)

Application to Discrete Math

Application to Discrete MathFind a formula in terms of n for the following

sum

Solution. Using the form we found on the previous slide

€

3+ 5k − 7k 2( )

k =0

n

∑

€

3+ 5k − 7k 2

k =0

n

∑ = 3C(k,0) − 2C(k,1) −14C(k,2)k =0

n

∑

= 3C(n +1,1) − 2C(n +1,2) −14C(n +1,3)

Aside: Why are these true?

http://webspace.ship.edu/deensley/DiscreteMath/flash/ch5/sec5_3/hockey_stick.html

€

C(k,2)k =0

n

∑ = C(n +1,3)€

C(k,0)k =0

n

∑ = C(n +1,1)

€

C(k,1)k =0

n

∑ = C(n +1,2)

Application to Discrete MathFind 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 MathFind a formula in terms of n for the following

sum

Solution. Continued…..

€

3+ 5k − 7k 2( )

k =0

n

∑

€

3+ 5k − 7k 2

k =0

n

∑ = 3C(k,0) − 2C(k,1) −14C(k,2)k =0

n

∑

= 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) + ___(n2) + ___(n3)