a vector space containing infinitely many vectors can be efficiently described by listing a set of...

A vector space containing infinitely many vectors can be efficiently described by listing a set of vectors that SPANthe space.

eg: describe the solutions to:

reduces to

zz

yy

zyx

zyw

zyxw

zyxw

1

1

43

11

04310

01101

03211

02112

032

022

A vector space containing infinitely many vectors can be efficiently described by listing a set of vectors that SPANthe space.

eg: describe the solutions to:

reduces to

zz

yy

zyx

zyw

zyxw

zyxw

1

1

43

11

04310

01101

03211

02112

032

022

A vector space containing infinitely many vectors can be efficiently described by listing a set of vectors that SPANthe space.

eg: describe the solutions to:

reduces to

zz

yy

zyx

zyw

zyxw

zyxw

1

1

43

11

04310

01101

03211

02112

032

022

A vector space containing infinitely many vectors can be efficiently described by listing a set of vectors that SPANthe space.

eg: describe the solutions to:

reduces to

zz

yy

zyx

zyw

zyxw

zyxw

1

1

43

11

04310

01101

03211

02112

032

022

A vector space containing infinitely many vectors can be efficiently described by listing a set of vectors that SPANthe space.

eg: describe the solutions to:

zz

yy

zyx

zyw

zyxw

zyxw

1

1

43

11

04310

01101

03211

02112

032

022

1

0

4

1

0

1

3

1

zy

A vector space containing infinitely many vectors can be efficiently described by listing a set of vectors that SPANthe space.

eg: describe the solutions to:

zz

yy

zyx

zyw

zyxw

zyxw

1

1

43

11

04310

01101

03211

02112

032

022

1

0

4

1

0

1

3

1

zy

These two vectors SPAN the set of solutions.Each of the infinitely many solutions is a linear combinationof these two vectors!

A spanning set can be an efficient way to describe a vectorspace containing infinitely many vectors.

1

1,

1

0,

0

1 SPANS R2 - but is it the most efficient wayto describe R2 ?

Why do we need this? It is a linear combination of (depends on) the other two.

A spanning set can be an efficient way to describe a vectorspace containing infinitely many vectors.

1

1,

1

0,

0

1 SPANS R2 - but is it the most efficient wayto describe R2 ?

Why do we need this? It is a linear combination of (depends on) the other two.This independent set still spans R2 , and is a more efficient

way to describe the vector space!

definition: An INDEPENDENT set of vectors that SPANS a vectorspace V is called a BASIS for V.

3

1,

2

1 =

SPANS R2 :

xyc

xyc

xy

xy

xy

x

y

x

ycc

xcc

ccy

x

2

3

210

301

210

11

32

11

32

11

3

1

2

1

2

1

21

21

21

Given any x and y thereexist c1 and c2 such that

3

1,

2

1 =

is INDEPENDENT:

0

0

010

001

010

011

032

011

032

011

0

0

3

1

2

1

2

1

21

21

21

c

c

cc

cc

cc

A linear combinationof these vectors producesthe zero vector ONLY IF c1 and c2 are both zero.

3

1,

2

1 =

is INDEPENDENT and SPANS R2 …. Therefore

is a BASIS for R2.

1

0,

0

1 is called the standard basis for R2

is a nonstandard basis - why do we need nonstandard bases?

Consider the points on the ellipse below:

Described relative to thestandard basis they aresolutions to:8x2 + 4xy + 5y2 = 1

Described relative to the basis they aresolutions to:9x2 + 4y2 = 1

basis =

52

51

,

51

52

1

1,

1

0,

0

1

1

10

1

04

0

15

4

5

1

12

1

02

0

13

4

5

1

12

1

06

0

17

4

5

There are lots of different ways to write v as a linear combination of the vectorsin the set

=

v =

not a BASIS for R2

example:

theorem: If = nvvvv ,,, 321 is a BASIS for a vector space V,

then for every vector v in V there are unique scalars

ncccc ,,, 321 nnvcvcvcvc 332211Such that:

the c’s exist because spans V

they are unique because is independent

v =

theorem: If = nvvvv ,,, 321 is a BASIS for a vector space V,

then for every vector v in V there are unique scalars

ncccc ,,, 321 nnvcvcvcvc 332211Such that:

nnvavavava 332211v

nnvbvbvbvb 332211v

nnn vbavbavbavba )()()()( 333222111 vv0 =

ONLY IF0 0 0 0

v =

theorem: If = nvvvv ,,, 321 is a BASIS for a vector space V,

then for every vector v in V there are unique scalars

ncccc ,,, 321 nnvcvcvcvc 332211Such that:

the coordinates of v

v =

relative to the basis

This is the vector v

Relative to the standard basis the coordinates of v are

1

0,

0

1

5

1

1

5

=

1

1,

1

2

Relative to the basis the coordinates of v are

3

2

1

1,

1

2

2

3

1

05

0

11

1

13

1

22 v = =

coordinates relative standard basis

coordinates relative basis

example: Suppose V is a vector space that is SPANNED by the two vectors 21 ,vv

Is it possible that this set of three vectorsis INDEPENDENT ?

321 ,, www

2211111 vavaw

2221122 vavaw

2231133 vavaw

example: Suppose V is a vector space that is SPANNED by the two vectors 21 ,vv

Is it possible that this set of three vectorsis INDEPENDENT ?

321 ,, www

2211111 vavaw

2221122 vavaw

2231133 vavaw

0332211 wcwcwc

0321 ccc 221111 vava 222112 vava 223113 vava

example: Suppose V is a vector space that is SPANNED by the two vectors 21 ,vv

Is it possible that this set of three vectorsis INDEPENDENT ?

321 ,, www

0332211 wcwcwc

0321 ccc 221111 vava 222112 vava 223113 vava

0332211 wcwcwc

example: Suppose V is a vector space that is SPANNED by the two vectors 21 ,vv

Is it possible that this set of three vectorsis INDEPENDENT ?

321 ,, www

0321 ccc 221111 vava 222112 vava 223113 vava

0321 ccc 11a 13a12a1v 321 ccc 21a 23a22a

2v+

0332211 wcwcwc

example: Suppose V is a vector space that is SPANNED by the two vectors 21 ,vv

Is it possible that this set of three vectorsis INDEPENDENT ?

321 ,, www

0321 ccc 221111 vava 222112 vava 223113 vava

0321 ccc 11a 13a12a1v 321 ccc 21a 23a22a

2v+

0332211 wcwcwc

example: Suppose V is a vector space that is SPANNED by the two vectors 21 ,vv

Is it possible that this set of three vectorsis INDEPENDENT ?

321 ,, www

0321 ccc 11a 13a12a1v 321 ccc 21a 23a22a

2v+

0332211 wcwcwc

IF=0 =0

321 ccc 11a 13a12a =0

321 ccc 21a 23a22a =0

IF

example: Suppose V is a vector space that is SPANNED by the two vectors 21 ,vv

Is it possible that this set of three vectorsis INDEPENDENT ?

321 ,, www

0332211 wcwcwc

321 ccc 11a 13a12a =0

321 ccc 21a 23a22a =0

IF

11a 13a12a

21a 23a22a0

3

2

1

c

c

c

The rank is less than the number of variables The solution is not unique

NO

3 VECTORS CANNEVER BE INDEPENDENTin a VECTOR SPACE that isSPANNED BY 2 VECTORS

The number of independent vectors in a vector space V can neverexceed the number of vectors that span V.

If

is INDEPENDENT in V and

SPANS V

then k m

kuuuu ,,, 321 k

mvvvv ..,,..,, 321 m

theorem:

Two different bases for the same vector space will contain the same number of vectors.

is a basis for V

theorem:If

and

then k = m

kuuuu ,,, 321 k

mvvvv ,,,, 321 m is a basis for V

proof:

kuuuu ,,, 321 k

vvvv ,,,, 321 mSPANS

IS INDEPENDENT

k < m

kuuuu ,,, 321 k

vvvv ,,,, 321 m

SPANS

IS INDEPENDENT

k > m

Two different bases for the same vector space will contain the same number of vectors.

is a basis for V

theorem:If

and

then k = m

kuuuu ,,, 321 k

mvvvv ,,,, 321 m is a basis for V

definition:

The number of vectors in a basis for V is called

the DIMENSION of V.

An independent set of vectors that does not span V can be “padded” to make a basis for V.

theorem:

kvvvv ,,, 321Suppose Is independent but does not span V.

Then there is at least one vector in V , call it w , such that

w Cannot be written as a linear combination of the vectors

in kvvvv ,,, 321. That is:

wvvvv k ,,,, 321 Is an independent set.

A spanning set that is not independent can be “weeded” tomake a basis.

theorem:

mvvvv ,,, 321Suppose spans V but is not independent.

mvuvvv ,,,,, 321

Then there is at least one vector in the set, call it u , such that u is a linear combination of the other vectors in the set.

Remove u and the remaining vectors in the set will still span V.

theorem:

theorem:If the dimension of V is n then the set vvvv ,,,, 321 n

Containing n vectors

is INDEPENDENT IF AND ONLY IF it SPANS V

theorem:If the dimension of V is n then the set

vvvv ,,,, 321 n

Containing n vectors

is INDEPENDENT IF AND ONLY IF it SPANS V

theorem:If the dimension of V is n then the set

vvvv ,,,, 321 n

Containing n vectors

is INDEPENDENT IF AND ONLY IF it SPANS V

nvvvv ,,,, 321 nvvvv ,,,, 321 If S = If S =spans V

then S is independent. is independent then S spans V.

If S = spans V nvvvv ,,,, 321

and is not independent then one vector can be removed leaving a spanningset containing n-1 vectors. Since dim V = n, there is in V a set of n independent vectors (basis ). This is impossible. You cannot have more independent vectors than spanning vectors

theorem:If the dimension of V is n then the set

vvvv ,,,, 321 n

Containing n vectors

is INDEPENDENT IF AND ONLY IF it SPANS V

nvvvv ,,,, 321 If S = spans V

then S is independent.

nvvvv ,,,, 321 If S =

is independent and does not span V, then a vector can be added to S making a set containing n+1independent vectors - impossible in a space spanned by n vectors-a basis for V contains n vectors

nvvvv ,,,, 321 If S =

is independent then S spans V.

theorem:If the dimension of V is n then the set vvvv ,,,, 321 n

Containing n vectors

is INDEPENDENT IF AND ONLY IF it SPANS V