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



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



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 =




nnvavavava 332211v

nnvbvbvbvb 332211v

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

ONLY IF0 0 0 0

v =




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



321 ,, www

2211111 vavaw

2221122 vavaw

2231133 vavaw

0332211 wcwcwc

0321 ccc 221111 vava 222112 vava 223113 vava



321 ,, www

0332211 wcwcwc


0332211 wcwcwc



321 ,, www


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

2v+

0332211 wcwcwc



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



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




vvvv ,,,, 321 n



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


vvvv ,,,, 321 n



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



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