CH6.1-6.4 Real Vector Space

7/23/2019 CH6.1-6.4 Real Vector Space

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Chapter 6

Real Vector Spaces


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Objective:

l Study many important properties of

vectors in

nR

l A carefully constructed generalization of

nR to many other important vector space.

l Application of vector space to the linearsystem and matrix.


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6.1. Vector Spaces

Definition of Vector Space

Definition:

A real vector space V is asetof elements

together with two operations, addition and

scalar multiplication , satisfying the

following properties:

Letu, v, andwbe vectors in V, and letcanddbe

scalars.


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Addition:

(a) Ifuandvare any elements ofV, then vu+

is inV(closed).

1.

uvvu +=+ .

2. wvuwvu ++=++ )()( .

3. V has a zero vector0such that for everyuin V, uuu =+=+ 00 .

4. For everyuinV, there is an element u- in

V, 0)( =-+ uu .


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Scalar multiplication:

(b) If uis any element ofVand cis any real

number, then cu is inV(closed).

5. ucdduc )()( = .

6. ducuudc +=+ )( .

7. cvcuvuc +=+

)(

.

8. uu=1 .


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Example :

nRV 1 together with standard vector addition and scalar

multiplication. Then, 1V

is a vector space since for any two vectors u and v in

n

and scalar c, both vu + and cu are innR . Therefore,

conditions (a) and (b) are satisfied. In addition, the

conditions (1) to (8) are satisfied (see the previoussubsection).


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Example :

2V the set consisting of all nm matrices together with

standard matrix addition and scalar multiplication. 2V is a

vector space since for any two nm

matrices u and vand scalar c, both vu+ and cu are nm matrices.

That is,both vu+ and cu are still in 2V .Therefore

conditions (a

) and (b

) are satisfied. In addition, theconditions (1) to (8) are satisfied (see section 2).


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Example:

3V the set consisting ofallpolynomials of degree 2 or less with

the form together with standard polynomial addition and scalar

multiplication. Is 3V a vector space?

We need to examine whether the conditions (a ), (b ), and the

conditions (1) to (8) are satisfied. Let

01

2

2 axaxau ++= ,

012

2 bxbxbv ++= ,

01

2

2 cxcxcw ++=

and let c and d be scalars. Then,


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Solution:

Addition:

(a):

30011

2

22

012

2012

2

)()()(

)()(

Vbaxbaxba

bxbxbaxaxavu

+++++=

+++++=+

since vu+ is a polynomial of degree 2 or less.


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Solution:

(1):

uvaxaxabxbxb

abxabxab

baxbaxba

bxbxbaxaxavu

+=+++++=

+++++=

+++++=

+++++=+

)()(

)()()(

)()()(

)()(

01

2

201

2

2

0011

2

22

00112

22

01

2

201

2

2


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Solution:

(2):

)(

)]()()[(

)()()(

)]()()[()()(

01

2

20011

2

22

000111

2

222

0011

2

2201

2

2

wvu

cxcxcbaxbaxba

cbaxcbaxcba

cbxcbxcbaxaxawvu

++=

++++++++=

++++++++=

++++++++=++


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Solution:

(3):

Let 00002 ++= xx . Then,

uaxaxa

uaxaxaaxaxau

=++=

+=+++++=+++++=+

01

2

2

01

2

201

2

2 0)0()0()0()0()0()0(0

(4):

Let ).()()( 012

2 axaxau -+-+-=- Then,

0000)]([)]([)]([ 200112

22 =++=-++-++-+=-+ xxaaxaaxaauu


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Solution:

Scalar multiplication:

(b):

301

2

2 )()()( Vcaxcaxcacu ++=

since cu is a polynomial of degree 2 or less.

(5):

cvcucbcaxcbcaxcbca

bacxbacxbacvuc

+=+++++=

+++++=+

)()()(

)]([)]([)]([)(

0011

2

22

0011

2

22


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Solution:

(6):

ducudaxdaxdacaxcaxca

adcxadcxadcudc

+=+++++=

+++++=+

)]()()[()]()()[(

])[(])[(])[()(

01

2

201

2

2

01

2

2

(7):

ucdacdxacdxacd

dacxdacxdacdaxdaxdacduc

)()()()(

)()()()]()()[()(

01

2

2

01

2

201

2

2

=++=

++=++=

(8):

uaxaxaaxaxau =++=++= 012

201

2

2 1111


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Note:

nP the set consisting ofallpolynomials of degree n

or less with the form together with standard

polynomial addition and scalar multiplication. Then,

nP is avector space. In addition,

the setconsisting ofallpolynomials with the form together

with standard polynomial addition and scalar

multiplication. Then, P is also avector space.


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Example:

4V the set consisting ofallreal-valuedcontinuous

unctionsdefined on the entire real line together with

standard addition and scalar multiplication. Is 4V a vector

space?

We need to examine whether the conditions (a), (b), andthe conditions(1) to (8) are satisfied. Let

)(xfu= , )(xgv= , )(xhw=

and letcanddbe scalars. Then,


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Solution:

Addition:

(a):

4)()( Vxgxfvu +=+

since )()( xgxf + is still a continuous function.

(1):

uvxfxgxgxfvu +=+=+=+ )()()()(

(2):

[ ] [ ]wvu

xhxgxfxhxgxfwvu

++=

++=++=++

)(

)()()()()()()(


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Solution:

(3):

Let the zero vector 00 = . Then,

uxfxfu ==+=+ )(0)(0

(4):

Let )(xfu -=-

. Then,

[ ] 0)()()()()( =-=-+=-+ xfxfxfxfuu


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Solution:

(b):

4)( Vxcfcu =

since)(xcf

is still a continuous function.(5):

[ ] cvcuxcgxcfxgxfcvuc +=+=+=+ )()()()()(

(6):

ducuxdfxcfxfdcudc +=+=+=+ )()()()()(


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Solution:

(7):

[ ]

ucdxfcdxcdfxdfcduc )()()()()()( ====

(8):

[ ]

uxfxfu === )()(1)(1


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Note:

Let *

4V the set of all differentiable functions

defined on the entire real line and

**4V the set o

all integrable functions defined on the entire real

line. Both*

4V and**

4V are vector space under

standard addition and scalar multiplication.


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Example:

5V the set consisting of all integers with standard

addition and scalar multiplication. Is 5V a vector

space?

5V is not a real vector space since for 51 Vu = , and

7.0=c ,

57.017.0 Vcu == ,

condition(b)is not satisfied.


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Example:

6V the set consisting of all vectors in2R with

standard addition and nonstandard scalar

multiplication defined by

=

0

1

2

1 cx

x

xc

. Is 6V a vector space?


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Solution:

6V is not a real vector space since for2

2

1R

c

cu

= ,

0,, 221 cRcc ,

uc

cc

c

cu =

=

=

2

11

2

1

011

,

condition(8)is not satisfied.


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Example:7V the set consisting of only second degree

olynomials with standard addition and scalar

multiplication. Is 7V a vector space?

7V isnota real vector space since for2xu = and

2xv -=

7

22 0)( Vxxvu =-+=+ ,

condition(a

)is not satisfied.


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Example:

8V the set consisting of all real-valued continuous

unctions such that 3)1( =f . Suppose the operations are

standard addition and scalar multiplication. Is 8V a vectorspace?


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Solution:

8V isnota real vector space since for

8)( Vxfu = , 8)( Vxgv =

8)()(3633)1()1(

Vxgxfvugf

+=+=+=+

,

condition(a)is not satisfied.

Example 1 9 Page 273 276


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Example1 p273

n Consider the set Rn together with the

operations of vector addition and

scalar multiplication as defined inSection 4.2. Theorem 4.2 in Section

4.2 established the fact that Rn is a

vector space under the operations ofaddition and scalar multiplications of n-

vector.


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Example2 p273

n Consider the set V of all ordered triples

of real number of the form (x,yx,0)and

define the operations and by

(x,y,0)(x,y,0)=(x+x,y+y,0)

c(x,y,0)=(cx,cy,0)


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Example3 p273

n Consider the set V of all ordered triples ofreal numbers (x,y,z) and define theoperations and by

1.(x,y,z)(x,y,z) = (x+x,y+y,z+z).

2.c(x,y,z)=(cx,y,z).

3.c[(x,y,z)(x,y,z)]=(c(x+x), y+y,z+z).

4.c (x,y,z)d(x,y,z)= ((c+d)x,2y,2z).


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Example4 p274

n Consider the set M23 of all 23matrices.

n Similarly the ser of all mn matricesunder the usual operations of matrix

addition and scalar multiplication is a

vector space will be denoted by Mmn.


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Example5 p274

n Let F[a,b] be the set of all real-valuedfunctions that are defined on the interval[a,b]. If f and g are in V, we define fg and

cf by (fg)(t)f(t)+g(t).(cf)(t)cf(t).

n A polynomial (in t) is a function that is

expressible asp(t)antn+ an-1t

n-1++a1t+a0a0,a0,a1,are real numbers.


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Example6 p274

n The follow functions are polynomials

n p1(t)3t42t25t1.

n p2(t)2t1.n p3(t)4.

n f4(t)2 6 and f5(t)1/t22t1 are

not polynomials.t


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Example7 p275

n p1(t) : degree 4

n p2(t) : degree 1

n p3(t) : degree 0


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Example8 p275

n If p(t)antn+ an-1t

n-1++a1t+a0 .

n q(t)bntn+ bn-1t

n-1++b1t+b0 .

n p(t) q(t)

(an+bn)tn+(an-1+bn-1)tn-1++(a1+b1)t+a0+b0

n cp(t) cantn+ can-1t

n-1++ca1t+ca0 .

n (c+d)p(t)cp(t)dp(t).


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Example9 p276

n Let uvu-v , cucu ,

n If u3,v2 ,

uv-1. vu1

n If u4,c2,d3

(c+d)(2+3)420

cudv812-4

Theorem 6 1:


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Theorem 6.1:

(Important Result)

IfVis a vector space and letube any element of a

real vector spaceV. Then,

(a) .00 =u

(b) .0,,00 VRcc =

(c) .0or0.0 === uccu

(d) .)1( uu -=-


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6.2. Subspace

Definition of subspace:

Wis called a subspace of a real vector spaceV if

1. Wis a subset of the vector spaceV.

2. W is a vector space with respect to the

operations in V.Example 1(Page 279)

Example 2(Page 279)


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Example1 p279

n Every vector space has at least two

subspaces, itself and subspace{0}.

n The subspace {0} is called the zerosubspace.


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Example2 p279

n Let u(a1,b1,0), v (a2,b2,0)

uv(a1+a2,b1+b2,0)

cu(ca1,cb1,0)

Theorem 6.2: (Important


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Theorem 6.2: (Important

Result)

Wis a subspace of a real vector spaceV

1.Ifuandvare any vectors inW, then

Wvu +.

2.If c is any real number and u is any vector in W,

then Wcu .


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Example:

1W the subset of3R consisting of all vectors of the

form,

Raa

,

0

0

,

together with standard addition and scalar multiplication.

Is 1W a subspace of3

R ?


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Solution:

We need to check if the conditions (1) and (2) are satisfied.Let

Rc

a

v

a

u

=

= ,00,

00

21

.

Then,

(1):

1

2121

0

0

0

0

0

0 W

aaaa

vu

+

=

+

=+.

.


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Solution:

(2):

1

1

0

0 W

ca

cu

=

.

1W is a subspace of 3R .


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Example:

Let the real vector space V be the set consisting of allnn matrices together with the standard addition and

scalar multiplication. Let

2W the subset of V consisting of all nn diagonalmatrices.

Is 2W a subspace of V?

Let

2

22

11

00

00

00

W

a

a

a

u

nn

=

L

MOMM

L

L

,2

22

11

00

00

00

W

b

b

b

v

nn

=

L

MOMM

L

L

, and

Rc .


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Solution:

(1):

2

2222

1111

00

00

00

W

ba

ba

ba

vu

nnnn

+

+

+

=+

LMOMM

L

L

since vu + is still a diagonal matrix.


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Solution:

(2):

2

22

11

00

00

00

W

ca

ca

ca

cu

nn

=

LMOMM

L

L

since cu is still a diagonal matrix.

2W is a subspace of V.


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Example:

nP the set consisting ofallpolynomials of degree n or

less with the form together with standard polynomial

addition and scalar multiplication. nP

is a vectorspace.

P the set consisting of allpolynomials withthe form together with standard polynomial addition and

scalar multiplication. Then, is also a vector space.

Then, nP is a subspace of P.


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Example:

4V the set consisting of all real-valued continuous

unctions defined on the entire real line together with

standard addition and scalar multiplication. Let

*4Vthe

set of all differentiable functions defined on the entire real

line together with standard

addition and scalar multiplication. Then, 4V

is a realvector space. Also.

*

4V is a subspace of 4V .


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Example:

3W the subset of 3R consisting of all vectors of the form,

Rba

b

a

a

,,2

,

together with standard addition and scalar multiplication. Is

3W a subspace of3R ?


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Solution:

(1):

( ) 3

21

2

21

21

21

2

2

2

1

21

2

2

2

2

1

2

1

1

W

bb

aa

aa

bb

aa

aa

b

a

a

b

a

a

vu

+

+

+

+

+

+

=

+

=+

.

Therefore, 3Wvu + .

3W isnota subspace of3R .


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Example:

3V the set consisting ofallpolynomials of degree 2 or less

with the form together with standard polynomial addition

and scalar multiplication. 3V is a vector space. Let

4W the subset of 3V consisting of all polynomials of the

form

2,2 =++++ cbacbxax .

Is 4W a subspace of 3V ?


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Solution:

Let

401

2

2 Waxaxau ++=

and

4012

2 Wbxbxbv ++= .

Then, 2012 =++ aaa and 2012 =++ bbb . Thus,

40011

2

22

01

2

201

2

2

)()()(

)()(

Wbaxbaxba

bxbxbaxaxavu

+++++=

+++++=+


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Solution:

since

( ) ( ) ( ) ( ) 422)( 012012001122 =+=+++++=+++++ bbbaaabababa .

4W isnota subspace of 3V .

Example 3--10(Page 280--282)


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Example3 p280

=

11

11

0

0

dc

bau

=

22

22

0

0

dc

bav

++

++=+

2121

2121

0

0

ddcc

bbaavu

=

11

11

0

0

kdkc

kbkaku


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Example5 p282

n Let u(a1,b1,1), v (a2,b2,1)

uv(a1+a2,b1+b2,2)


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n We let Pn denote the vector space

consisting of all polynomials of defree

n and the zero polynomial. And let Pdenote the vector space of akkpolynomials. It is easy to verify that P2is a subspace of P3 and in general

that Pn is a subspaces of Pn+1. Pn is a

subspace of P.

Example6 p282


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Example7 p282

n Let V be the set all polynomials of

degree exactly2.V is a subset P2 but

it is not a subspace of P2 , since thesum of polynomials 2t2+3t+1 and

-2t2+t+2 , a polynomial of degree 1, is

not in V.

E l 8 282


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Example8 p282

n Let C[a,b] denote the set of all real-

values continuous functions that are

defined on the interval [a,b]. If f and gare in C[a,b] then f+g is in C[a,b].

E l 9 282


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Example9 p282

n Consider the homogeneous system

Ax=0, A is an mn matrix.A solution consists of avector x in Rn. Let W be the subset of Rn consistingof all solutions to the homogeneous system. Since

A0=0, we conclude that W is not empty. To checkthat W us a subspace of Rn, we verify properties ()and () of Them 6.2. Thus let x and y be solutions.

Ax=0 and Ay=0

A(x+y)=Ax+Ay=0A(cx)=c(Ax)=0

E l 10 282


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Example10 p282

n Let v1 and v2 be fixed vectors in a

vector space V and let W be the set of

all linear combinations of v1 and v2.w1=a1v1+a2v2

w2=b1v1+b2v2

w1+w2= (a1+b1)v1+ (a2+b2)v2

cw1=(ca1)v1+ (ca2)v2

D fi iti f li bi ti


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Definition of linear combination:

Let kvvv ,,, 21 K be vectors in a real vector space V. A

vector v inV is called a linear combination o

kvvv ,,, 21 K

if

kkvcvcvcv +++= L2211 ,

where kccc ,,, 21 K are real numbers.

E l


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Example:

Let

=

-=

-=

=

12

80,

31

02,

21

31,

01

20321 vvvv

be vectors in the vector space consisting of all 22

matrices. Then,

321 231

02)1(

21

312

01

201

12

80vvvv -+=

--+

-+

=

= .

That is, v is a linear combination of 321 ,, vvv .

E ample


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Example:

For linear system

b

x

x

x

Ax =

=

-

=

1

1

1

123

012

101

3

2

1

.

-=1

3

2

x is a solution for the above linear system

Thus,

Example:


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Example:

bAcolAcolAcol

A

=

=+-=

-+

-

=

-

-=

-

1

11

)()(3)(2

1

0

1

1

2

1

0

3

3

2

1

2

1

3

2

123

012

101

1

3

2

121

That is, b is a linear combination of the column vectorsofA,

)(),(),( 321 AcolAcolAcol

Note:


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Note:

For a linear system 11 = mnnm bxA , the linear

system has solution or solutions

b is a linear

combination of the column vectors ofA,)(,),(),( 21 AcolAcolAcol nL .

For example, if

=

nc

c

c

c M

2

1

is a solution of

11 = mnnm bxA ,

Note:


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Note:

then

bAcolcAcolcAcolc nn =+++ )()()( 2211 L .

On the other hand, the linear system has no solution

b isnota linear combination of the columnvectors ofA

Example:


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Example:

Is the vector

=

5

5

4

va linear combination of the vectors

=

-

=

=

2

3

3

,

4

1

1

,

3

2

1

321 vvv

.

Solution:


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Solution:

We need to find the constants 321 ,, ccc such that

332211321

2

3

3

4

1

1

3

2

1

5

5

4

vcvcvccccv ++=

+

-

+

=

=

.

we need to solve for the linear system

=

-

=

5

5

4

243

312

311

3

2

1

3

2

1

c

c

c

c

c

c

A.

Solution:


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Solution:

The solutions are

Rttctctc =-=+-= ,,1,32 321 .

Thus,

( ) ( ) Rttvvtvtv +-++-= ,132 321v is a linear combination of 321 ,, vvv withinfinite number

of expressions.

Example:


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Example:

Is the vector

-

-=

6

4

3

va linear combination of the vectors

=

-

-

-

=

=

5

4

1

,

2

1

1

,

3

2

1

321 vvv

.

Solution:


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Solution:

We need to find the constants 321 ,, ccc such that

332211321

5

4

1

2

1

1

3

2

1

6

4

3

vcvcvccccv ++=

+

-

-

-

+

=

-

-=

.

Solution:


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Solution:


-

-=

-

-

-

=

6

4

3

523

412

111

3

2

1

3

2

1

c

c

c

c

c

c

A.

The linear system hasnosolution.

v isnota linear combination of 321 ,, vvv

Note:


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Note:

Let nvvv ,,, 21 K and v be vectors in m and let Am*nbe

the matrix with column vectors njvAcol jj ,,2,1,)( K== .

Thus,

vAx= has solution or solutions v is a linea

combinationof nvvv ,,, 21 K .

vAx= has no solution v is not a linearcombinationof nvvv ,,, 21 K .

Definition of spanning set:


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Definition of spanning set:

Let { }

kvvvS ,,, 21 K= be a set of vectors in a real vector

space V. Then, the span of S, denoted by )( Sspan , is

the set consisting of all the vectors that are linear

combinations of kvvv ,,, 21 K . That is,

{ }RcccvcvcvcSspan kkk +++= ,,,|)( 212211 KL .

If VSspan =)( , it is said that V is spanned by S or S

spans V.

Example:


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Example:

Let

{ }321321 ,,and,

1

0

0

,

0

1

0

,

0

0

1

eeeSeee =

=

=

=.

Then,

3

321

3

2

1

332211 ,,|)( RRccc

c

c

c

ecececSspan =

=++=,

Example: Example 1(Page 292)


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Example: Example 1(Page 292)

Let

{ }321321 ,,and,01

1

,20

1

,12

1

vvvSvvv =

=

=

= .

Does3)( RSspan = ?

Solution:


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Solution:

3)( RSspan = For any vector3R

c

b

a

v

=

, there

exist real numbers 321 ,, ccc such that

332211321

0

1

1

2

0

1

1

2

1

vcvcvcccc

c

b

a

v ++=

+

+

=

=.


Solution:


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Solution:

=

c

b

a

c

c

c

3

2

1

021

102

111

.

The solution is

3

24,

3,

3

22321

cbac

cbac

cbac

--=

+-=

++-=

.

3213

2433

22 vcbavcbavcbav

--+

+-+

++-= .

That is, every vector in 3R can be a linear combination of

321 ,, vvv


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Example 10(Page 282)Example 11(Page 283)

Example 12(Page 284)

Theorem 6.3: (Important result)

Let { }kvvvS ,,, 21 K= be a set of vectors in a real vector

space V. Then, )( Sspan is a subspace of V.

Example 13(Page 285)

Example11 p283


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p p

n v1=(1,2,1), v2=(1,0,2), v3=(1,1,0), v=(2,1,5)

n If c1v1+c2v2 +c3v3 =v ,find c1 c2 c3 ?

n c1

(1,2,1) + c2

(1,0,2) + c3

(1,1,0) =(2,1,5)

c1+c2 +c3=2

2c1+0c2 +c3=1

c1+2c2 +0c3=5

So c1=1,c2=2,c3=-1

Example12 p284


82/167

p p

=

100

000,

010

000,

000

010,

000

001S

=

+

+

+

dc

badcba

0

0

100

000

010

000

000

010

000

001

Where a,b,c ,and d are real number

Section 6.3


83/167

Linear Independence

Definition:


84/167

The vectors 1, 2, , k in a vector space are said tospan if

every vector in is a linear combination of 1, 2,, k. Moreover,

ifS={1, 2, , k}, then we also say that the setSspans , or that

{1, 2, , k}spans , or that isspannedbyS, or in the

language of Section 6.2, spanS=.

The procedure to check if the vectors 1, 2, , k spanthe vector

space is as follows.

Definition:


85/167

Step 1. Choose an arbitrary vector in .

Step 2. Determine if is a linear combination of the given

vectors. If it is, then the given vectors span . If it is not,

they do not span .

Example 2--6(Page 292--294)


86/167

Question:

Let { }kvvvS ,,, 21 K= and WSspan =)( . Is it possible to

find a smaller (or even smallest) set, for example,

{ }121 ,,, -* = kvvvS K ,such that

)()( *== SspanWSspan ?To answer this question, we need to introduce the concept o

linear independence and linear dependence.

Linear Independence


87/167

Definition of linear dependence and linear

independence:

The vectors

kvvv ,,, 21K

in a vector space V arecalled linearly dependent if there exist constants,

kccc ,,, 21 K , not all 0, such that

02211 =+++ kkvcvcvc L .

Definition of linear dependence

and linear independence:


88/167

kvvv ,,, 21 K are linearly independent if

00 212211 =====+++ kkk cccvcvcvc LL

The procedure to determine if kvvv ,,, 21 K

are

linearly dependent or linearly independent:

Definition of linear dependence

and linear independence:


89/167

1.Form equation 02211 =+++ kkvcvcvc L , which lead to a

homogeneous system.

2.If the homogeneous system has only the trivial solutionthen the given vectors are linearly independent if it

has a nontrivial solution, then the vectors are linearly

dependent.

Example:


90/167

{ }321321 ,,and,

1

0

0

,

0

1

0

,

0

0

1

eeeSeee =

=

=

=. Are

21 , ee and 3e linearly independent?

Solution:


91/167

0

100

010

001

1

0

0

0

1

0

0

0

1

3

2

1

321332211 =

=

+

+

=++

c

c

c

cccececec

=

0

0

0

3

2

1

c

c

c

.

Therefore, 21 , ee

and 3e

are linearly

independent.

Example:


92/167

.

10

6

8

,

1

1

2

,

3

2

1

321

=

-

=

= vvv. Are 21 , vv and 3v

linearly independent?

Solution:


93/167

0

1013

612

821

10

6

8

1

1

2

3

2

1

3

2

1

321332211 =

-=

+

-+

=++

c

c

c

cccvcvcvc

Rtt

c

c

c

-

-=

,

1

2

4

3

2

1

.

Therefore, 21 , vv

and 3v

are linearly dependent.

Example:


94/167

Determine whether the following set of vectors in

the vector space consisting of all 22 matrices is

linearly independent or linearly dependent.

{ }

==

02

01,

12

03,

10

12,, 321 vvvS

.

Solution:


95/167

=

+

+

=++

00

00

02

01

12

03

10

12321332211 cccvcvcvc

.

Thus,

0

022

0

032

21

32

1

321

=+=+

=

=++

cc

cc

c

ccc

=

+

+

0

0

0

0

0

2

0

1

1

2

0

3

1

0

1

2

321 ccc

.

Solution:


96/167

The homogeneous system is

=

0

0

0

0

011

220

001

132

3

2

1

c

c

c

.

The associated homogeneous system has only the trivial

solution

=

0

0

0

3

2

1

c

c

c

.

Therefore, 21 , vv and 3v are linearly independent.

Example:


97/167

Determine whether the following set of vectors in

the vector space consisting of all polynomials o

degree n is linearly independent or linearlydependent.

{ } 223,2,2,, 222321 +++++== xxxxxxvvvS .

Solution:


98/167

( ( ( 022322 232221332211 =+++++++=++ xxcxxcxxcvcvcvc .Thus,

022

02

032

31

321

321

=++

=++

=++

cc

ccc

ccc

=

+

+

0

0

0

2

2

3

0

1

2

2

1

1

321 ccc.

Solution:


99/167

The associated homogeneous system is

=

0

0

0

202

211

321

3

2

1

c

c

c

.

The homogeneous system has infinite number of solutions,

.,

1

1

1

3

2

1

Rtt

c

c

c

-

=

Therefore, 21 , vv and 3v

are linearly dependent since

Rttvtvtv =-+ ,0321 .

Example 7--12(Page 295--296)

Example7 p295


100/167

-

-

1

1

02

0

0

11

and

=

-

+

-

0

0

00

1

1

02

0

0

11

21 cc

c1=c2=0. the vectors arelinear independent.

Example8 p295


101/167

n Are the vectors v1=(1,0,1,2),

v2=(0,1,1,2) , v3=(1,1,1,3) independent?

c1+c

3=0

c2+c3=0

c1+ c2+c3=0

2 c1+2 c2+3c3=0

The onlysolutions

c1=c2=c3 =0. the

vectors are

linear

independent.

Example9 p295


102/167

n Are the vectors v1=(1,2,-1), v2=(1,-2,1) ,

v3=(-3,2,-1), v4=(2,0,0), independent?

c1+ c2+-3c3+2c4=0

2c1+-2c2+2c3=0

-c1+ c2-c3=0

The only solutionsc1=1,c2=2, c3 =1.

c4 =0 or

c1=1,c2=1, c3 =0.

c4 =-1 so thevectors are linear

dependent.

Example10 p296


103/167

n The vector e1 and e2 in R2 , defined in EX4, are

linearly independent since

c1(1,0)+c2(0,1)=(0,0)

only if c1=c2=0

n We form the matrix A, whose columns are the

given n vectors. Then the given vectors arelinearly independent if and only if det(A)0.

=10

01A

Example11 p296


104/167

n Consider the vectors

n p1(t)=t2+t+2, p2(t)=2t

2+t, p3(t)=3t2+2t+2

n c1+2 c2+3c3=0n c1+ c2+2c3=0

n 2 c1+2c3=0

c1=1,c2=1, c3=-1.

Linear dependent

Example12 p296


105/167

n If v1, v2, v3, vk are k vectors in andy

vector space and vi is the zero vector.

n Then S={v1

, v2

, v3

, vk

} is linearly

dependent.

Note:


106/167

In the examples with,

10

68

,

1

12

,

3

21

321

=

-=

= vvv

or with

{ } 223,2,2,, 222321 +++++== xxxxxxvvvS, 21

, vv

an

3v are linearly dependent. Observe that 3v in both

examples are linear combinations of 21 , vv ,

Note:


107/167

233 24

1

12

2

3

21

4

10

68

vvv -=

--

=

=

and

21

222

3 22223 vvxxxxxxv +=++++=++= .

As a matter of fact, we have the following generalresult.

Theorem 6.4: (Importantresult)


108/167

The nonzero vectors kvvv ,,, 21 K in a vector space V

are linearly dependent if and only if one of the

vectors 2, jvj , is a linear combination of the

preceding vectors 121 ,,, -jvvv K .

Note:


109/167

Every set of vectors containing the zero vector is

linearly dependent. That is, kvvv ,,, 21 K are k vectors

in any vector space and iv is the zero vector, then

kvvv ,,, 21 K are linearly dependent.

Example 13 on page 298

Example 14 on page 299

Example13 p298


110/167

n If v1, v2, v3, v4 are as in Ex9,then we

find that v1+ v2+ 0v3- v4=0

n So v1

, v2

, v3

, v4

are linear dependent.

We then hace v4= v1+ v2

Example14 p299


111/167

2144321 vSinceS.spanlet w

0

1

1

2

,

0

1

1

0

,

0

1

0

1

,

0

0

1

1

vvvvvv +==

=

=

=

=

We conclude thar W=span S1 where

S1={ v1,v2,v3 }

6.4. Basis and Dimension

vvv


112/167

The vectors

kvvv ,,,

21K

in a vector space V are said to

form abasisof V if

(a) kvvv ,,, 21 K span V (i.e., Vvvvspan k =),,,( 21 K ).

(b)

kvvv ,,, 21 K are linearly independent.

Example 1(Page 303)

Natural basis or standard basis

Example1 p303


113/167

n The vector e1=(1,0) and e2=(0,1) from

a basis for R2.Each of these set of

vectors is called the natural basis or

standard basis.

Example:


114/167

{ }321321 ,,and,

1

00

,

0

10

,

0

01

eeeSeee =

=

=

=

. Are 21 , ee

and 3e a basis in3

R ?

Solution:


115/167

21 , ee and 3e form a basis in 3R since

(a)3

321 ),,()( ReeespanSspan == (see the example in the

previous section).

(b) 21 , ee and 3e are linearly independent (also see the

example in the previous section).

Example:


116/167

=

=

=

43,

10,

01

321 vvv . Are 21 , vv and 3v a basis in2?

[solution:]

21 , vv and 3v arenota basis of 2 since 21 , vv and 3v are

linearly dependent,

043 321 =-+ vvv .

ote that2

321 ),,( Rvvvspan = .

Example:


117/167

.

10

68

,

1

12

,

3

21

321

=

-=

= vvv

. Are 21 , vv and 3v a

basis in3

R ?

Solution:


118/167

21 , vv and 3v are not a basis in3R since 21 , vv and 3v

are linearly dependent,

233 24

1

1

2

2

3

2

1

4

10

6

8

vvv -=

-

-

=

=.

Example 2(Page 303)

Example 3(Page 304)

Example 4(Page 304)

Example2 p303


119/167

n S={ v1,v2,v3,v4} where

n v1=(1,0,1,0), v2=(0,1,-1,2)

n v3=(0,2,2,1) ,v

4=(1,0,0,1)

n Show that S is a basis for R4

part1


120/167

n c1+c4=0

n c2+2c3 =0

n c1-c

2+2c

3=0

n 2 c2+c3 +c4=0

The only solutions

c1=c2=c3=c4=0,

Showing that S islinear indepeddent

part2


121/167

n Let v =(a,b,c,d) be any vector in R4.

n We now seek constants k1,k2,k3,k4

such that k1v

1+k

2v

2+ k

3v

3+ k

4v

4=v

n Substituting for v1 v2 v3 v4 and v we find

a solution for k1,k2,k3,k4 to the resulting

linear system for any a,b,c,d . Hence S

spans R4 and is a vasis for R4.

Example3 p304 part 1


122/167

n To show S={t2+1,t-1,2t+2} is a basisfor the vectorspace P2.

n at2+bt+c=a1t2+(a2+

a3)t+(a1-a2+2a3)

n a1=a

n a2+2a3=bn a1-a2+2a3=c

4,

2, 321

abca

cbaaaa

-+=

-+==

13622

++ ttgiven

a1=2 , a2=-5/2 ,a3=17/4

a1(t2+1)+a2(t-1)+a3(2t+2)=0

Example3 p304 part 2

2 ( 2 ) ( 2 ) 0


123/167

Then a1t2+(a2+2a3)t+(a1-a2+2a3)=0

a1=0

a2+2a3=0

a1

-a2

+2a3

=0

The only solutions a1=a2=a3=0,

S is linear indepeddent

Example4 p304

Fi d b i f th b V f


124/167

n Find a basis for the subspace V ofP2 ,consisting of all vector of the format2+bt+c ,where c=a-b.

n at2+bt+a-b

n a(t2+1)+b(t-1)+cn So the t2+1and t-1 span V.

n a(t2+1)+b(t-1)=0

n Since this equation id to hold all values of t,

we must have a1=a2=0.

Example:

L


125/167

Let

{ }321321 ,,and,

0

1

1

,

2

0

1

,

1

2

1

vvvSvvv =

=

=

=.

Are S a basis in3R ?

Solution:

( )


126/167

(a)

3)( RSspan = For any vector3R

c

b

a

v

=, there

exist real numbers 321 ,, ccc such that

332211321

0

1

1

2

0

1

1

2

1

vcvcvcccc

c

b

a

v ++=

+

+

=

=

.

Solution:

d t l f th li t


127/167


=

c

b

a

c

c

c

3

2

1

021

102

111

.

The solution is

3

24

,3,3

22321

cba

c

cba

c

cba

c

--

=

+-

=

++-

= .

Solution:

Th


128/167

Thus,

321 3

24

33

22

v

cba

v

cba

v

cba

v

--+

+-+

++-=

.

That is, every vector in3R can be a linear

combination of321 ,, vvv

and

3)( RSspan =.

Solution:

(b)


129/167

(b)

Since

0

0

0

0

2

2 321

21

31

321

332211 ===

=

+

+

++

=++ ccc

cc

cc

ccc

vcvcvc,

321 ,, vvv are linearly independent.

By (a) and (b), 321 ,, vvv are a basis of3R .

Theorem 6.5:(Important result)

If { }

kvvvS 21 i b i f t V th


130/167

If

{ }kvvvS ,,, 21 K= is a basis for a vector space V, thenevery vector in V can be written in an unique(one

and only one) way as a linear combination of the

vectors in S.

Example:

{ }

001


131/167

{ }321321 ,,and,

1

0,

0

1,

0

0 eeeSeee =

=

=

=. S is a basis o

3R . Then, for any vector

=

c

b

a

v

,

321

1

0

0

0

1

0

0

0

1

cebeaecba

c

b

a

v ++=

+

+

=

=

is uniquely determined.

Theorem 6.6:(Importantresult)


132/167

Let { }kvvvS ,,, 21 K= be a set of nonzero vectors in a

vector space V and let { }kvvvspanW ,,, 21 K= . Then,

some subset of S is a basis of W.

Example:

Let


133/167

Let

{ }

==

2

3

1

,

1

0

0

,

0

3

2

,

0

1

0

,

0

0

1

,,,, 23121 aeaeeS

and ( )3RSspan = .Please find subsets of S which

form a basis of 3R .

Solution:


134/167

We first check if 1e and 2e are linearly independent.

Since they are linearly independent, we continue to check i

1e

, 2e

and 1a

are linearly independent.Since032 121 =-+ aee ,

Solution:

we delete 1a from S and form a new set 1S , { }23211 ,,, aeeeS = .

Th i h k if d li l


135/167

Then, we continue to check if 1e, 2e and 3e are linearly

independent.They are linearly independent. Thus, we finally

check if 1e, 2e 3e and 2a are linearly independent.Since

023 2321 =-++ aeee ,we delete

1a from S and form a new set 2S , { }3212 ,, eeeS = .

Therefore,

{ }3212 ,, eeeS =

is the subset of S which form a basis of form a basis of3R .

How to find a basis (subset of S) of W?

There are two methods:


136/167

There are two methods:ethod 1:

The procedure based on the proof of the above

important result.

ethod 2:

Step 1: Form equation

02211 =+++ kk vcvcvc L .


Step 2: Construct the augmented matrix associated with the

eq ation in step 1 and transform this a gmented matri


137/167

equation in step 1 and transform this augmented matrix

to the reduced row echelon form.

Step 3: The vectors corresponding to the columnscontaining the leading 1s form a basis. For example, i

6=k and the reduced row echelon matrix is


01


138/167

0000000

0000000

01000

010001

MMMMMMM ,

then the 1st, the 3nd, and the 4th columns contain

a leading 1 and thus

431 ,, vvv are a basis of { }621 ,,, vvvspanW K= .

Example:


139/167

Let

{ }

==

2

3

1

,

1

0

0

,

0

3

2

,

0

1

0

,

0

0

1

,,,, 23121 aeaeeS

and

( )

3

RSspan =

.Please find subsets of S whichform a basis of 3R .

Solution:

ethod1:


140/167

We first check if 1e and 2e are linearly independent.Since they are linearly independent, we continue to check i

1e, 2e and 1a are linearly independent.Since

032 121 =-+ aee ,

Solution:

we delete 1a from S and form a new set 1S , { }

23211 ,,, aeeeS = .


141/167

we delete a from S and form a new set ,

{ }.Then, we continue to check if 1e, 2e and 3e are linearly

independent.They are linearly independent. Thus, we

finally check if 1e, 2e 3e and 2a are linearly

independent.Since

023 2321 =-++ aeee ,

Solution:

we delete 1a from 1S and form a new set2S ,


142/167

we delete from and form a new set ,

{ }3212 ,, eeeS = . Therefore,

{ }3212 ,, eeeS =

is the subset of S which form a basis of form a

basis of3R .

Solution:

ethod 2:


143/167

Step 1:

The equation is

0

3

2

1

1

0

0

0

3

2

0

1

0

0

0

1

54321 =

+

+

+

+

ccccc

.

Solution:

Step 2:

The augmented matrix and its reduced row echelon matrix is


144/167

The augmented matrix and its reduced row echelon matrix is

031000

020310

010201

.

The 1st, the 2nd and 4th columns contain the leading 1s.

Thus,

{ }321 ,, eee forms a basis.Example 5(Page 309)

Example5 p309

n S={ v1,v2,v3,v4,v5}, v1= (1,2,-2,1) ,


145/167

S { v1,v2,v3,v4,v5}, v1 (1,2, 2,1) ,v2=(3-,0,-4,3), v3=(2,1,1,-1), v4=(-3,3,-9,6), v5=(9,3,7,-6).

n Find a subset of S that is a basis forW=span S.

n Step1:

n c1

(1,2,-2,1) + c2

(3-,0,-4,3)+ c3

(2,1,1,-1) + c4 (-3,3,-9,6) + c5(9,3,7,-6)=0

Step2:

331


146/167

--

000000

000000

025

23

2110

02

3

2

3

2

101

Step3:

n The leading 1s appear in columns 1


147/167

g ppand 2, so {v1,v2}is a basis for W= span

S.


Let { }

nvvvS ,,, 21K=

be a basis for a vector space V and{ }


148/167

{ } plet { }rwwwT ,,, 21 K= is a linear independent set o

vectors in V.Then, nr .

.

Corollary 6.1:

Let { }

nvvvS ,,, 21K=

and { }

mwwwT ,,, 21K=

be two


149/167

{ },,,

{ }bases for a vector space V. Then, mn = .

Note:

For a vector space V, there are infinite bases. But


150/167

pthe number of vectors in two different bases are

the same.

Example:

For the vector space

3

R ,111


151/167

{ }321321 ,,,

0

1

1

,

2

0

1

,

1

2

1

vvvSvvv =

=

=

=is a basis

for3R (9094).

Example:

previous example). Also,


152/167

{ }321321 ,,,

1

0

0

,

0

1

0

,

0

0

1

eeeTeee =

=

=

=is basis for

3R .

There are 3 vectors in both S and T.

Definition of dimension:


153/167

The dimension of a vector space V is the number

of vectors in a basis for V. We often write dim(V

for the dimension of V. dim({0}) = 0.

Example:

{ }

001


154/167

{ }321321 ,,,

1

0,

0

1,

0

0 eeeTeee =

=

=

=is basis for 3R .

The dimension of

3

R is 3.

Example 6--8(Page 310)

Example6 p310

n The dimension of R2 is 2


155/167

n The dimension of R3 is 3

n The dimension of Rn is n.

Example7 p310

n The dimension of P2 is 3


156/167

n The dimension of P3 is 4

n The dimension of Pn is n+1

Example8 p310

n The subspace w of R4 considered inE 5 h di i 2


157/167

Ex5 has dimension 2.


If S is a linearly independent set of vectors in a


158/167

IfS is a linearly independent set of vectors in a

finite-dimensional vector space V, then there is a

basisTforV, which containsS.

Example 9(Page311)

Example9 p311

n Find the basis for R4 that contains thevector v =(1 0 1 0) and v =(-1 1 -1 0)


159/167

vector v1=(1,0,1,0) and v2=(-1,1,-1,0).

n Let {e1, e2, e3, e4, } be the natural basis

for R

4

wheren e1=(1,0,0,0), e2=(0,1,0,0)

n e3=(0,0,1,0) e4=(0,0,0,0)

n

c1v1+ c2v2+ c3e1+ c4e2+ c5e3+ c6e4=0

Example9 p311

00010100011001


160/167

-

01000000010100

0001010

Since leading 1s appear in columns 1,2,3 and 6 we

conclude that {v1,v2,e1.e4} is a basis for R4 containing

v1and v2


Let V be an n-dimensional vector space, and let


161/167

{ }nvvvS ,,, 21 K= be a set of n vectors in V.

(a) If S is linearly independent, then S is a basis for

V.

(b) If S spans V, then S is a basis for V.

Example:

Is{ }321321 ,,,1

1

,0

1

,2

1

vvvSvvv =

=

=

=a basis


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Is 021

a basis

for

3

R ?

Note:

n

.


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.

Solution:

Since

3

R is a 3-dimensional vector space, not likein the previous example we only need to examine


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in the previous example, we only need to examine

whether S is linearly independent or S spans3R .

We dont need to examine S being both linearly

independent and spans V.

Example:

421


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Is

{ }321321 ,,,

1

2,

0

1,

1

3 vvvSvvv =

=

=

-

=a basis for

3R ?

Solution:

Since3R is a 3-dimensional vector space, we only need to

examine whether S is linearly independent or S spans 3.Because


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0

0

0

0

23

42

321

31

321

321

332211 ===

=

+-

++

++

=++ ccc

cc

ccc

ccc

vcvcvc

,

321 ,, vvv are linearly independent. Therefore, 321 ,, vvv are a

basis of

3

R .Example 10(Page 312)

Example10 p312

n In Ex5. W= span S is a subspace of R4,sodim W4. Since S contains five vectors we

l d b C ll 6 1 th t S i t


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conclude by Corollary 6.1 that S is not abasis for W. In Ex2, since dim R4 =4, and

the set S contains four vectors, it is possiblefor S to be a basis for R4.If S is linearlyindependent or spans R4,it is a basisotherwise it is not a basis. Thus we needonly check one of the conditions in Thm 9.6not both.

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CH6.1-6.4 Real Vector Space