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4

4.3

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Vector Spaces

LINEARLY INDEPENDENTSETS; BASES

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Slide 4.3- 2 2016 Pearson Education, Inc.

LINEAR INDEPENDENT SETS; BASES

An indexed set of vectors {v1, , vp} in Vis said to

be linearly independentif the vector equation

(1)

has onlythe trivial solution, . he set {v1, , vp} is said to be linearly dependent

if (1) has a nontrivial solution, i.e., if there are so!e

"ei#hts, c1, , cp, not all zero, such that (1) holds. $n such a case, (1) is called a linear dependence

relationa!on# v1, , vp.

1 1 % %v v ... v &

p pc c c+ + + =

1 &,..., &pc c= =

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Slide 4.3- 3

LINEAR INDEPENDENT SETS; BASES

Theorem 4:An indexed set {v1, , vp} of t"o or!ore vectors, "ith , is linearl' dependent if

and onl' if so!e vj("ith ) is a linear

co!bination of the precedin# vectors, .

Definition:etHbe a subspace of a vector space V.

An indexed set of vectors B in Vis a

basis forHif(i) Bis a linearl' independent set, and

(ii) he subspace spanned b' Bcoincides "ithH

that is,

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1v &

1j>

1{b ,...,b }

p=

1

Span{b ,...,b }p

H=

1 1v ,...,v

j

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Slide 4.3- 4

LINEAR INDEPENDENT SETS; BASES

he definition of a basis applies to the case "hen ,

because an' vector space is a subspace of itself.

hus a basis of Vis a linearl' independent set thatspans V.

*hen , condition (ii) includes the require!ent that

each of the vectors b1, , bp!ust belon# toH, because

Span {b1, , bp} contains b1, , bp.

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H V=

H V

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Slide 4.3- 5

STANDARD BASIS

et e1, , enbe the colu!ns of the !atrix,In. hat is,

he set {e1, , en} is called the standard basisfor . See

the follo"in# fi#ure.

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n n

1 %

1 & &

& 1e ,e ,...,e

&& & 1

n

= = =

M

M M

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Slide 4.3- 6

THE SPANNING SET THEOREM

Theorem 5:et be a set in V, andlet .

a. $f one of the vectors in S+sa', vk+is a linear

co!bination of the re!ainin# vectors in S,

then the set for!ed fro! Sb' re!ovin# vkstillspansH.

b. $f , so!e subset of Sis a basis forH.

Proof:a. ' rearran#in# the list of vectors in S, if

necessar', "e !a' suppose that vpis a linear

co!bination of +sa',

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1{v ,...,v }

pS=

1Span{v ,...,v }

pH=

{&}H

1 1v ,...,v

p

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THE SPANNING SET THEOREM

(-)

iven an' xinH, "e !a' "rite

(/)

for suitable scalars c1, , cp.

Substitutin# the expression for vpfro! (-) into(/), it is eas' to see that xis a linear

co!bination of .

hus spansH, because x"as anarbitrar' ele!ent ofH.

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1 1 1 1

v v ... vp p p

a a = + +

1 1 1 1x v ... v v

p p p pc c c

= + + +

1 1

v ,...vp

1 1{v ,...,v }

p

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THE SPANNING SET THEOREM

b. $f the ori#inal spannin# set Sis linearl'independent, then it is alread' a basis forH.

0ther"ise, one of the vectors in Sdepends on

the others and can be deleted, b' part (a).

So lon# as there are t"o or !ore vectors inthe spannin# set, "e can repeat this process

until the spannin# set is linearl' independent

and hence is a basis forH.

$f the spannin# set is eventuall' reduced to

one vector, that vector "ill be nonero (and

hence linearl' independent) because .

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{&}H

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THE SPANNING SET THEOREM

Example 7:et , ,

and .

2ote that , and sho" that

. hen find a basisfor the subspaceH.

Solution:3ver' vector in Span {v1, v%} belon#s toHbecause

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1

&v %

1

=

%

%v %

&

=

-

4v 14

5

=

1 % -Span{v ,v ,v }H=

- 1 %v 5v -v= +

1 % - 1 %Span{v , v , v } Span{v ,v }=

1 1 % % 1 1 % % -v v v v &vc c c c+ = + +

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THE SPANNING SET THEOREM

2o" let xbe an' vector inH+sa',

.

Since , "e !a' substitute

hus xis in Span {v1, v%}, so ever' vector inHalread'

belon#s to Span {v1, v%}.

*e conclude thatHand Span {v1, v%} are actuall' the

set of vectors.

$t follo"s that {v1, v%} is a basis ofHsince {v1, v%} is

linearl' independent. 2016 Pearson Education, Inc.

1 1 % % - -x v v vc c c= + +

- 1 %v 5v -v= +

1 1 % % - 1 %

1 - 1 % - %

x v v (5v -v )

( 5 )v ( - )v

c c c

c c c c

= + + +

= + + +

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BASIS FOR COL B

Example 8:6ind a basis for 7olB, "here

Solution:3ach nonpivot colu!n ofBis a linear

co!bination of the pivot colu!ns. $n fact, and .

' the Spannin# Set heore!, "e !a' discard b%and

b/, and {b1, b-, b5} "ill still span 7olB.

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1 % 5

1 / & % &

& & 1 1 &b b b

& & & & 1

& & & & &

B

= =

L

% 1b /b=

/ 1 -b %b b=

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BASIS FOR COL B

et

Since and no vector in Sis a linearco!bination of the vectors that precede it, Sis

linearl' independent. (heore! /).

hus Sis a basis for 7olB.

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1 - 5

1 & && 1 &

{b ,b ,b } , ,& & 1

& & &

S

= =

1b &

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BASES FOR NUL AAND COL A

Theorem :he pivot colu!ns of a !atrixAfor! a

basis for 7olA.

Proof:etBbe the reduced echelon for! ofA.

he set of pivot colu!ns ofBis linearl' independent,for no vector in the set is a linear co!bination of thevectors that precede it.

SinceAis ro" equivalent toB, the pivot colu!ns of

Aare linearl' independent as "ell, because an' lineardependence relation a!on# the colu!ns ofAcorresponds to a linear dependence relation a!on#the colu!ns ofB.

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Slide 4.3- 14

BASES FOR NUL AAND COL A

6or this reason, ever' nonpivot colu!n ofAis a linear

co!bination of the pivot colu!ns ofA.

hus the nonpivot colu!ns of a !a' be discarded fro!

the spannin# set for 7olA, b' the Spannin# Set

heore!. his leaves the pivot colu!ns ofAas a basis for 7olA.

!arnin":he pivot colu!ns of a !atrixAare evident

"henAhas been reduced onl' to echelon for!.

ut, be careful to use the pivot colu!ns ofAitself for

the basis of 7olA.

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Slide 4.3- 15

BASES FOR NUL AAND COL A

8o" operations can chan#e the colu!n space of a!atrix.

he colu!ns of an echelon for!BofAare often notin the colu!n space ofA.

T#o $ie#s of a %asis

*hen the Spannin# Set heore! is used, the deletionof vectors fro! a spannin# set !ust stop "hen the set

beco!es linearl' independent. $f an additional vector is deleted, it "ill not be a

linear co!bination of the re!ainin# vectors, andhence the s!aller set "ill no lon#er span V.

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Slide 4.3- 16

TWO VIEWS OF A BASIS

hus a basis is a spannin# set that is as s!all aspossible.

A basis is also a linearl' independent set that is aslar#e as possible.

$f Sis a basis for V, and if Sis enlar#ed b' one vector

+sa', #+fro! V, then the ne" set cannot be linearl'independent, because Sspans V, and #is therefore alinear co!bination of the ele!ents in S.

2016 Pearson Education, Inc.