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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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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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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.