Objetie: Study mny importnt properties of etors in R n A refully onstruted generliztion of R n to mny other importnt etor spe. Applition of etor spe t

Chpter 6 Rel Vetor Spes

Objetie: Study mny importnt properties of etors in R n A refully onstruted generliztion of R n to mny other importnt etor spe. Applition of etor spe to the liner system nd mtrix.

6.. Vetor Spes Definition of Vetor Spe Definition: A rel etor spe V is set of elements together with two opertions ddition " nd slr multiplition. " stisfying the following properties: Let u nd w be etors in V nd let nd d be slrs.

Addition: (α If u nd re ny elements of V then u is in V(losed.. u u.. u w ( u w (.. V hs zero etor suh tht for eery u in V u u u. 4. For eery u in V there is n element u V ( u u. in

Slr multiplition: ( β If u is ny element of V nd is ny rel number then u is in V(losed. 5. du ( d u (. 6. ( d u u du. 7. ( u u. 8. u u.

Exmple : n V R together with stndrd etor ddition nd slr multiplition. Then V is etor spe sine for ny two etors u nd in nd slr both u nd u re in n R. Therefore onditions (α nd ( β re stisfied. In ddition the onditions ( to (8 re stisfied (see the preious subsetion. R n

Exmple : V the set onsisting of ll m n mtries together with stndrd mtrix ddition nd slr multiplition. V is etor spe sine for ny two m n mtries u nd nd slr both u nd u re m n mtries. Tht is both u nd u re still in V. Therefore onditions (α nd ( β re stisfied. In ddition the onditions ( to (8 re stisfied (see setion.

Exmple: V the set onsisting of ll polynomils of degree or less with the form together with stndrd polynomil ddition nd slr multiplition. Is V etor spe? We need to exmine whether the onditions ( α ( β nd the onditions ( to (8 re stisfied. Let u x x b x b x b w nd let nd d be slrs. Then x x

Solution: Addition: (α : ( ( ( ( ( V b x b x b b b x x b x x u sine u is polynomil of degree or less.

Solution: (: u x x b b x x b b x b x b b x b x b b b x x b x x u ( ( ( ( ( ( ( ( ( (

Solution: (: ( ] ( ( [( ( ( ( ] ( ( [( ( ( w u x x b x b x b b x b x b b x b x b x x w u

Solution: (: Let x x. Then u x x u x x x x u ( ( ( ( ( ( (4: Let. ( ( ( x x u Then ] ( [ ] ( [ ] ( [ x x x x u u

Solution: Slr multiplition: ( β : u ( V x ( x ( sine u is polynomil of degree or less. (5: ( u [ ( ( b b ] x x [ ( ( b ] x [ ( b x ( b b ] u

Solution: (6: du u d x d x d x x d x d x d d u ] ( ( [( ] ( ( [( ] [( ] [( ] [( ( (7: d u d d x x d d x d x d d x d x d du ( ( ( ( ( ( ( ] ( ( [( ( (8: u x x x x u

Note: P n the set onsisting of ll polynomils of degree n or less with the form together with stndrd polynomil ddition nd slr multiplition. Then P n is etor spe. In ddition P the set onsisting of ll polynomils with the form together with stndrd polynomil ddition nd slr multiplition. Then P is lso etor spe.

Exmple: V 4 the set onsisting of ll rel lued ontinuous funtions defined on the entire rel line together with stndrd ddition nd slr multiplition. Is V 4 spe? etor We need to exmine whether the onditions (α ( β nd the onditions ( to (8 re stisfied. Let u f (x g (x w h (x nd let nd d be slrs. Then

Solution: Addition: (α : 4 ( ( V x g x f u sine ( ( x g x f is still ontinuous funtion. (: u x f x g x g x f u ( ( ( ( (: [ ] [ ] w u x h x g x f x h x g x f w u ( ( ( ( ( ( ( (

Solution: (: Let the zero etor. Then u f ( x f ( x u (4: Let u f (x. Then [ ( ] ( ( u ( u f ( x f x f x f x

( β : Solution: sine f (x (5: (6: u f ( x V is still ontinuous funtion. [ f ( x g ( x ] f ( x g ( x u ( u ( d u ( d f ( x f ( x df ( x u 4 du

Solution: (7: (8: [ x ] df x d f x d u ( du df ( ( ( ( ( ( [ f x ] f x u u ( (

Let Note: * V 4 the set of ll differentible funtions defined on the entire rel line nd ** V 4 the set of ll integrble funtions defined on the entire rel line. Both * V 4 nd ** V 4 re etor spe under stndrd ddition nd slr multiplition.

V 5 Exmple: the set onsisting of ll integers with stndrd ddition nd slr multiplition. Is V 5 spe? etor V 5 is not rel etor spe sine for u V 5. 7 u. 7. 7 V 5 ondition ( β is not stisfied. nd

Exmple: V 6 the set onsisting of ll etors in R with stndrd ddition nd nonstndrd slr multiplition defined by x x x. Is V 6 etor spe?

Solution: 6 V is not rel etor spe sine for R u R u u ondition (8 is not stisfied.

V 7 Exmple: the set onsisting of only seond degree polynomils with stndrd ddition nd slr multiplition. Is V 7 etor spe? V 7 is not rel etor spe sine for u x nd x u x ( x V 7 ondition ( α is not stisfied.

Exmple: V 8 the set onsisting of ll rel lued ontinuous funtions suh tht f (. Suppose the opertions re stndrd ddition nd slr multiplition. Is V 8 etor spe?

Solution: 8 V is not rel etor spe sine for 8 ( V x f u 8 ( V g x 8 ( ( 6 ( ( V x g x f u g f ondition ( α is not stisfied. Exmple 9 Pge 7 76

Exmple p7 Consider the set R n together with the opertions of etor ddition nd slr multiplition s defined in Setion 4.. Theorem 4. in Setion 4. estblished the ft tht R n is etor spe under the opertions of ddition nd slr multiplitions of n etor.

Exmple p7 Consider the set V of ll ordered triples of rel number of the form (xyxnd define the opertions nd by (xy (x y (xx yy (xy(xy

Exmple p7 Consider the set V of ll ordered triples of rel numbers (xyz nd define the opertions nd by.(xyz (x y z (xx yy zz.. (xyz(xyz.. [(xyz (x y z ]((xx yy zz. 4. (xyz d (x y z ((dxyz.

Exmple4 p74 Consider the set M of ll mtries. Similrly the ser of ll m n mtries under the usul opertions of mtrix ddition nd slr multiplition is etor spe will be denoted by M mn.

Exmple5 p74 Let F[b] be the set of ll rel lued funtions tht re defined on the interl [b]. If f nd g re in V we define f g nd f by (f g(tf(tg(t. ( f(tf(t. A polynomil (in t is funtion tht is expressible s p(t n t n n t n t re rel numbers.

Exmple6 p74 The follow funtions re polynomils p (tt 4 -t 5t-. p (tt. p (t4. f 4 (t t -6 nd f 5 (t /t -t re not polynomils.

Exmple7 p75 p (t : degree 4 p (t : degree p (t : degree

Exmple8 p75 If p(t n t n n t n t. q(tb n t n b n t n b tb. p(t q(t ( n b n t n ( n b n t n ( b t b p(t n t n n t n t. (d p(t p(t d p(t.

Exmple9 p76 Let u u uu If u u. u If u4d (d ( 4 u d 8 4

Theorem 6.: (Importnt Result If V is etor spe nd let u be ny element of rel etor spe V. Then ( u. (b R V. ( u. or u. (d ( u u.

6.. Subspe Definition of subspe: W is lled subspe of rel etor spe V if. W is subset of the etor spe V.. W is etor spe with respet to the opertions in V. Exmple (Pge 79 Exmple (Pge 79

Exmple p79 Eery etor spe hs t lest two subspes itself nd subspe{}. The subspe {} is lled the zero subspe.

Exmple p79 Let u( b ( b u( b b u( b

Theorem 6.: (Importnt Result W is subspe of rel etor spe V.If u nd re ny etors in W then u W..If is ny rel number nd u is ny etor in W then u W.

W form Exmple: the subset of R onsisting of ll etors of the together with stndrd ddition nd slr multiplition. Is W subspe of R? R

Solution: We need to hek if the onditions ( nd ( re stisfied. Let R u. Then (: W u..

Solution: (: u W. W is subspe of R.

Exmple: Let the rel etor spe V be the set onsisting of ll n n mtries together with the stndrd ddition nd slr multiplition. Let W the subset of V onsisting of ll n n digonl mtries. Is W subspe of V? Let u M M L L O L M nn W b M R. b M L L O L b M nn W nd

Solution: (: W b b b u nn nn L M O M M L L sine u is still digonl mtrix.

Solution: (: u M sine u is still digonl mtrix. M L L O L M nn W W is subspe of V.

P n Exmple: the set onsisting of ll polynomils of degree n or less with the form together with stndrd polynomil ddition nd slr multiplition. P n is etor spe. P the set onsisting of ll polynomils with the form together with stndrd polynomil ddition nd slr multiplition. Then P is lso etor spe. Then P n is subspe of P.

Exmple: V 4 the set onsisting of ll rel lued ontinuous funtions defined on the entire rel line together with stndrd ddition nd slr multiplition. Let * V 4 the set of ll differentible funtions defined on the entire rel line together with stndrd ddition nd slr multiplition. Then V 4 is rel etor spe. Also. * V 4 is subspe of V 4.

Exmple: W the subset of R onsisting of ll etors of the form R b b together with stndrd ddition nd slr multiplition. Is W subspe of R?

Solution: (: ( W b b b b b b u. Therefore W u. W is not subspe of R.

Exmple: V the set onsisting of ll polynomils of degree or less with the form together with stndrd polynomil ddition nd slr multiplition. V is etor spe. Let W the subset of 4 form Is W 4 subspe of V? V onsisting of ll polynomils of the x bx b.

Solution: Let 4 W x x u nd 4 W b b x b x. Then nd b b b. Thus 4 ( ( ( ( ( W b x b x b b b x b x x x u

Solution: sine ( b ( b ( b ( b b b 4 (. W 4 is not subspe of V. Exmple (Pge 8 8

Exmple p8 d b u d b d d b b u kd k kb k ku

Exmple5 p8 Let u( b ( b u( b b

Exmple6 p8 We let P n denote the etor spe onsisting of ll polynomils of defree n nd the zero polynomil. And let P denote the etor spe of kk polynomils. It is esy to erify tht P is subspe of P nd in generl tht P n is subspes of P n. P n is subspe of P.

Exmple7 p8 Let V be the set ll polynomils of degree extly.v is subset P but it is not subspe of P sine the sum of polynomils t t nd t t polynomil of degree is not in V.

Exmple8 p8 Let C[b] denote the set of ll rellues ontinuous funtions tht re defined on the interl [b]. If f nd g re in C[b] then fg is in C[b].

Exmple9 p8 Consider the homogeneous system Ax A is n m n mtrix.a solution onsists of etor x in R n. Let W be the subset of R n onsisting of ll solutions to the homogeneous system. Sine A we onlude tht W is not empty. To hek tht W us subspe of R n we erify properties (α nd (β of Them 6.. Thus let x nd y be solutions. Ax nd Ay A(xyAxAy A(x(Ax

Exmple p8 Let nd be fixed etors in etor spe V nd let W be the set of ll liner ombintions of nd. w w b b w w ( b ( b w ( (

Let Definition of liner ombintion: K k be etors in rel etor spe V. A etor in V is lled liner ombintion of K k if where L k k K k re rel numbers.

Exmple: Let 8 be etors in the etor spe onsisting of ll mtries. Then ( 8. Tht is is liner ombintion of.

Exmple: For liner system b x x x Ax. x is solution for the boe liner system. Thus

Exmple: b A ol A ol A ol A ( ( ( Tht is b is liner ombintion of the olumn etors of A ( ( ( A ol A ol A ol.

Note: For liner system m n n m A x b the liner system hs solution or solutions b is liner ombintion of the olumn etors of A ol ( A ol ( A L ol n ( A. For exmple if M n A m n x n b m is solution of

Note: then ol ( A ol ( A L n ol n ( A b. On the other hnd the liner system hs no solution b is not liner ombintion of the olumn etors of A

Exmple: Is the etor 5 5 4 liner ombintion of the etors 4.

Solution: We need to find the onstnts suh tht 4 5 5 4. we need to sole for the liner system 5 5 4 4 A.

Solution: The solutions re Thus t t t t R. ( t ( t t t R is liner ombintion of with infinite number of expressions.

Exmple: Is the etor 6 4 liner ombintion of the etors 5 4.

Solution: We need to find the onstnts suh tht 5 4 6 4.

Solution: we need to sole for the liner system 6 4 5 4 A. The liner system hs no solution. is not liner ombintion of

Note: Let K n nd be etors in m the mtrix with olumn etors Thus Ax ombintion of R nd let A m*n be ol j ( A j j K n. hs solution or solutions is liner K n. Ax hs no solution is not liner ombintion of K n.

Definition of spnning set: { } Let S K k be set of etors in rel etor spe V. Then the spn of S denoted by spn (S is the set onsisting of ll the etors tht re liner ombintions of K k. Tht is { R } spn ( S k k K L k. If spn ( S V it is sid tht V is spnned by S or S spns V.

Exmple: Let { } nd e e e S e e e. Then ( R R e e e S spn

Exmple: Exmple (Pge 9 Let { } nd S. Does ( R S spn?

Solution: ( R S spn For ny etor R b there exist rel numbers suh tht b. we need to sole for the liner system

Solution: b. The solution is 4 b b b. 4 b b b. Tht is eery etor in R n be liner ombintion of

Exmple (Pge 8 Exmple (Pge 8 Exmple (Pge 84 Theorem 6.: (Importnt result Let S { } K k be set of etors in rel etor spe V. Then spn (S is subspe of V. Exmple (Pge 85

Exmple p8 ( ( ( (5 If find? ( ( ( (5 5 So

Exmple p84 S d b d b Where b nd d re rel number

Setion 6. Liner Independene

Definition: The etors k in etor spe V re sid to spn V if eery etor in V is liner ombintion of k. Moreoer if S{ k } then we lso sy tht the set S spns V or tht { k } spns V or tht V is spnned by S or in the lnguge of Setion 6. spn S V. The proedure to hek if the etors k spn the etor spe V is s follows.

Definition: Step. Choose n rbitrry etor in V. Step. Determine if is liner ombintion of the gien etors. If it is then the gien etors spn V. If it is not they do not spn V.

Exmple 6(Pge 9 94 Question: { } Let S K k nd spn S W (. Is it possible to find smller (or een smllest set for exmple { } S K k suh tht spn ( S W spn ( S? To nswer this question we need to introdue the onept of liner independene nd liner dependene.

Liner Independene Definition of liner dependene nd liner independene: The etors K k in etor spe V re lled linerly dependent if there exist onstnts K k not ll suh tht L k k.

Definition of liner dependene nd liner independene: K k re linerly independent if L k k L k The proedure to determine if k K re linerly dependent or linerly independent:

Definition of liner dependene nd liner independene:.form eqution L k k whih led to homogeneous system..if the homogeneous system hs only the triil solution then the gien etors re linerly independent; if it hs nontriil solution then the etors re linerly dependent.

Exmple: { } nd e e e S e e e. Are e e nd e linerly independent?

Solution: e e e. Therefore e e nd e re linerly independent.

Exmple:. 6 8. Are nd linerly independent?

Solution: 6 8 6 8 R t t 4. Therefore nd re linerly dependent.

Exmple: Determine whether the following set of etors in the etor spe onsisting of ll mtries is linerly independent or linerly dependent. { } S.

Solution:. Thus.

Solution: The homogeneous system is. The ssoited homogeneous system hs only the triil solution. Therefore nd re linerly independent.

Exmple: Determine whether the following set of etors in the etor spe onsisting of ll polynomils of degree n is linerly independent or linerly dependent. { } { x x x x x } S x.

Solution: ( ( ( x x x x x x. Thus.

Solution: The ssoited homogeneous system is. The homogeneous system hs infinite number of solutions. R t t Therefore nd re linerly dependent sine R t t t t. Exmple 7 (Pge 95 96

Exmple7 p95 nd. the etors re liner independent.

Exmple8 p95 Are the etors ( ( ( independent? The only solutions. the etors re liner independent.

Exmple9 p95 Are the etors ( ( ( 4 ( independent? 4 The only solutions. 4 or. 4 so the etors re liner dependent.

Exmple p96 The etor e nd e in R defined in EX4 re linerly independent sine only if ( (( A We form the mtrix A whose olumns re the gien n etors. Then the gien etors re linerly independent if nd only if det(a.

Exmple p96 Consider the etors p (tt t p (tt t p (tt t. Liner dependent

Exmple p96 If k re k etors in ndy etor spe nd i is the zero etor. Then S{ k } is linerly dependent.

Note: In the exmples with 8 6 or with { } { x x x x x } S x re linerly dependent. Obsere tht exmples re liner ombintions of nd in both

Note: 4 4 6 8 nd ( ( x x x x x x. As mtter of ft we he the following generl result.

Theorem 6.4: (Importnt result The nonzero etors K k in etor spe V re linerly dependent if nd only if one of the etors j j is liner ombintion of the K. preeding etors j

Note: Eery set of etors ontining the zero etor is linerly dependent. Tht is in ny etor spe nd K k re linerly dependent. Exmple on pge 98 Exmple 4 on pge 99 i K k re k etors is the zero etor then

Exmple p98 If 4 re s in Ex9then we find tht 4 So 4 re liner dependent. We then he 4

Exmple4 p99 4 4 S. Sine spn let w We onlude thr Wspn S where S { }

6.4. Bsis nd Dimension The etors K k in etor spe V re sid to form bsis of V if ( K k spn V (i.e. spn ( K k V. (b K k re linerly independent. Exmple (Pge Nturl bsis or stndrd bsis

Exmple p The etor e ( nd e ( from bsis for R.Eh of these set of etors is lled the nturl bsis or stndrd bsis.

Exmple: { } nd e e e S e e e. Are e e nd e bsis in R?

Solution: e e nd e form bsis in R sine ( spn ( S spn ( e e e R (see the exmple in the preious setion. (b e e nd e re linerly independent (lso see the exmple in the preious setion.

Exmple: 4. Are nd bsis in R? [solution:] nd re not bsis of R sine nd linerly dependent 4. Note tht spn ( R. re

Exmple:. 6 8. Are nd bsis in R?

Solution: nd re not bsis in R sine nd re linerly dependent 4 4 6 8. Exmple (Pge Exmple (Pge 4 Exmple 4(Pge 4

Exmple p S{ 4 } where ( ( ( 4 ( Show tht S is bsis for R 4

prt 4 4 The only solutions 4 Showing tht S is liner indepeddent

prt Let (bd be ny etor in R 4. We now seek onstnts k k k k 4 suh tht k k k k 4 4 Substituting for 4 nd we find solution for k k k k 4 to the resulting liner system for ny bd. Hene S spns R 4 nd is sis for R 4.

Exmple p4 prt To show S{t t t} is bsis for the etor spe P. t bt t ( t( b b b 4 gien t 6 t 5/ 7/4 (t (t (t

Exmple p4 prt Then t ( t( The only solutions S is liner indepeddent

Exmple4 p4 Find bsis for the subspe V of P onsisting of ll etor of the form t bt where b. t bt b (t b(t So the t nd t spn V. (t b(t Sine this eqution id to hold ll lues of t we must he.

Exmple: Let { } nd S. Are S bsis in R?

Solution: ( ( R S spn For ny etor R b there exist rel numbers suh tht b.

Solution: we need to sole for the liner system b. The solution is 4 b b b.

Solution: Thus 4 b b b. Tht is eery etor in R n be liner ombintion of nd ( R S spn.

Solution: (b Sine re linerly independent. By ( nd (b re bsis of R.

Theorem 6.5:(Importnt result If S { } K k is bsis for etor spe V then eery etor in V n be written in n unique(one nd only one wy s liner ombintion of the etors in S.

Exmple: { } nd e e e S e e e. S is bsis of R. Then for ny etor b e be e b b is uniquely determined.

Theorem 6.6:(Importnt result Let S { } K k be set of nonzero etors in etor spe V nd let W spn { } K k some subset of S is bsis of W.. Then

Exmple: Let { } e e e S nd ( R S spn. Plese find subsets of S whih form bsis of R.

Solution: We first hek if e nd e re linerly independent. Sine they re linerly independent we ontinue to hek if e e nd re linerly independent. Sine e e

Solution: we delete from S nd form new set S S { e e e }. Then we ontinue to hek if e e nd e re linerly independent. They re linerly independent. Thus we finlly hek if e e e nd re linerly independent. Sine e e e we delete from S nd form new set Therefore S { e e e } S { e e e } S. is the subset of S whih form bsis of form bsis of R.

How to find bsis (subset of S of W? There re two methods: Method : The proedure bsed on the proof of the boe importnt result. Method : Step : Form eqution L k k.

How to find bsis (subset of S of W? Step : Construt the ugmented mtrix ssoited with the eqution in step nd trnsform this ugmented mtrix to the redued row ehelon form. Step : The etors orresponding to the olumns ontining the leding s form bsis. For exmple if k 6 nd the redued row ehelon mtrix is

How to find bsis (subset of S of W? M M M M M M M then the st the nd nd the 4 th olumns ontin leding nd thus 4 re bsis of { } 6 spn W K.

Exmple: Let { } e e e S nd ( R S spn. Plese find subsets of S whih form bsis of R.

Method: Solution: We first hek if e nd e re linerly independent. Sine they re linerly independent we ontinue to hek if e e nd re linerly independent. Sine e e

Solution: we delete from S nd form new set S { e e e } S. Then we ontinue to hek if e e nd e re linerly independent. They re linerly independent. Thus we finlly hek if e e e nd re linerly independent. Sine e e e

Solution: we delete from S nd form new set S S { e e e } S. Therefore { e e e } is the subset of S whih form bsis of form bsis of R.

Solution: Method : Step : The eqution is 5 4.

Solution: Step : The ugmented mtrix nd its redued row ehelon mtrix is The st the nd nd 4 th olumns ontin the leding s. Thus { e e } e forms bsis. Exmple 5(Pge 9.

Exmple5 p9 S{ 4 5 } ( ( 4 ( 4 ( 96 5 (97 6. Find subset of S tht is bsis for Wspn S. Step: ( ( 4 ( 4 ( 96 5 (97 6

Step: 5

Step: The leding s pper in olumns nd so { }is bsis for W spn S.

Theorem 6.7: (Importnt result Let S { } K n be bsis for etor spe V nd let T { w w w K r } is liner independent set of etors in V. Then r n. 獨立向量集的向量個數必定小於或等於基底的向量個數. 多於基底向量個數的向量集合必定不為獨立向量集

Corollry 6.: Let S { } K n nd T { w w w } bses for etor spe V. Then K m be two n m.

Note: For etor spe V there re infinite bses. But the number of etors in two different bses re the sme.

Exmple: For the etor spe R { } S is bsis for R ( 見投影片 9 94.

Exmple: preious exmple. Also { } e e e T e e e is bsis for R. There re etors in both S nd T.

Definition of dimension: The dimension of etor spe V is the number of etors in bsis for V. We often write dim(v for the dimension of V. dim({}.

Exmple: { } e e e T e e e is bsis for R. The dimension of R is. Exmple 6 8(Pge

Exmple6 p The dimension of R is ; The dimension of R is ; The dimension of R n is n.

Exmple7 p The dimension of P is ; The dimension of P is 4; The dimension of P n is n;

Exmple8 p The subspe w of R 4 onsidered in Ex5 hs dimension.

Theorem 6.8: (Importnt result If S is linerly independent set of etors in finite dimensionl etor spe V then there is bsis T for V whih ontins S. Exmple 9(Pge

Exmple9 p Find the bsis for R 4 tht ontins the etor ( nd (. Let {e e e e 4 } be the nturl bsis for R 4 where e ( e ( e ( e 4 ( e 4 e 5 e 6 e 4

Exmple9 p Sine leding s pper in olumns nd 6 we onlude tht { e.e 4 } is bsis for R 4 ontining nd

Theorem 6.9: (Importnt result Let V be n n dimensionl etor spe nd let { } S K n be set of n etors in V. ( If S is linerly independent then S is bsis for V. (b If S spns V then S is bsis for V.

Exmple: Is { } S bsis for R?

Note: 定理總結獨立向量集的向量個數必定小於或等於基底的向量個數. 多於基底向量個數的向量集合必定不為獨立向量集生成且獨立之向量集基底形成一組的少於基底向量個數的向量集合必定不為生成向量集生成向量集的向量個數必定大於或等於基底的向量個數.

Sine Solution: R is dimensionl etor spe not like in the preious exmple we only need to exmine whether S is linerly independent or S spns R. We don t need to exmine S being both linerly independent nd spns V.

Exmple: Is { } 4 S bsis for R?

Solution: Sine R is dimensionl etor spe we only need to exmine whether S is linerly independent or S spns Beuse 4 re linerly independent. Therefore bsis of R. Exmple (Pge R. re

Exmple p In Ex5. W spn S is subspe of R 4 so dim W 4. Sine S ontins fie etors we onlude by Corollry 6. tht S is not bsis for W. In Ex sine dim R 4 4 nd the set S ontins four etors it is possible for S to be bsis for R 4.If S is linerly independent or spns R 4 it is bsis otherwise it is not bsis. Thus we need only hek one of the onditions in Thm 9.6 not both.