4 Basis

Definition A subset peu is a basis ifit is i a spanningset

ii linearly independent

Ex in i futV vespant

7 Vi UnE B and Cc icnEF.SIV C V 1 Chun

Ii If for Vi i Unef Cc i Cuffthen Cie Cn O

TEAMS Il Fh e _11,0 so a en lo ion

e en is a standard basis for Fn

Check i A V Iv vn Eff we have

V Vie t f Un en C Spanfee eu

il If Cien ten eu Cci cn o then

Ce Cn 0

1 Mmm th Eis if o

Eiji Kien lejen is the standard basis

PnlF I 1 x inn is the standardbasis

141 Ifg Yo to is not a basis

because it s not linear independent

151 l too is not a basis

because it's not a spanning set

Need to be enough and not redundant

Theorem algorithm Suppose SEV is a

finite spanning set Then I basis PES

Proof i 5 0 or 303 V 303 f 4Ii Sff Us S lui un

Go through the elements onebyoneon lecture notes

Examples

l S 2 3,5 Cf 12,20 Clio 27

0,2 1 C7,40 f 1123

Ui C2 3,5 p luiUseCJ 12,207 4 Ui skipwya 4 o 2 d spank p Lui.usuy 0,2 1 spanking f Lununuceuse 7.2.07 C spanlununky skip

p Lunar my is a basis

2 Xel X x4xtl Et 2x V IBAR

Ur XH p luiUri X f Lui.nuU3 x4xH Cspan lunar skip

UyexHx I spank nu

i f lui.ua Uu is a basis

theorem Replacement theorem

Suppose a is a spanning set of vector

Space V L is a linearly independentset sit G m and ten then

N EM

7 HE G s't H m n and

spankUH V

G TITHEL 1 2

Proof By induction Con n

i n o K fo take H G

Ii Suppose that is true for some n zowe prove it's true for n 4

Suppose L Lu mm G Lui Um

L is linearly independent 3

Vi in is linearly independent

By induction hypothesis Rsm and

7 HE L H m n sit LUHgenerates

Suppose H Lui mm n then

Unt E span Vi Un Ui Um a

7 scalars Ai An bi bm n such that

Unt A Vit Aunt b Uit bma Umm

If n m then Vue a v t aavnespaylui.vn

contradicts with L linearlyindependencein m nZmt1

and bi burn cannot all be O's for the

same reason suppose b toa Ur fbiadv.tl bi agua Cbi'an Un

bi't Urry t C bi bullet C bibn.in Um

nGSfpanIV1iVny U2 Um n

H Luz um n and then LUH generates

vector space V The

Exampled V L lx.y.tlElks ritzy132 03

find a basis for vector space V

Wion VF I l 17 VE l l 4 3

then f Wi un is

a linearly independenta spanning set of V

c n p is a basis for V

We don't need to first find a

spanning set and then conductthe

algorithm We can just find theexact number of vectors and show theysatisfy the conditions