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Linear Algebra I Gefterjmir2021 Fall 2021

E Spank veEf ve t vs f V 2k vs 0 v.v v3 are

linearlydependent

i h v.v are linearly independent

spank in v3 span v.v V

9 4m

22 K 03 da

Proof I let vi Ne and war wie be bases of V

U re eV are ein indepandV spanfwn.intdem

emWir WmEU are lin indep and V Spanky ve Mel

i If wir um are lin indep we are doneand wir was is a basis Ew dt Wm

If we Wm are hin dependent then by

Thm 9.3 there exists aj h.im with

V span w im spanSw Wie Wie Wm

Now repeat withhm wie wie Wm

Iii Assume hin re eV are lin indep

If V spank v3 we are done and

vi vel is a basis of V

If V span un ve then we can

find ne V with u spanky viSet an U By Lemma 9 Tun venee

are linearly independent Rspansee

Repeat until V spank Nein ME

I If V 03 then d I is abasis

span 3 spang 03

If V GOT then there exist well withFO V is linearly independent

By iii we can construct a basis of VA

This makes sense because of Tha 10.2 1

Example 21 it dim R _n because Sei enis a basis of R

i If V span Y k V then dink 2 because

Sy v22 is a basis of V vi f v f vs

Proof i ii If u um are hin indep but

V spank um then Tha 10.2there would exist a basis of Uwith more than m elements

This can not be the case becausedime m V Spank um

i i If V spanEu _um but k um are

ein dependent then by Thm9.3 we

can find a basis with less thanm elements But dim r m

and therefore un Um are hin indep

it and i i by definition A

Example 22 Determine bases for KerlF and imof the following linearmap

F IR RE

x KKerne KF EXER I FCH03

FIX O

c 9 41 48 64000011

rief FSolutions to F f 0 are

Zt tzt

tut ERX3 tz

Xy tz

X t G tz f Kerftspannis

Y V2

Are u u lin indep

If t v Eu O t GO

EE have are ein indep

und is a basis

of Ker F

ImageF

im F span d E Ix GEU Uz Us My

Xu Kantons ka o F f O

E e KerlSaw before

f t E taff fortatien

tiltro f Y Zuturo Ur zu

Ustadt 0,41 U 43 44 0 UE 4 43

gespanfußim F span u U span 4,433

Area us lin indep

If 4,4 4343 0

Mail Roll f t.noUnd lin

indep

General calculation of a basis for KerlF andimIFLet F R R be a linearmap

x AX AERLet rief A have pivotelements in columns a er

Then the columns at a at form a basis

of im F dim im V VK A

piratelements A

The vectors obtained in the standartparametrisation

ie for each free variable we have a

parameter ti and a vector Vi

of the solutions of F Xo forma basis

of Kerl F

dimlkerl n r n dim im

free variablesRTRMAX Note

A Ed riefen4 Un U Ut UT

im F Span u uXE KerlF KEE 24

3 3

YI YE.EE

I

tl t f t.LYV V2 V3

Ker F spanEu iv v33

Fatsof KerlF

Proof D columns of IFTdim Kerl columns without pivotelements

dir im F columns with pivot elements

R Ä R