Linear Algebra I Fall 2020

9 Linear independence

In Example 16 we considered the linear map

6 Rt IR

l 9 x

and calculated its image

im 6 IR span 9

But we also learned that

im 6 span of columns of 6

span911,19 il B

This gives

spansel 1913 span Hd He

We just need Too muchvectors Can2 vectors How can we remove some

weshowthis

Lemma 9.1 Let v ve ER tfveespanEY.n.vethen

span94 ve span 4 ve

T W

Proof clearly we have Wc V Want to showKW

If V E span v ve V then there exist HEIRwith

Xiv t.nl eVOe H

since Vee span94 ve there also existX hey ER with

d Yt t deVey I

Combining HI and gives

X Yt Xe.ie the 44 t.ntde.it

the Yt leitete vet

and therefore VE span v Ve i e KWÜ A

Example 18

For the linear map G in Example 16 we get

im G span t 14,191,43 spannteB

Lemma 9 I since G H K

span t fIH span 1B

Lemma 9 I IH 19

span I'd HI 3

Lemma 9.1 H d 9span I'd 1913 Ri

General question When is it possible to remove elements

from span94 ve without changingit

Definition 9.2 Vectors 4 ve C Rmare called linearly independe

if the equation

X V t t here 0 t HERjust has the nice solution X L te 0

Otherwise v ve are called linearly dependent

Example 19 Are the vectors

HK uff ufflinearly independent

The equation X v Kurts 0 is equivalent to

Iii HHHH.tl L

Ei iL

L

Solutions X 2TX 3T 4,4 V are

XE t linearly dependent

For El we get X 2 tz 3 73 1

24 13kt 2 3 t f 0

24 3h GE span 4,43Lemma9.1

span viii vs span 4,43

But V.v are linearly independent since

KUHNE 0 ma f

µX XE 0

Theorem 9.3 let v ve ER Thefollowingstatementsare equivalent

i V ve are linearly dependent

ii There exists a j l such that v is alinear combination of the other vectors

iii There exists a H 1 with

span vi vii ve Span 4 ve

Proof ii iii is Lemma 9 I

iii ii span94 ve spank n.ve

v E span9K Ve

ii i Suppose Vj Iv tut vi thinkinThem

0 1 v kj.it vjttjtiftittevl.nuXiv withXE I

h ve are linearly dependent

i ii Suppose Iv tut here 0 with dito

Thenvi f v t.it ht Hütte

Vj E Span Vi Vitti Vl