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
