Lecture Ce :

• Vectors aren't always " vectors"

• Linear combinations

Idea is a sum of rescaled vectors

F =

"there exist

"

Def : Let X = {x , , Kz , - . .

,an } be a set

of vectors in a v. s . I . (X CI)we call we V a linear combination

- -"scalars

"

of X if and only if I 2 , .dz .. - . ,2n

• 7 . ( such that)C- IF

UV = L,7C

,t 22 >Cz t - - - t Ankh

-

n

= Eau .cn C"2%riAb- = . (Vit , t ,

x

TC , Xz

Ex : X={ ( i. o, i ) ,( o ,

I,l)}

,V -- 1123

⑥ IF = 112

2. ( 1,0 , I ) t 3 ( o , I , I)-- -

-

L.

"I Lz "

2

2 46= ( 2 , 3,5 )

LG : Q1 pw : tin- com

• Span

Def : Given a v.s. V'

g a set (family) ofvectors XCV , X = {x , ,

- - - , an } , we

denote by Ex > or span{x} the"

span"

of X,which is the set of

all linear combinations of X.

Intuition : 422

¥•-avg.us

- ZV,t Uz

L,

= - 2, Lz = I

in 1123

¥÷÷¥ae• Notation

,if there are two sets X

,Y ,

they use ex , Y S to mean the

span of X u Y = all elements ofX and all elements of Y .

Thm: Let V be a v.s.,XCV a family of

vectors . Then Ex> CV ,and

ex > is a subspace .scalar mutts

n f w/ IF

Pf : Let x. E ex> ,then a = ¥

,

dunk

[ vector t or ⑦

where each are EX. Since V is closed

under t and x,x EV

⇒ ex> cv . F

Recall that if a subset of a v. s .is closed under t Ix ,

then that

subset is a subspace .

⑦ : let x, y C- ex >

,then we can

write

K = ⇐ Lukic , y -

- € Puke

xty = ( Iancu) + ( Ethan)= EGuxutpu.ae) ①

= E (dat Bu.) xx ⑤-

n.

= { Fescue ⇒ xty c- ex >

① : let a be as above, BE IF

Bx =p ( Edu. xx )

= ? BH.eu) ⑤

=ECpcu ⑦2k

= E Kock ⇒ pace ex>

Bar

26Q2 pw : span

• Linear Independence I dependenceMain idea : If given some vectorsif any of those vectors are a

linear combination of the others,

then He set is linearly dependent .

Ex : X = { ( o ,I

,o)

, ( o ,o

,I)

,( o ,I, l) }

X is ↳D . since

( o,I , I ) = ( o

, I ,o ) t ( o ,

O,I )

span X = E X ) = span { (o , 1,07 .( o

,o, 1B

implicitly a v.s .

goXcv

Def : Given a family X =3 K , .. - -

,an 3 of

Vectors,we say X is ↳ Do if

F scalars 2, . . . .

,Ln E IF a 7 •

n

d,se

,t - - - t 2n In = E dyKk

b- = i

= O

a⇐d at least one Li ¥0 .

X is L . I . if X is not L- D.

Anotherway to think of Lal .

X is L. I . ifn

E dy kn = O

K-- i

only if 2,= 22 -- - - - = Ln =

{ ( 1,0 co) ,Costco)

,( O co ,

I ) , ( 0,1 ,1) 3

z 3I ee e

Ex : { (1,0 ,o )

,( o , I co) , ( 0,0 ,

I)} = E

I claim ESpz3 ,and E is

L-l . Las FE> = 1123

linear combo ,

a.b. c C- 112

ae't be"

t ee's

= ( a ,b

,c )

⇒ X E > = 1123

( a,b. c)

'- O = ( o ,

O,o )

⇒ a -- o -- b = c ⇒ L. I .

Ex : X -- { ( I

, ooo ) , ( o , I ,o ) , ( I , 1,0) }7C,

Kz 7cg

Claim : X is L. D . (not c. la) , do not

span 1123

a ,b, c

C- 112

ax , tb>Cz t CK, = (ate ,

b. t C,O)

Doesn't span ,since Coco ,

i ) C- Ex>

=O = ( O ,O co)

at C = o ⇒ a = - C

b t c = O b - a = O

o -

- o b = a

choose a = I,b = I ,

C = - l

a 7C,t b>Cz t CKz

= 0

Got a ,b ,c to ⇒ L

. D.

Ex : n

::" "= ( o

,o )

b = a EIR

⇒ a = - ab = - a T

⇒ a -- o

⇒ b -- O-

SO L. le

Ki l"Z '"3

gassume a

, bee to

Caz = ax ,t baz

{⇒ ax, t b>Cz - C>Cz = O

• Brute Force Method :

Given so, cuz . - -

, 7cm E IR"

E Mn, , ( IR)

A If only trivial so) (2=0 )then Lal .

A = ( K ,Kz -

- - km ) E Mn +mate)

t÷÷" ÷.

"

: )L Emma , ( "2) = ( &'m)

c " c :o){ ( i , o ,

o ),( o , yo ) ,

( i , I ,o ) }

A- = ( too % !) Aa -- o

x -

- A-'

o

= 0

I

7Yton - trivial( : ' o 'o ! :) Elution( 1,1 , - D

-

RREF 7C = - Z choose

y =- Z 2C EEL

⇒ x -- y

Trivial solution = (o 10,0 )

X :{ ( I , Ico ) , C- 1,0 , I ) ,( 2 , I ,

- l ) }

c :÷:ii:LE.⇒ ( iii. ÷ )"" '

c : :'

mi-m-r.ci.

-

i. ÷ )-

Not Lala not equiv nto I

TC,I ( K , , , Kzn , 131 )

:

"z= ( Riz , Kazi €33 )

ax , t b >Cz tc×3

= 0

ax, ,t b >c. z t CK , ,

= o

Cruz , t b >czz t CK 23= 0

A 7cg , t buzz '- Caz> I

("

ik :) .

- C :o)40-63 pw : Ind

Def : A basis of a vector space is aC- le spanning , set.So X is a basis of V if and

only if① X is c. i .

② Ex> = ✓ ← spanning set

X a spanning setif 1×7 -

-V

-

END OF CONTENT FORMT

Quiz 1 :7C, Xz

( I ⑨ c ) ( o ⑧ i )

any linear combo

will have2nd component

zero

( a o - G )

= Goc,- 1227oz

Quiz 2 : span 1122 ✓ZC ! .p.

( 1,0 ) j (2 ,o ) ←

Tia THIEF( 1,1 ) I C - 1,4 )

.

-

C- I , - 1) = - ( 1,1 )

¥¥" ""

③To

¢ all , 1) t BC -1,1 )

(a - b ,atb ) = 0,0

a - b -

- c

aa ? .by ⇒ b= - batb =D

⇒ b=o

C,

' -

ill ; ) -

- C 'd ) ⇒ a -- o

60 Lola

nc : :'

; :)au.on.ca?

Lol. =

Quiz 3 : f Dd f- l

Cl ) ( Ico ) I ( I,- l ) H

CZ ) ( 1,0 ,c ) ,

( oil , i ) I ( I , - 1,0)-

( I, -1,0 ) = ( 1,0 ,

I ) - ( oil ,))

SO L- Do

c : :"i'T ! ! !:) . . - z

g y - Z

- f ! !!) ⇒ c. D .