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 .