MAM2084F
Linear Algebra
, eme
Chapter One
eme
Vector Spaces
What is a vector ? - An element of R
that follows certain rules
* We call the set V =
EU
,V ,
w ... 3 a
Vector space and elements its vectors
.
·
V is closed under vector addition
ie
. If 4 ,V EV then
(u + 1) EV
V closed under multiplication
·
is scalar
If
i
.e
. EV and a is a scalar
then (av) ev
°
Show R is vector space
eg
. a
4923c2(
R= , 3)/6i E e
((
Let U =
and v =
(ii (
: + Vi
u v =
, eme
Vector Spaces
numbers
Using real :
Let u
=
(())mav (e) =
(3)
: n + v =
just
This
:
results in another Vector
in Since U + V is just a
real number
For scalar multiplication :
u
(iie and ceEIR than
=
(ie
qu
=
Which is since is
again
in V qui
a real number
, eme
Vector Spaces
Using real numbers :
Let
(b ( and a 3
=
u =
ou
(is (
=
eg
.
Show that S =
&P(x) :
P(x) is
a
polynomial of
degree 23
is NOT a vector
x + x Es and -x + xES
2
↑
not polynomial deg
(x (-x + x)
"
x) + =
2x : S
: S is not a vector space
=> not closed under addition
, eme
Vector Spaces
eg .
p =
4(j)/x0 and
y4, 03
You can show P is closed under
Vector addition
(2) Ep a = -
3
-
3(2) =
(3) ... P
..
P is not closed under scalar multiplication
Examples of Vector Spaces
· RU
·
R
*
=
Ef : R +
12/-
> is function 3
-
V =
Eax" + bx + c : a
,
b
,
c
, EIR3
You
eg
.
can
buy bag I with I
apple
2 3 bananas
bannanas , bag ↓ 2
apples
Can 17 29 bannanas
get apples
(i) =
> +
Su
= (2) + s(5)
=
(2)
, reme
Linear Combination
If V is a vector
space and V , V2 ... Un
are vectors in V then a LINEAR
COMBINATION of these vectors is
.
av ,
=
9 V ,
+ R2V ... a n
eg
. Give possible linear combinations
(2) ER
s =
Ea(i) :
delR3
)eo) e
ER
=
:
=
-.
-
S < &R
..
This is a
straight line in 1R
through -
(8)
=
origin
, reme
Linear Combination
Give possible linear combinations
eg
.
(i) (i) E -
s =
Ea(i) B(i) +
:
a
; Be1R3
xY
# Does & of Y
& B of Uz
-
↑ =
>
-
: S =
12" the whole ID plane
eg .
(i) =
4(i) + 3(i)
, reme
Linear Combination
Give all possible linear combinations
eg
.
6(8) , (! ) EiR
= 2 Scalars
2 vectors
w =
Ea(i) + B(i) ; a
,
Be13
=
(aBB) :
W >- R
& now ID plane
3
W is a XY PLANE in IR that
passes through origine
↓ if take I nonlinear vectors
generates plane
eg
.
Take linear
LINE
comb
(i) (3) ,
you
get a in R3
, reme
Linear Combination
+
Give all linear camb of f(x) =
x xc + 1
x EIRR
g(x) x
=
;
-
* IRP is set all functions IRPR
Let c, < be scalars and
,
any
n(x) f(x) azg(x)
=
0 ,
+
=
a , (x + x + 1) +
a(x -
x)
* h(x) is form ax + bx + C
=
x ,
x +
4x + x ,
+ 0x -
ex
=
ax + bx + C
We want this to be solvable for
, CRz
!
a
C
-
=
x(a ,
+
az) + x(a ,
az) +
,
=
ax +
3+ >
, reme
Linear Combination
de Es
I (
z
(
I I a
I b + Rz
& -
R2 -
RI
I O C
R3 +
R3 -
RI
( ..
↑ ↑
2
-0
-
R 4
R3
·
( :I
n
-
I
O
-
Z R3 + R3-2 R2
( -I se
d
N
O O
Linear Algebra
, eme
Chapter One
eme
Vector Spaces
What is a vector ? - An element of R
that follows certain rules
* We call the set V =
EU
,V ,
w ... 3 a
Vector space and elements its vectors
.
·
V is closed under vector addition
ie
. If 4 ,V EV then
(u + 1) EV
V closed under multiplication
·
is scalar
If
i
.e
. EV and a is a scalar
then (av) ev
°
Show R is vector space
eg
. a
4923c2(
R= , 3)/6i E e
((
Let U =
and v =
(ii (
: + Vi
u v =
, eme
Vector Spaces
numbers
Using real :
Let u
=
(())mav (e) =
(3)
: n + v =
just
This
:
results in another Vector
in Since U + V is just a
real number
For scalar multiplication :
u
(iie and ceEIR than
=
(ie
qu
=
Which is since is
again
in V qui
a real number
, eme
Vector Spaces
Using real numbers :
Let
(b ( and a 3
=
u =
ou
(is (
=
eg
.
Show that S =
&P(x) :
P(x) is
a
polynomial of
degree 23
is NOT a vector
x + x Es and -x + xES
2
↑
not polynomial deg
(x (-x + x)
"
x) + =
2x : S
: S is not a vector space
=> not closed under addition
, eme
Vector Spaces
eg .
p =
4(j)/x0 and
y4, 03
You can show P is closed under
Vector addition
(2) Ep a = -
3
-
3(2) =
(3) ... P
..
P is not closed under scalar multiplication
Examples of Vector Spaces
· RU
·
R
*
=
Ef : R +
12/-
> is function 3
-
V =
Eax" + bx + c : a
,
b
,
c
, EIR3
You
eg
.
can
buy bag I with I
apple
2 3 bananas
bannanas , bag ↓ 2
apples
Can 17 29 bannanas
get apples
(i) =
> +
Su
= (2) + s(5)
=
(2)
, reme
Linear Combination
If V is a vector
space and V , V2 ... Un
are vectors in V then a LINEAR
COMBINATION of these vectors is
.
av ,
=
9 V ,
+ R2V ... a n
eg
. Give possible linear combinations
(2) ER
s =
Ea(i) :
delR3
)eo) e
ER
=
:
=
-.
-
S < &R
..
This is a
straight line in 1R
through -
(8)
=
origin
, reme
Linear Combination
Give possible linear combinations
eg
.
(i) (i) E -
s =
Ea(i) B(i) +
:
a
; Be1R3
xY
# Does & of Y
& B of Uz
-
↑ =
>
-
: S =
12" the whole ID plane
eg .
(i) =
4(i) + 3(i)
, reme
Linear Combination
Give all possible linear combinations
eg
.
6(8) , (! ) EiR
= 2 Scalars
2 vectors
w =
Ea(i) + B(i) ; a
,
Be13
=
(aBB) :
W >- R
& now ID plane
3
W is a XY PLANE in IR that
passes through origine
↓ if take I nonlinear vectors
generates plane
eg
.
Take linear
LINE
comb
(i) (3) ,
you
get a in R3
, reme
Linear Combination
+
Give all linear camb of f(x) =
x xc + 1
x EIRR
g(x) x
=
;
-
* IRP is set all functions IRPR
Let c, < be scalars and
,
any
n(x) f(x) azg(x)
=
0 ,
+
=
a , (x + x + 1) +
a(x -
x)
* h(x) is form ax + bx + C
=
x ,
x +
4x + x ,
+ 0x -
ex
=
ax + bx + C
We want this to be solvable for
, CRz
!
a
C
-
=
x(a ,
+
az) + x(a ,
az) +
,
=
ax +
3+ >
, reme
Linear Combination
de Es
I (
z
(
I I a
I b + Rz
& -
R2 -
RI
I O C
R3 +
R3 -
RI
( ..
↑ ↑
2
-0
-
R 4
R3
·
( :I
n
-
I
O
-
Z R3 + R3-2 R2
( -I se
d
N
O O
