CHAPTER 4 : ORTHONOMALITY :
DOT PRODUCTS AND ORTHONORMAL BASES :
0
DEALT WITH VECTORS IN ☐2h 0 VECTOR WITH N COMPONENTS AND THEIR
COMPONENTS ARE REAL NUMBERS .
TO FIND THE DOT PRODUCT
IN 12h : : MULTIPLY RESPECTIVE COMPONENTS
GIVEN TWO VECTORS AND ADD THE PRODUCTS
0 HOW DO WE PERFORM THE DOT PRODUCT IF GIVEN VECTORS WITH COMPONENTS
IN THE SET OF COMPLEX NUMBERS ?
" ^
% EXTEND THE DOT PRODUCT FROM VECTORS IN DZ TO VECTORS ②
.
DEFINITION : THERE ARE TWO VECTORS A AND b
• "
¥wo•÷÷i¥: Let (a b ( b , ; bz ; bn)
a
; Az ; ; an ;
= =
, . . . .
; . . .
h COMPONENTS OF EACH VECTOR
A b E ② (
0
WHERE
j ARE COMPLEX NUMBERS
THEN THE DOT PRODUCT OF VECTORS A AND b IS DENOTED
BY THE USUAL NOTATION A • b AND DEFINED BY :
J¥¥Ñ"
IN SUMMATION
v
V V L
TO COMPUTE DOT
CONJUGATION
PRODUCT
BETW . 2 VECTORS THAT INTRODUCE
LIE ☒ CONJUGATE N
!
IN
②
• i.
THE COMPONENTS IN THE SECOND SINCE
VECTOR !
EXAMPLE : LET VECTOR a =
(1; i ;3 j b = ( i ;it1 ; 1
COMPUTE THE DOT PRODUCT .
i. aob =
( I ;i ; 3) ( i ;i+1;1 •
* WHEN CONJUGATING A complex
= (1) ( j ) + ( i)( it 1) + (3) (1) NUMBER
OF THE
JUST CHANGE
IMAGINARY PART !
THE SIGN
=
(1) C-i) till i ) + -
3
= i ti- IZ + 3 - i2= -1
= 4
,
NB : IF A ; b E ☐2h :O EACH COMPONENT OF VECTORS 9 and b ARE REAL
DEFINITION DOT PRODUCT
con "
"YÉ☐n NUMBERS OF
'
. .
. .
REDUCES TO THE USUAL ONE KNOW .
h
.
'
.
a •
b = aibi BUT since bi =
bi
i=1
n
= aibi
i=1
D
, IF
°
NOTE : TAKE DOT PRODUCT OF VECTOR A BY ITSELF :
n
ao a =
aiai
[ =L COMPLEX NUMBER
MULTIPLIED BY ITS
D- CONJUGATE
2
BUT KNOW FROM COMPLEX NUMBERS : 2 =
Z
n
2
÷ a o a =
, ai
i=1
I.
COMPONENT OF
A VECTOR !
% CAN NOW DEFINE THE
LENGTHOF THE VECTOR
!
^
DEFINITION : THE LENGTH OF A VECTOR A WHERE A E ② IS
GIVEN BY
• + VE SQUARE ROOT
OF DOT PRODUCT
=
a aoaa
EXAMPLE : VECTOR a = (i; it 1 ; -1
FIND THE LENGTH OF VECTOR A .
1 A / =
A- •
a ( BY DEFINITION
( i ;i+1 ; 1) ( i ;i+1 ; -1) SHORTCUT :
2 a. a
•
= -
n 2
( i)( T) + (it 1) ( its )t C- 1)C- a. a =
=
ai
i=1
= it ( it 1) ( 1- i) + 1
-
n =3 :
=
It i i 2+1 i +1 -
-
=
4
=
il 't it , 2+112
, ANY COMPLEX
NUMBER CAN BE
WRITTEN IN THIS FORM
:
ÉÉEAÉNEN
2
SQUARED !
✗
tiy A MODULUS :
(Re)2t(Im)Z
LENGTH OF
g. -12 xzty !
: a 1 (2) Zt I
=
=
z a = 4 = +
p
=
4
,
NB PROPERTY SUPPOSE A MEANS THAT IT 'S
'
: : = 0 . .
LENGTH IS ZERO .
IF GIVEN VECTOR SUCH THAT ITS LENGTH IS ZERO THEN SURELY THE
,
VECTOR IS ZERO :
LENGTH :
To A = 0 IF AND ONLY IF THE VECTOR ITSELF A = 0 .
( WORKS BOTH WAYS !)
DOT PRODUCTS AND ORTHONORMAL BASES :
0
DEALT WITH VECTORS IN ☐2h 0 VECTOR WITH N COMPONENTS AND THEIR
COMPONENTS ARE REAL NUMBERS .
TO FIND THE DOT PRODUCT
IN 12h : : MULTIPLY RESPECTIVE COMPONENTS
GIVEN TWO VECTORS AND ADD THE PRODUCTS
0 HOW DO WE PERFORM THE DOT PRODUCT IF GIVEN VECTORS WITH COMPONENTS
IN THE SET OF COMPLEX NUMBERS ?
" ^
% EXTEND THE DOT PRODUCT FROM VECTORS IN DZ TO VECTORS ②
.
DEFINITION : THERE ARE TWO VECTORS A AND b
• "
¥wo•÷÷i¥: Let (a b ( b , ; bz ; bn)
a
; Az ; ; an ;
= =
, . . . .
; . . .
h COMPONENTS OF EACH VECTOR
A b E ② (
0
WHERE
j ARE COMPLEX NUMBERS
THEN THE DOT PRODUCT OF VECTORS A AND b IS DENOTED
BY THE USUAL NOTATION A • b AND DEFINED BY :
J¥¥Ñ"
IN SUMMATION
v
V V L
TO COMPUTE DOT
CONJUGATION
PRODUCT
BETW . 2 VECTORS THAT INTRODUCE
LIE ☒ CONJUGATE N
!
IN
②
• i.
THE COMPONENTS IN THE SECOND SINCE
VECTOR !
EXAMPLE : LET VECTOR a =
(1; i ;3 j b = ( i ;it1 ; 1
COMPUTE THE DOT PRODUCT .
i. aob =
( I ;i ; 3) ( i ;i+1;1 •
* WHEN CONJUGATING A complex
= (1) ( j ) + ( i)( it 1) + (3) (1) NUMBER
OF THE
JUST CHANGE
IMAGINARY PART !
THE SIGN
=
(1) C-i) till i ) + -
3
= i ti- IZ + 3 - i2= -1
= 4
,
NB : IF A ; b E ☐2h :O EACH COMPONENT OF VECTORS 9 and b ARE REAL
DEFINITION DOT PRODUCT
con "
"YÉ☐n NUMBERS OF
'
. .
. .
REDUCES TO THE USUAL ONE KNOW .
h
.
'
.
a •
b = aibi BUT since bi =
bi
i=1
n
= aibi
i=1
D
, IF
°
NOTE : TAKE DOT PRODUCT OF VECTOR A BY ITSELF :
n
ao a =
aiai
[ =L COMPLEX NUMBER
MULTIPLIED BY ITS
D- CONJUGATE
2
BUT KNOW FROM COMPLEX NUMBERS : 2 =
Z
n
2
÷ a o a =
, ai
i=1
I.
COMPONENT OF
A VECTOR !
% CAN NOW DEFINE THE
LENGTHOF THE VECTOR
!
^
DEFINITION : THE LENGTH OF A VECTOR A WHERE A E ② IS
GIVEN BY
• + VE SQUARE ROOT
OF DOT PRODUCT
=
a aoaa
EXAMPLE : VECTOR a = (i; it 1 ; -1
FIND THE LENGTH OF VECTOR A .
1 A / =
A- •
a ( BY DEFINITION
( i ;i+1 ; 1) ( i ;i+1 ; -1) SHORTCUT :
2 a. a
•
= -
n 2
( i)( T) + (it 1) ( its )t C- 1)C- a. a =
=
ai
i=1
= it ( it 1) ( 1- i) + 1
-
n =3 :
=
It i i 2+1 i +1 -
-
=
4
=
il 't it , 2+112
, ANY COMPLEX
NUMBER CAN BE
WRITTEN IN THIS FORM
:
ÉÉEAÉNEN
2
SQUARED !
✗
tiy A MODULUS :
(Re)2t(Im)Z
LENGTH OF
g. -12 xzty !
: a 1 (2) Zt I
=
=
z a = 4 = +
p
=
4
,
NB PROPERTY SUPPOSE A MEANS THAT IT 'S
'
: : = 0 . .
LENGTH IS ZERO .
IF GIVEN VECTOR SUCH THAT ITS LENGTH IS ZERO THEN SURELY THE
,
VECTOR IS ZERO :
LENGTH :
To A = 0 IF AND ONLY IF THE VECTOR ITSELF A = 0 .
( WORKS BOTH WAYS !)
