CHAPTER 4
ALGEBRA
ORTHO NORMALITY
, Dot products and orthonormal bases
In this section want to investigate the geometrical aspect of vectors
°
we
especially lengths and angles ,
in higher dimensions
°
We
generate.se the dot product to n dimensions and define :
A-
ok =
( a ,
Az
,
.
. .
an ) o ( bi .bz ,
. . .
bn )
BTt BI bi where b- the conjugate of b
=
a + a
. . . t an is
,
if bi
=
ai
,
( -3 ) find
example :
If =
( Iti ,
2-
Ii 3 )
be , and =
ti
,
H2i ,
-
i :
II , b- a-Cda '
,
and b-ok
fo I = ( Hi
,
2 -
is 3) of -3 ti
,
It Li
,
-
i )
=
( it ill -3 -
i ) t
( 2. i ) ( I -
zi ) t (3) Li )
= -
3
-
i -
3 i + I t 2 -
4 i -
i -
2 -13 i
= -
2 - 6 i
b-eat =
I -
3 ti ,
It 2 is -
i ) o
( Hi ,
2 -
i , 3)
=
( -3 ti ) (I -
i ) t It -12 i ) .
12 ti ) t fi ) (3)
*
the of a ok
2+6 i
conjugate
= -
b-ok 16
=
16
Eobe =
,For complex numbers the factor essential
conjugate in the second are
because the dot product of a vector with itself must be the
square of its magnitude
'
.
Is
"
a- a . = Ex Akai =
19kt which is real and non -
negative
, ,
i. e
Ibet =
EE
orthogonal ( ! )
n
If ok 0 Freiman
a then vectors and be
=
a are
If a
and b- are raw vectors then the dot product is equal to a
f bi I
-
matrix product i.e a. be
=
.
a
[ transpose of b conjugate
example :
I
=
I Hi
,
2 ti
,
3 ) and be =
f- 3 ti
,
It 2i ,
-
i )
a-of
= -
2 -
Gi from above
so a b-
a =
[ Hi ,
zti .
3 ] of :{if ? mtaitriglg
multiplication
=
2 Gi
-
-
.
-
It follows from matrix algebra that
I i) Ia +
d) . I =
Eohe + do b-
( ii ) I Ce ) a
be =
C Ceobe )
} 778¥
Iii it be a
=
a y
ate ,
b.
.
For lil and Iii ) show that the dot product is a linear operator
different
For column vectors A and B the dat product is given slightly
.
a
matrix product : A o
B =
ATE
, Root iiis
Required to
prove
that :
q.ie =
Ee
LHS =
la ,
,
as
.
.
.
an ) o
(b ,
,
ba -
- -
bn )
=
AT + asbT +
. . .
+ Anton
=
a- , b t a-zbz t - a .
t AT bn
be a- RHS
=
=
.
F④ ,
Jefinition : An orthonormal basis of linear is basis
p
a space a
g.
consisting of mutually perpendicular unit vectors
example : if t =
GO ,
o
) , j =
I 0,1 ,
o ) and I =
6,0 ,
I ) show that
bi ) forms
ii. j .
an orthonormal basis .
I oj =
( I , 0,0 ) s
( a ,
I
, a) =
0+0+0=0
I °
bi =
( I ,
o ,
a ) o
( a ,
a ,
I ) - a -10+0=0
Jabi =
( oil ,
o ) .
( o ,
o
, 1)
.
-
O -10+0=0
lil =
III =
Ik I =
EE = I
'
I i. j bi ) orthonormal
. .
,
is an basis
Ba
