Math 1b Prac — Orthogonal bases, orthogonal projection
January 29, 2009
[This is an outline of material I want to cover before starting Chapter 4 of the text.
Examples and motivation will be given in class.]
We will write
h
x
,
y
i
for the dot product of vectors
x
,
y
∈
R
n
in this section. This is
one type of
inner product
; see page 307. Properties of inner products include
h
c
a
,
b
i
=
c
h
a
,
b
i
and
h
a
1
+
a
2
,
b
i
=
h
a
1
,
b
i
+
h
a
2
,
b
i
.
Recall that two vectors
x
,
y
are orthogonal when
h
x
,
y
i
=0
. Wh
enw
esaytha
t
v
1
,
v
2
,...,
v
k
are orthogonal, we mean that every pair
v
i
and
v
j
,
i
6
=
j
, are orthogonal.
The
length
of
v
is the square-root of the inner product of
v
with itself:
||
v
||
=
p
h
v
,
v
i
.
This is a positive number (and in particular, not zero) as long as
v
6
=
0
.The
(Euclidean)
distance
between
u
and
w
is the length of
u
−
w
, i.e.
||
u
−
w
||
.
Note that if
a
and
b
are orthogonal,
||
a
+
b
||
2
=
||
a
||
2
+
||
b
||
2
.
We may call this the Phythagorean Theorem. It holds because
||
a
+
b
||
2
=
h
a
+
b
,
a
+
b
i
=
h
a
,
a
i
+
h
a
,
b
i
+
h
b
,
a
i
+
h
b
,
b
i
=
||
a
||
2
+0+0+
||
b
||
2
.
Theorem 1.
Nonzero orthogonal vectors
v
1
,
v
2
,...,
v
k
are linearly independent.
Proof:
Suppose
v
1
,
v
2
,...,
v
k
are nonzero orthogonal vectors and that
c
1
v
1
+
c
2
v
2
+
...
+
c
k
v
k
=
0
.
Given
i
, take the inner product of both sides with
v
i
to get
c
1
h
v
1
,
v
i
i
+
...
+
c
i
h
v
i
,
v
i
i
+
...
+
c
k
h
v
k
,
v
i
i
=
h
0
,
v
i