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Math 1b Prac — The GramSchmidt process; orthogonal projection
January 24, 2011 — extended/revised January 28, 2011
To warm up, we start with
Theorem 1.
Nonzero pairwise orthogonal vectors
u
1
,
u
2
,...,
u
k
are linearly independent.
Proof:
Suppose
u
1
,
u
2
u
k
are nonzero orthogonal vectors and that
c
1
u
1
+
c
2
u
2
+
...
+
c
k
u
k
=
0
.
Given
i
, take the inner product of both sides with
u
i
to get
c
1
h
u
1
,
u
i
i
+
+
c
i
h
u
i
,
u
i
i
+
+
c
k
h
u
k
,
u
i
i
=
h
0
,
u
i
i
=0
.
Most of the inner products on the left are zero, and the only surviving term is
c
i
h
u
i
,
u
i
i
=
c
i

u
i

2
,
and because

u
i

2
6
= 0, it must be that
c
i
= 0. This holds for all
i
=1
,
2
,...,k
,andth
is
means that
u
1
,
u
2
u
k
are linearly independent.
ut
To save typing, we borrow the description of of the GramSchmidt process from the
Wikipedia article.
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View Full DocumentThe defnition oF proj
u
(
v
) should be modifed to include the case when
u
=
0
,where
we understand proj
0
(
v
)=
0
.E
x
e
r
c
i
s
e
:I
F
x
−
c
u
is orthogonal to
u
6
=0
,th
en
c
=
h
x
,
u
i
/
h
u
,
u
i
, i.e.
c
u
=proj
u
(
x
).
Example.
Let
v
1
=(1
,
1
,
1
,
1
,
1)
,
v
2
=(0
,
1
,
2
,
3
,
4)
,
v
3
,
1
,
4
,
9
,
16)
.
Then
u
1
,
1
,
1
,
1
,
1)
,
u
2
=(
−
2
,
−
1
,
0
,
1
,
2)
,
u
3
=(2
,
−
1
,
−
2
,
−
1
,
2)
.
Theorem 2.
The vectors
u
1
,
u
2
,
u
3
,...
produced by the GramSchmidt process are pair
wise orthogonal.
Proof:
We need to check that
h
u
i
,
u
j
i
=0Fo
r
i
6
=
j
. We proceed by induction on
k
.
Suppose we know
h
u
i
,
u
j
i
=0For1
≤
i<j
≤
k
−
1. Then For
i<k
,
h
u
i
,
u
k
i
=
h
u
i
,
v
k
−
(
proj
u
1
(
v
k
)+proj
u
2
(
v
k
)+
...
+proj
u
k
−
1
(
v
k
)
)
i
=
h
u
i
,
v
k
i−
(
h
u
i
,
u
1
(
v
k
)
i
+
+
h
u
i
,
u
k
−
1
(
v
k
)
i
)
.
Since proj
u
j
(
x
) is a scalar multiple oF
u
j
, all the terms
h
u
i
,
u
j
(
v
k
)
i
are 0 except
possibly when
i
=
j
, and we have
h
u
i
,
u
k
i
=
h
u
i
,
v
k
i−h
u
i
,
u
i
(
v
k
)
i
=
h
u
i
,
v
k
u
i
,
h
u
i
,
v
k
i
h
u
i
,
u
i
i
u
i
i
=
h
u
i
,
v
k
h
u
i
,
v
k
i
h
u
i
,
u
i
i
h
u
i
,
u
i
i
.
ut
Note that iF
u
1
,
u
2
,
u
3
are nonzero orthogonal vectors, and we let
e
i
=
u
i
/

u
i

,
then
e
1
,
e
2
,
e
3
are orthonormal vectors.
Orthonormal
means that the vectors are or
thogonal and that they each have length 1. The standard basis in
R
n
is only one example
oF an orthonormal basis For
R
n
.
Theorem 3.
Let
u
1
,
u
2
,...,
u
k
be the vectors produced from
v
1
,
v
2
v
k
by the Gram
Schmidt orthogonalization process; let
U
and
V
be the matrices whose rows are, respec
tively,
u
1
,
u
2
u
k
and
v
1
,
v
2
v
k
. Then there are (square) lower triangular matrices
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 Winter '08
 Aschbacher
 Vectors

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