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Projections and the GramSchmidt Procedure

7
...
Similarly, we saw
how to find the projection of y onto a plane, which is just a 2-dimensional subspace
...

Let y E JRn and let S be a subspace of JR11 To match what we did in Chapter 1, we
•

want to write y as
y = proj5 y + perp5 y
where proj 5 y is a vector in S and perp5 y is a vector orthogonal to S
...


Definition

Let S be a subspace of ]Rn
...
L
...
l

= {1E]Rn11·s=0 for all SES}

Remark

Note that if :B = {v1,


...
l

=

{1E]Rn11
...
On the other hand,

we saw in Chapter 1 that the plane is the set of all vectors orthogonal to n (or any

scalar multiple of

plane
...
4 such that v

1

�

·

= 0 and v

V4

·

�

= 0
...
W" =span

{i ;}
·

-�


...


Theorem 1

Let S be ak-dimensional subspace of ll11• Then S is a subspace of JR" and:

(1) s n SJ_ = {O}
(2) dim(SJ_) = n k
-

(3) If {v 1

vk} is an orthonormal basis for Sand Wk+l
, vn} is an orthonormal
1
basis for SJ_, then {i11, •••, vk> Vk+I
...



...



...



...


We are now able to return to our goal of defining the projection of a vector 1

ffi?
...
, i111} for SJ_
...
11• Therefore, from our work in
1

onto a subspaces of

•

•


...



...



...
1, we can find the coordinates of

x

with respect to this orthonormal basis
...
In particular, we have

(x · v1)111 + · ·· + (x vk)Vb
(x · vk+t)Vk+l +·· · + (x · v11 )111 , which is a vector in§j_
...


Definition
Projection onto

Let§ be ak-dimensional subspace of IR
...
If x is any vector in IR
...
We could, of course, make a similar definition for the projection if
we have only an orthogonal basis
...

We have defined perp8 x so that we do not require an orthonormal basis for §j_
...

For any 1 � i � k, we have

v; · perp8 x= v; · [x - ((x · i11)v1 + · · · + (x · vk)Vk)]
=

v;
...
ccx · v1)111 +
...
1

= v;
=

·

·

=0
since '13 is an orthonormal basis and the dot product is symmetric
...
, vk}

of §
...


EXAMPLE3
Let§= Span

{l : : }
,l

1

Solution:

-

-1

and let X=

_

�


...


3

An orthonormal basis for § is '13

=

W1
...


Recall that we showed in Chapter 1 that the projection of a vector

a plane in JR3 is the vector in the plane that is closest to
projection of

Theorem 2

4
4

9/2

-5/2

3

-5/2

1

E

1
...
11 is the vector in§ that is closest to 1
...


Let § be a subspace of lR
...
11, the unique vector s

E § that

{i11,
...
, v,,} be an or­

thonormal basis for§j_
...
But this means that

The Gram-Schmidt Procedure
For many of the calculations in this chapter, we need an orthonormal (or orthogonal)
basis for a subspace § of JR
...
11, it is certainly
possible to use the methods of Section 4
...
,wk} for§
...
,vk} for§
...

The construction is inductive
...
Then, given an orthogonal basis {V1,
...
,w;_J}, we want to find a vector v; such that {V1,
...
,w;}
...
,vd for
Span{w1,
...
To do this, we will use the following theorem
...
, Vk

E

JR
...
, tk-1 ER

You are asked to prove Theorem 3 in Problem D5
...

First step: Let v1 = w1• Then the one-dimensional subspace spanned by v1 is
obviously the same as the subspace spanned by w1• We will denote this subspace
as S1= Span{Vi}
...
w2}
...
We
denote the two-dimensional subspace by§2 = Span{il\,v2}
...
,v;-d is
orthogonal, and S;-1= Spanfv1,
...
,w;-d
...
,v;} is orthogonal and Span{v1,
...
,w;}
...
,vk}= Sp

...
,wk}=§
and an orthogonal basis has been produced for the original subspace§
...
It is an important feature of the construction that §;_ 1 is a subspace of the
next§;
...
Since it is really only the direction of v; that is important in this procedure, we
can rescale each v; in any convenient fashion to simplify the calculations
...
Notice that the order of the vectors in the original basis has an effect on the
calculations because each step takes the perpendicular part of the next vector
...


4
...
The procedure will actually detect a linearly
dependent vector by returning the zero vector when we take the perpendicular
part
...


EXAMPLE4

Use the Gram-Schmidt Procedure to find an orthonormal basis for the subspace of JR5

defined by §

=

Span

1

-1

0

1

2

1

0

1

1

0

1
2

Solution: Call the vectors in the basis w1, Wz, and w3, respectively
...


Second step: Determine perp51 w2:
-3/2
3/2
1
-1/2
1/2

(It is wise to check your arithmetic by verifying that v1

·

perp51 Wz

=

0
...


EXAMPLE4

Third step: Determine perp32 w3:

(continued)

-1/4
-3/4
1/2
1/4
3/4
(Again, it is wise to check that perps2 w3 is orthogonal to V1 and Vz
...
To obtain an orthonormal
basis for §, we divide each vector in this basis by its length
...
-1 2
...


Solution: Call the vectors in the spanning set w1, w2, and w3, respectively
...


Second step: Determine perps1 w2:

We take v2

=

�1

-2

and S2

=

Span{V1, v2}
...


E

Span{v1, v2}

=

Span{w1, w2}
...
2
Practice Problems
Al Each of the following sets is orthogonal but not
2
3
orthonormal
...

(a ) � =

Cb ) '13 =

( c) C =

( c)

(d )

1 ' 0 ' -1

0

1

1

,

-

(c) s =Span

,

,

,

orthonormal basis for the subspace spanned by

:

each set
...


(b) S=Span

{[�]·[i] ·[;]}
{[;]
...
-I}

(a)

(b)

( c)

{[i]·[;]
...
[:J
...


(d)

0
1

0

-

' -1

1

-1

1

0

1

1

-1
1

0

JPi
...
x for any x E JPi
...
Determine the projection of x = _
onto the subspace spanned by each set
...
l=

{ j _;}
1

-2

(b) � =

(c) C =

p ;)
{I -� �)
{ : ' -� ' _: )
,

,

(c)

(d)

-

(d) v

=

0

-1

l

B2 Find a basis for the orthogonal complement of each

{[�]}
j]
fi
]
}
{[
{[J
...


7
...

(a)

(b)

{[�l-l= :J
...
Ul

Conceptual Problems

11
Dl Prove that if § is a k-dimensional subspace of JR
...
,
then§ n §J_ = {0}
...


'

B4 Use the Gram-Schmidt Procedure to produce an
orthonormal basis for the subspace spanned by
each set
...

(a) S=Span

{-! ; ' n
P
...


(a) Find an orthonormal basis :B for§
...


(c) Find an orthonormal basis C for §J_
...
x
...
'1,
then (§J_ )J_ =§
...
, vk} is an orthonormal basis for
§ and {vk+I•
...
1•
then {V 1,

D4 Prove that if

•

•

•

,

DS Suppose thatv1,
...
, tk t ER

Prove that
Span{v1,
...
,v-

+
...
,v}
k be an orthogonal basis of
§
...


D7 If {V 1,

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