5.2 Orthogonal Complements and Projections

7/29/2019 5.2 Orthogonal Complements and Projections

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$ 5.2 # 1

5.2 Orthogonal Complements and projections

Orthogonal complements

Definition Let Wbe a subspace ofRRRRn and let x RRRRn.

(a) x is orthogonal to W x W w Wx w w Wx w 0.(b) W the orthogonal complement of W

the set of all vectors in RRRRn that are orthogonal to W. x RRRRn : x W x RRRRn : w Wx w x RRRRn : w Wx w 0.

Theorem (5.9) Let Wbe a subspace ofRRRRn. Then:

(a) W is also a subspace ofRRRRn.(b) W W, 0 .

(c) ifW spanw 1, w 2, , wk thenW v RRRRn : v w i 0 for i 1, 2, ,k .

1. Proof

(a) . W,

is not empty:w W0 w 0 0 W W

. W, is closed under addition:

x Wand y W

w Wx w 0and w Wy w 0

w W x y w x w y w 0 0 0

x y W.

. W, is closed under scalar multiplication:


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k RRRR and x W

w Wx w 0

w W k x

w

kx

w

k0

0 k x W.

(b) . 0 W W, :

0 W and 0 W,The trivial space 0 is a subspace

for every subspace ofW

0 W W,.

. W W, 0 :

x W W,

x Wand x W,

x Wand w Wx w 0

x x 0

x 0

x 0 .

(c) W spanw 1, w 2, , wk W v RRRRn : i v w i 0 :

x W

spanw 1, w 2, , wk

,

x , W spanw 1, w 2, , wk

x , w 1, w 2, , wk

x w i 0 for i 1, 2, ,k

x v RRRRn : v w i 0 for i 1, 2, ,k


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$ 5.2 # 3

Notation Let A be an m n matrix. Then:

(a) RSA rowA the row space of A.(b) CSA colA the column space of A.(c) NSA nullA the null space ofA.

Theorem (5.10) Let A be an m n matrix. Then:(a) RSA, NSA.(b) CSA, NSAT.

Proof(a)

x RSA

,

x , RSA

x , every row ofA

A x 0

x NSA.

(b)

CSA, RSA, NSAT, by part (a).

Example 1. Given that


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A

1 1 3 1 6

2 1 0 1 1

3 2 1 2 1

4 1 6 1 3

1 0 1 0 1

0 1 2 0 3

0 0 0 1 4

0 0 0 0 0

RREFA

AT

1 2 3 4

1 1 2 1

3 0 1 6

1 1 2 1

6 1 1 3

1 0 0 1

0 1 0 6

0 0 1 3

0 0 0 0

0 0 0 0

RREFAT

a. Determine the dependency equation(s) of the columns in A.

Solution The linear dependence or independence in RREFA

correspondingly determines the linear dependence or independence

in A. All the columns in RREFA containing leading 1s are linearlindependent, that is, c 1, c 2 and c 4 in RREFA are linearly

independent. The columns c 3 and c 5 are linearly dependent on

c 1, c 2 and c 4 as follows:

c 3 1c 1 2c 2 0c 4

c 5 1c 1 3c 2 4c 4.

Therefore the columns dependency equations in A are:

col3A col1A 2col2A

col5A col1A 3col2A 4col4A .

Check:


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col1A 2col2A

1

2

3

4

2

1

1

2

1

3

0

1

6

col3A

col1A 3col2A 4col4A

1

2

3

4

3

1

1

2

1

4

1

1

2

1

6

1

1

3

col5A.

b. Determine the dependency equation(s) of the rows in A.

Solution

c 4 c 1 6c 2 3c 3 in RREFAT

col4AT col1AT 6col2AT 3col3AT

row4A row1A 6row2A 3row3A

Check:


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row1A 6row2A 3row3A

1 1 3 1 6 6 2 1 0 1 1 3 3 2 1 2

4 1 6 1 3 row4A.


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c. Determine a basis in RREFA for RSA.

Solution RSA RSRREFA row1,row2,row3 is a basi

for RSA in RREFA.d. Determine a basis in A for RSA.

Solution col1,col2,col3 are linearly independent in

RREFAT col1AT,col2AT,col3AT is a basis for CSAT

in AT row1A,row2A,row3A is a basis for RSA in A sinc

RSA CSAT.

e. Determine a basis in A for CSA.

Solution col1A,col2A,col4A

f. Determine a basis in RREFAT for CSA.

Solution row1,row2,row3 is a basis for CSA in RREFAT.

g. Determine a basis for NSA.

Solution

NSA

x3 x5

2x3 3x5

x3

4x5

x5

span

1

2

1

0

0

,

1

3

0

4

1

h. Determine a basis for NSAT.

Solution


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NSAT

x4

6x4

3x4

x4

span

1

6

3

1

i. Show that RSA, NSA.

Solution It is enough to show that each basis vector of RSA is

orthogonal to each basis vector of NSA :

1 0 1 0 1

0 1 2 0 3

0 0 0 1 4

1 1

2 3

1 0

0 4

0 1

0 0

0 0

0 0

or

1 1 3 1 6

2 1 0 1 1

3 2 1 2 1

1 1

2 3

1 0

0 4

0 1

0 0

0 0

0 0

j. CSA, NSAT.

Solution It is enough to show that each basis vector of CSA is

orthogonal to each basis vector of NSAT :


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1 2 3 4

1 1 2 1

1 1 2 1

1

6

3

1

0

0

0

or

1 0 0 1

0 1 0 6

0 0 1 3

1

6

31

0

0

0


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$ 5.2 # 10

Example 2. Let Wbe the subspace ofR5 spanned by

w 1

1

3

5

0

5

, w 2

1

1

2

2

3

w 3

0

1

4

1

5

Find a basis for W.

Example Solution Let

A w 1 w 2 w 3

1 1 0

3 1 1

5 2 4

0 2 1

5 3 5

Now,

W CSA RSAT

W CSA, RSAT, NSAT NS RREFAT .

Therefore,

AT

1 3 5 0 5

1 1 2 2 3

0 1 4 1 5

1 0 0 3 4

0 1 0 1 3

0 0 1 0 2

RREFAT

so that


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W

3x4 4x5

x4 3x5

2x5

x4

x5

span

3

1

01

0

,

4

3

20

1


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Orhtogonal projections

Definition Let v and u 0 in n. Then:

the component v parallel to u

the projection of v onto u

proj u v v uu u

u

and

the component of v orthogonal to u

perp u v

Remark Since

v proj u v perp u v

it follows that

perp u v v proj u v .

IfW span u , then w proj u v W and

w, proj u v ,

perp u v W,. Therefore, there is a decompostion

of v into the sum v w w, such that w W and w, W,.

Definition Let S u 1, u 2, , u k be an orthogonal basis for thesubspace Win n. For any v in n,

the component of v in W

projW v

v u 1

u 1 u 1u

1

v u 2

u 2 u 2u

2

v u k

u k u ku

k

proj u 1 v proj u 2 v proj uk v

and


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the component of v orthogonal to W

perpW v

projW,

v

projW v

,

v projW v .

Example (1.) PPPP

x

y

z

: x y 2z 0 is a plane in 3 and let

v 31

2

. Find the orthogonal prjection of v onto PPPPand the

component of v orthogonal to PPPP.

Solution (1.)

PPPP,n

1

1

2

Therefore,


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the component of v orthogonal to PPPP

projPPPP, v perpPPPP v projn v v n

n nn

3

1

2

1

1

2

1

1

2

1

1

2

1

1

2

43

1

1

2

43

43

83

and

the component of v in PPPP

projPPPP v v perpPPPP v

3

1

2

43

43

83

53

13

23

13

5

1

2

Solution (2.)

PPPP

y 2z

yz

span u 1

1

10

, u 2

1

11

,

hence, PPPPhas an orthogonal basis Look at Example 5.3, page 367-36

Therefore,


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the component of v in PPPP

projPPPP v proj u 1 v proj u 2 v

v u 1u 1 u 1 u 1

v u 2u 2 u 2 u 2

3

1

2

1

1

0

1

1

0

1

1

0

1

1

0

3

1

2

1

1

1

1

1

1

1

1

1

1

1

1

1

1

0

23

1

1

1

53

13

23

13

5

1

2

and

the component of v orthogonal to PPPP

projPPPP, v perpPPPP v v projPPPP v

3

1

2

53

13

23

43

43

83

43

1

1

2

.

Check:


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w w, projPPPP v perpPPPP v 13

5

1

2

43

1

1

2

3

1

2

Theorem (5.11) The Orthogonal Decomposition Theorem Let Wbe a

subspace ofn and let v n. Then there are unique vectors w in Wandw, in W, such that

v w w,.

Proof

(a) . Show that the decomposition exists, that is,w W w, W, v w w, : Let S u 1, u 2, , u k be anorthogonal basis for the subspace Win n and let v n. Let

w projW v v u 1

u 1 u 1u 1

v u 2u 2 u 2

u 2 v u k

u k u ku k

i1

k

v u iu i u i

u i

and let

w, perpW v v projW v .

Then

w w, projW v perpW v projW v v projW v v .

. w W :

w projW v

v u 1

u 1 u 1u 1

v u 2u 2 u 2

u 2 v u k

u k u ku k spanS

. w, W, : i,


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w , u i v projW v u i

vv u 1

u 1 u 1u 1

v u ku k u k

u k u i

v u iv u i

u i u iu i u i,

u i uj 0 if i j

u i2 uj

20 if i j

v u i v u i 0,

which implies thatw S w spanS W w W, Theorem 5.

(b) Show that the uniqueness of the decomposition exists, that is,

x W y W, v x y x w and y w :

a Wand b W, such that v a b

w w, a b

w a b w,, where w a Wand b w, W,

w a b w, 0 , since W W, 0

a w and b w,.

Theorem (5.12)

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5.2 Orthogonal Complements and Projections