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1 Inner Product, Length, and Orthogonality

# 1 Inner Product, Length, and Orthogonality

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6.1

Inner Product, Length, and Orthogonality

331

2

3

2

3

2

3

EXAMPLE 1 Compute u v and v u for u D 4 5 5 and v D 4 2 5.

1

3

SOLUTION

2

3

3

5

1 4 2 5 D .2/.3/ C . 5/.2/ C . 1/. 3/ D 1

u v D uT v D Œ 2

3

2

3

2

3 4 5 5 D .3/.2/ C .2/. 5/ C . 3/. 1/ D 1

v u D vT u D Œ 3 2

1

ItisclearfromthecalculationsinExample 1why u v D v u. Thiscommutativity

oftheinnerproductholdsingeneral. Thefollowingpropertiesoftheinnerproduct

areeasilydeducedfrompropertiesofthetransposeoperationinSection 2.1. (See

Exercises 21and22attheendofthissection.)

THEOREM 1

Let u, v, and w bevectorsin Rn , andlet c beascalar. Then

a.

b.

c.

d.

u vDv u

.u C v/ w D u w C v w

.c u/ v D c.u v/ D u .c v/

u u 0, and u u D 0 ifandonlyif u D 0

Properties(b)and(c)canbecombinedseveraltimestoproducethefollowinguseful

rule:

.c1 u1 C

C cp up / w D c1 .u1 w/ C

C cp .up w/

The Length of a Vector

If v isin Rn , withentries v1 ; : : : ; vn , thenthesquarerootof v v isdeﬁnedbecause v v

isnonnegative.

DEFINITION

(a, b)

⎯⎯⎯⎯⎯

a2 + b2

|a|

0

x1

FIGURE 1

Interpretationof kvk aslength.

Ä

a

. Ifweidentify v withageometricpointinthe

b

plane, asusual, then kvk coincideswiththestandardnotionofthelengthoftheline

segmentfromtheoriginto v. ThisfollowsfromthePythagoreanTheoremappliedtoa

trianglesuchastheoneinFig. 1.

A similarcalculationwiththediagonalofarectangularboxshowsthatthedeﬁnition

oflengthofavector v in R3 coincideswiththeusualnotionoflength.

Foranyscalar c , thelengthof c v is jcj timesthelengthof v. Thatis,

Suppose v isin R , say, v D

2

x2

|b|

The length (or norm)of v isthenonnegativescalar kvk deﬁnedby

q

p

kvk D v v D v12 C v22 C C vn2 ; and kvk2 D v v

kc vk D jcjkvk

(Toseethis, compute kc vk2 D .c v/ .c v/ D c 2 v v D c 2 kvk2 andtakesquareroots.)

332

CHAPTER 6

Orthogonality and Least Squares

A vectorwhoselengthis1iscalleda unitvector. Ifwe divide anonzerovector v

byitslength—thatis, multiplyby 1=kvk—weobtainaunitvector u becausethelength

of u is .1=kvk/kvk. Theprocessofcreating u from v issometimescalled normalizing

v, andwesaythat u is inthesamedirection as v.

Severalexamplesthatfollowusethespace-savingnotationfor(column)vectors.

EXAMPLE 2 Let v D .1; 2; 2; 0/. Findaunitvector u inthesamedirectionas v.

SOLUTION First, computethelengthof v:

kvk2 D v v D .1/2 C . 2/2 C .2/2 C .0/2 D 9

p

kvk D 9 D 3

Then, multiply v by 1=kvk toobtain

x2

3 2

3

1

1=3

6

7

1

1

16 2 7

7 D 6 2=3 7

uD

vD vD 6

4

5

4

2

2=3 5

kvk

3

3

0

0

W

2

Tocheckthat kuk D 1, itsufﬁcestoshowthat kuk2 D 1.

1

x

D

x1

1

C

D1

C

2 2

3

C .0/2

vector z thatisabasisfor W .

x2

y

z

x1

1

1

9

2 2

3

EXAMPLE 3 Let W bethesubspaceof R2 spannedby x D . 32 ; 1/. Findaunit

(a)

1

1 2

C

3

4

4

C 9 C0

9

kuk2 D u u D

(b)

FIGURE 2

Normalizingavectortoproducea

unitvector.

SOLUTION W consistsofallmultiplesof x, asinFig. 2(a). Anynonzerovectorin W

isabasisfor W . Tosimplifythecalculation, “scale” x toeliminatefractions. Thatis,

multiply x by3toget

Ä

2

yD

3

p

Nowcompute kyk2 D 22 C 32 D 13, kyk D 13, andnormalize y toget

Ä

Ä p

1

2

2=p13

zD p

D

3

3= 13

13

p

p

SeeFig. 2(b). Anotherunitvectoris . 2= 13; 3= 13/.

Distance in Rn

arerealnumbers, thedistanceonthenumberlinebetween a and b isthenumber ja bj.

in Rn .

a

1

2

b

3

4 5 6

6 units apart

7

8

a

9

|2 – 8| = |– 6| = 6 or |8 – 2| = |6| = 6

FIGURE 3 Distancesin R.

–3 – 2 –1

b

0 1 2

7 units apart

3

4

5

|(– 3) – 4| = |– 7| = 7 or |4 – (–3)| = |7| = 7

Inner Product, Length, and Orthogonality

6.1

DEFINITION

333

For u and v in Rn , the distancebetweenuandv, writtenas dist .u; v/, isthe

lengthofthevector u v. Thatis,

dist .u; v/ D ku

vk

In R2 and R3 , thisdeﬁnitionofdistancecoincideswiththeusualformulasforthe

Euclideandistancebetweentwopoints, asthenexttwoexamplesshow.

EXAMPLE 4 Computethedistancebetweenthevectors u D .7; 1/ and v D .3; 2/.

SOLUTION Calculate

u

ku

Ä

Ä

Ä

7

3

4

D

1

2

1

p

p

vk D 42 C . 1/2 D 17

vD

Thevectors u, v, and u v areshowninFig. 4. Whenthevector u v isadded

to v, theresultis u. NoticethattheparallelograminFig. 4showsthatthedistancefrom

u to v isthesameasthedistancefrom u v to 0.

x2

v

||u – v||

u

1

x1

1

u–v

–v

FIGURE 4 Thedistancebetween u and v is

thelengthof u

v.

EXAMPLE 5 If u D .u1 ; u2 ; u3 / and v D .v1 ; v2 ; v3 /, then

p

dist .u; v/ D ku vk D .u v/ .u v/

p

D .u1 v1 /2 C .u2 v2 /2 C .u3

v3 /2

Orthogonal Vectors

||u – v||

u

v

||u –(– v)||

0

–v

FIGURE 5

Therestofthischapterdependsonthefactthattheconceptofperpendicularlinesin

ordinaryEuclideangeometryhasananaloguein Rn .

Consider R2 or R3 andtwolinesthroughtheorigindeterminedbyvectors u and v.

ThetwolinesshowninFig. 5aregeometricallyperpendicularifandonlyifthedistance

from u to v isthesameasthedistancefrom u to v. Thisisthesameasrequiringthe

squaresofthedistancestobethesame. Now

2

Œ dist .u; v/  D ku . v/k2 D ku C vk2

D .u C v/ .u C v/

D u .u C v/ C v .u C v/

Du uCu vCv uCv v

D kuk C kvk C 2u v

2

2

Theorem 1(b)

Theorem 1(a), (b)

Theorem 1(a)

(1)

334

CHAPTER 6

Orthogonality and Least Squares

Thesamecalculationswith v and v interchangedshowthat

Œdist .u; v/2 D kuk2 C k

vk2 C 2u . v/

D kuk2 C kvk2

2u v

Thetwosquareddistancesareequalifandonlyif 2u v D 2u v, whichhappensifand

onlyif u v D 0.

Thiscalculationshowsthatwhenvectors u and v areidentiﬁedwithgeometric

points, thecorrespondinglinesthroughthepointsandtheoriginareperpendicularif

andonlyif u v D 0. Thefollowingdeﬁnitiongeneralizesto Rn thisnotionofperpendicularity(or orthogonality, asitiscommonlycalledinlinearalgebra).

DEFINITION

Twovectors u and v in Rn are orthogonal (toeachother)if u v D 0.

Observethatthezerovectorisorthogonaltoeveryvectorin Rn because 0T v D 0

forall v.

TherighttriangleshowninFig. 6providesavisualizationofthelengthsthatappearin

thetheorem.

THEOREM 2

u+v

The Pythagorean Theorem

Twovectors u and v areorthogonalifandonlyif ku C vk2 D kuk2 C kvk2 .

||v||

u

||u + v||

Orthogonal Complements

Toprovidepracticeusinginnerproducts, weintroduceaconceptherethatwillbeofuse

inSection 6.3andelsewhereinthechapter. Ifavector z isorthogonaltoeveryvector

inasubspace W of Rn , then z issaidtobe orthogonalto W . Thesetofallvectors z

thatareorthogonalto W iscalledthe orthogonalcomplement of W andisdenotedby

v

||u||

0

FIGURE 6

EXAMPLE 6 Let W beaplanethroughtheoriginin R3 , andlet L betheline

w

0

z

L

W

FIGURE 7

A planeandlinethrough 0 as

orthogonalcomplements.

throughtheoriginandperpendicularto W . If z and w arenonzero, z ison L, and

w isin W , thenthelinesegmentfrom 0 to z isperpendiculartothelinesegmentfrom 0

to w; thatis, z w D 0. SeeFig. 7. Soeachvectoron L isorthogonaltoevery w in W .

Infact, L consistsof all vectorsthatareorthogonaltothe w’sin W , and W consistsof

allvectorsorthogonaltothe z’sin L. Thatis,

LDW?

and

W D L?

Thefollowingtwofactsabout W ? , with W asubspaceof Rn , areneededlater

inthechapter. ProofsaresuggestedinExercises 29and30. Exercises 27–31provide

excellentpracticeusingpropertiesoftheinnerproduct.

1. A vector x isin W ? ifandonlyif x isorthogonaltoeveryvectorinasetthat

spans W .

2. W ? isasubspaceof Rn .

Inner Product, Length, and Orthogonality

6.1

335

thesubspacesshowninFig. 8. (AlsoseeExercise 28inSection 4.6.)

A

0

Nu

0

T

lA

Nu

lA

Co

wA

Ro

lA

FIGURE 8 Thefundamentalsubspacesdetermined

byan m

THEOREM 3

n matrix A.

Let A bean m n matrix. Theorthogonalcomplementoftherowspaceof A is

thenullspaceof A, andtheorthogonalcomplementofthecolumnspaceof A is

thenullspaceof AT :

.Row A/? D Nul A

and

.Col A/? D Nul AT

PROOF Therow–columnruleforcomputing Ax showsthatif x isin Nul A, then x is

orthogonaltoeachrowof A (withtherowstreatedasvectorsin Rn /. Sincetherows

of A spantherowspace, x isorthogonalto Row A. Conversely, if x isorthogonalto

Row A, then x iscertainlyorthogonaltoeachrowof A, andhence Ax D 0. Thisproves

theﬁrststatementofthetheorem. Sincethisstatementistrueforanymatrix, itistrue

for AT . Thatis, theorthogonalcomplementoftherowspaceof AT isthenullspaceof

AT . Thisprovesthesecondstatement, because Row AT D Col A.

Angles in R2 and R3 (Optional)

If u and v arenonzerovectorsineither R2 or R3 , thenthereisaniceconnectionbetween

theirinnerproductandtheangle # betweenthetwolinesegmentsfromtheorigintothe

pointsidentiﬁedwith u and v. Theformulais

(2)

u v D kuk kvk cos #

Toverifythisformulaforvectorsin R2 , considerthetriangleshowninFig. 9, withsides

oflengths kuk, kvk, and ku vk. Bythelawofcosines,

ku

vk2 D kuk2 C kvk2

2kuk kvk cos #

(u1, u2)

||u – v||

||u||

(v1, v2)

||v||

FIGURE 9 Theanglebetweentwovectors.

336

CHAPTER 6

Orthogonality and Least Squares

whichcanberearrangedtoproduce

1

kuk2 C kvk2 ku vk2

2

1 2

D

u C u22 C v12 C v22 .u1

2 1

D u1 v1 C u2 v2

Du v

kuk kvk cos # D

v1 /2

.u2

v2 /2

Theveriﬁcationfor R3 issimilar. When n > 3, formula (2)maybeusedto deﬁne the

anglebetweentwovectorsin Rn . Instatistics, forinstance, thevalueof cos # deﬁned

by(2)forsuitablevectors u and v iswhatstatisticianscalla correlationcoefﬁcient.

PRACTICE PROBLEMS

2

3

2

3

4=3

5

3

Let a D

, c D 4 1 5, and d D 4 6 5.

1

2=3

1

Â

Ã

a b

a b

1. Compute

and

a.

a a

a a

2. Findaunitvector u inthedirectionof c.

3. Showthat d isorthogonalto c.

4. UsetheresultsofPracticeProblems 2and3toexplainwhy d mustbeorthogonalto

theunitvector u.

Ä

2

,bD

1

Ä

6.1 EXERCISES

ComputethequantitiesinExercises1–8usingthevectors

3

3

2

2

Ä

Ä

6

3

1

4

uD

, vD

, w D 4 1 5, x D 4 2 5

2

6

3

5

v u

1. u u, v u, and

u u

3.

5.

1

w

w w

u vÁ

v v

x w

2. w w, x w, and

w w

1

u

u u

4.

v

7. kwk

x wÁ

x

x x

6.

8. kxk

InExercises9–12, ﬁndaunitvectorinthedirectionofthegiven

vector.

2

3

Ä

6

30

10. 4 4 5

9.

40

3

2

3

7=4

11. 4 1=2 5

1

12.

Ä

13. Findthedistancebetween x D

10

3

DeterminewhichpairsofvectorsinExercises15–18areorthogonal.

3

3

2

2

Ä

Ä

2

12

8

2

15. a D

,bD

16. u D 4 3 5, v D 4 3 5

5

3

5

3

3

3

2

3

2

2

3

2

1

3

4

3

6 87

6 77

6 17

6 27

7

7

6

7

6

7

6

17. u D 6

4 5 5, v D 4 2 5 18. y D 4 4 5, z D 4 15 5

7

6

0

0

InExercises19and20, allvectorsarein Rn . Markeachstatement

19. a. v v D kvk2 .

b. Foranyscalar c , u .c v/ D c.u v/.

c. Ifthedistancefrom u to v equalsthedistancefrom u to

v, then u and v areorthogonal.

8=3

2

Ä

3

3

2

4

0

14. Findthedistancebetween u D 4 5 5 and z D 4 1 5.

8

2

2

and y D

Ä

d. Forasquarematrix A, vectorsin Col A areorthogonalto

vectorsin Nul A.

1

.

5

e. If vectors v1 ; : : : ; vp span a subspace W and if x is

orthogonaltoeach vj for j D 1; : : : ; p , then x isin W ? .

6.1

20. a. u v

v u D 0.

b. Foranyscalar c , kc vk D ckvk.

c. If x isorthogonaltoeveryvectorinasubspace W , then x

isin W ? .

d. If kuk2 C kvk2 D ku C vk2 , then u and v areorthogonal.

e. Foran m n matrix A, vectorsinthenullspaceof A are

orthogonaltovectorsintherowspaceof A.

21. Usethetransposedeﬁnitionoftheinnerproducttoverify

parts(b)and(c)ofTheorem 1. Mentiontheappropriatefacts

fromChapter 2.

22. Let u D .u1 ; u2 ; u3 /. Explain why u u 0. When is

u u D 0?

2

3

2

3

2

7

23. Let u D 4 5 5 and v D 4 4 5. Computeandcompare

1

6

u v, kuk2 , kvk2 , and ku C vk2 . Donotuse thePythagorean

Theorem.

24. Verifythe parallelogramlaw forvectors u and v in Rn :

ku C vk2 C ku vk2 D 2kuk2 C 2kvk2

Ä

Ä

a

x

25. Let v D

. Describetheset H ofvectors

thatare

b

y

orthogonalto v. [Hint: Consider v D 0 and v Ô 0.]

3

2

5

26. Let u D 4 6 5, andlet W bethesetofall x in R3 suchthat

7

u x D 0. Whattheorem inChapter 4canbeusedtoshowthat

W isasubspaceof R3 ? Describe W ingeometriclanguage.

27. Supposeavector y isorthogonaltovectors u and v. Show

that y isorthogonaltothevector u C v.

28. Suppose y isorthogonalto u and v. Showthat y isorthogonaltoevery w in Span fu; vg. [Hint: Anarbitrary w

in Span fu; vg hastheform w D c1 u C c2 v. Showthat y is

orthogonaltosuchavector w.]

w

u

0

v

}

v

u,

n{

y

a

Sp

29. Let W D Span fv1 ; : : : ; vp g. Showthatif x isorthogonalto

each vj , for 1 Ä j Ä p , then x isorthogonaltoeveryvector

in W .

Inner Product, Length, and Orthogonality

337

30. Let W beasubspaceof Rn , andlet W ? bethesetofall

vectorsorthogonalto W . Showthat W ? isasubspaceof Rn

usingthefollowingsteps.

a. Take z in W ? , andlet u representanyelementof W .

Then z u D 0. Takeanyscalar c andshowthat c z is

orthogonalto u. (Since u wasanarbitraryelementof W ,

thiswillshowthat c z isin W ? .)

b. Take z1 and z2 in W ? , andlet u beanyelementof W .

Showthat z1 C z2 isorthogonalto u. Whatcanyou

concludeabout z1 C z2 ? Why?

c. Finishtheproofthat W ? isasubspaceof Rn .

31. Showthatif x isinboth W and W ? , then x D 0.

32. [M] Constructapair u, v ofrandomvectorsin R4 , andlet

2

3

:5 :5 :5 :5

6 :5 :5 :5 :5 7

7

4 :5 :5 :5 :5 5

:5 :5 :5 :5

a. Denote the columns of A by a1 ; : : : ; a4 .

Compute the length of each column, and compute a1 a2 ,

a1 a3 ; a1 a4 ; a2 a3 ; a2 a4 , and a3 a4 .

b. Computeandcomparethelengthsof u, Au, v, and Av.

c. Useequation(2)inthissectiontocomputethecosineof

theanglebetween u and v. Comparethiswiththecosine

oftheanglebetween Au and Av.

d. Repeatparts(b)and(c)fortwootherpairsofrandom

vectors?

33. [M] Generaterandomvectors x, y, and v in R4 withinteger

entries(and v Ô 0), andcomputethequantities

x v

y v Á .x C y/ v .10x/ v

v;

v;

v;

v

v v

v v

v v

v v

Repeatthecomputationswithnewrandomvectors x and

y. Whatdoyouconjectureaboutthemapping x 7! T .x/ D

x v

v (for v Ô 0)? Verifyyourconjecturealgebraically.

v v

3

2

6

3

27

33

13

6 6

5

25

28

14 7

6

7

8

6

34

38

18 7

34. [M] Let A D 6

6

7. Construct

4 12

10

50

41

23 5

14

21

49

29

33

a matrix N whose columns form a basis for Nul A, and

constructamatrix R whose rows formabasisfor Row A (see

Section4.6fordetails). Performamatrixcomputationwith

N and R thatillustratesafactfromTheorem3.

338

CHAPTER 6

Orthogonality and Least Squares

SOLUTIONS TO PRACTICE PROBLEMS

Â

Ã

Ä

a b

7

a b

7

14=5

1. a b D 7, a a D 5. Hence

D , and

.

7=5

a a

5

a a

5

3

2

4

p

2. Scale c, multiplyingby3toget y D 4 3 5. Compute kyk2 D 29 and kyk D 29.

2

p 3

2

4=p29

1

Theunitvectorinthedirectionofboth c and y is u D

y D 4 3=p29 5.

kyk

2= 29

3. d isorthogonalto c, because

2

3 2

3

5

4=3

2

20

d c D 4 65 4 1 5 D

6

D0

3

3

1

2=3

4. d isorthogonalto u because u hastheform k c forsome k , and

d u D d .k c/ D k.d c/ D k.0/ D 0

6.2 ORTHOGONAL SETS

A setofvectors fu1 ; : : : ; up g in Rn issaidtobean orthogonalset ifeachpairofdistinct

vectorsfromthesetisorthogonal, thatis, if ui uj D 0 whenever i Ô j .

EXAMPLE 1 Showthat fu1 ; u2 ; u3 g isanorthogonalset, where

2 3

3

u1 D 4 1 5;

1

x3

u3

2

3

1

u 2 D 4 2 5;

1

2

3

1=2

u3 D 4 2 5

7=2

SOLUTION Considerthethreepossiblepairsofdistinctvectors, namely, fu1 ; u2 g,

fu1 ; u3 g, and fu2 ; u3 g.

u1 u2 D 3. 1/ C 1.2/ C 1.1/ D 0

u2

u1 u 3 D 3

u1

x1

x2

FIGURE 1

THEOREM 4

u2 u 3 D

1

2

1

1

2

C 1. 2/ C 1

7

2

C 2. 2/ C 1

7

2

D0

D0

Eachpairofdistinctvectorsisorthogonal, andso fu1 ; u2 ; u3 g isanorthogonalset. See

Fig. 1; thethreelinesegmentstherearemutuallyperpendicular.

If S D fu1 ; : : : ; up g isanorthogonalsetofnonzerovectorsin Rn , then S is

linearlyindependentandhenceisabasisforthesubspacespannedby S .

PROOF If 0 D c1 u1 C

C cp up forsomescalars c1 ; : : : ; cp , then

0 D 0 u1 D .c1 u1 C c2 u2 C C cp up / u1

D .c1 u1 / u1 C .c2 u2 / u1 C C .cp up / u1

D c1 .u1 u1 / C c2 .u2 u1 / C C cp .up u1 /

D c1 .u1 u1 /

because u1 isorthogonalto u2 ; : : : ; up . Since u1 isnonzero, u1 u1 isnotzeroandso

c1 D 0. Similarly, c2 ; : : : ; cp mustbezero. Thus S islinearlyindependent.

6.2

DEFINITION

Orthogonal Sets

339

An orthogonalbasis forasubspace W of Rn isabasisfor W thatisalsoan

orthogonalset.

Thenexttheoremsuggestswhyanorthogonalbasisismuchnicerthanotherbases.

Theweightsinalinearcombinationcanbecomputedeasily.

THEOREM 5

Let fu1 ; : : : ; up g beanorthogonalbasisforasubspace W of Rn . Foreach y in

W , theweightsinthelinearcombination

y D c1 u1 C

aregivenby

cj D

y uj

uj uj

C cp u p

.j D 1; : : : ; p/

PROOF Asintheprecedingproof, theorthogonalityof fu1 ; : : : ; up g showsthat

y u1 D .c1 u1 C c2 u2 C

C cp up / u1 D c1 .u1 u1 /

Since u1 u1 isnotzero, theequationabovecanbesolvedfor c1 . Toﬁnd cj for

j D 2; : : : ; p , compute y uj andsolvefor cj .

3

EXAMPLE 2 Theset

2 S D3fu1 ; u2 ; u3 g inExample 1isanorthogonalbasisfor R .

6

Expressthevector y D 4 1 5 asalinearcombinationofthevectorsin S .

8

SOLUTION Compute

ByTheorem 5,

y u1 D 11;

u1 u1 D 11;

yD

y u2 D 12;

u2 u2 D 6;

y u3 D 33

u3 u3 D 33=2

y u2

y u3

y u1

u1 C

u2 C

u3

u1 u1

u2 u 2

u3 u3

11

12

33

u1 C

u2 C

u3

11

6

33=2

D u1 2u2 2u3

D

Noticehoweasyitistocomputetheweightsneededtobuild y fromanorthogonal

basis. Ifthebasiswerenotorthogonal, itwouldbenecessarytosolveasystemoflinear

equationsinordertoﬁndtheweights, asinChapter 1.

Weturnnexttoaconstructionthatwillbecomeakeystepinmanycalculations

An Orthogonal Projection

Givenanonzerovector u in Rn , considertheproblemofdecomposingavector y in Rn

intothesumoftwovectors, oneamultipleof u andtheotherorthogonalto u. Wewish

towrite

y D yO C z

(1)

340

CHAPTER 6

z = y – yˆ

0

Orthogonality and Least Squares

y

yˆ = ␣u

u

FIGURE 2

Finding ˛ tomake y

orthogonalto u.

yO

where yO D ˛ u forsomescalar ˛ and z issomevectororthogonalto u. SeeFig. 2. Given

anyscalar ˛ , let z D y ˛ u, sothat(1)issatisﬁed. Then y yO isorthogonalto u ifand

onlyif

0 D .y ˛ u/ u D y u .˛ u/ u D y u ˛.u u/

y u

y u

Thatis, (1)issatisﬁedwith z orthogonalto u ifandonlyif ˛ D

and yO D

u.

u u

u u

Thevector yO iscalledthe orthogonalprojectionofyontou, andthevector z iscalled

the componentofyorthogonaltou.

If c isanynonzeroscalarandif u isreplacedby c u inthedeﬁnitionof yO , thenthe

orthogonalprojectionof y onto c u isexactlythesameastheorthogonalprojectionof y

onto u (Exercise 31). Hencethisprojectionisdeterminedbythe subspace L spanned

by u (thelinethrough u and 0). Sometimes yO isdenotedby projL y andiscalledthe

orthogonalprojectionofyonto L. Thatis,

yO D projL y D

y u

u

u u

(2)

Ä

7

4

and u D

. Findtheorthogonalprojectionof y

6

2

onto u. Thenwrite y asthesumoftwoorthogonalvectors, onein Span fug andone

orthogonalto u.

EXAMPLE 3 Let y D

Ä

SOLUTION Compute

y uD

u uD

Ä

Ä

Ä

7

6

Ä

4

2

4

2

D 40

4

2

D 20

Theorthogonalprojectionof y onto u is

Ä

y u

40

4

yO D

uD

uD2

2

u u

20

andthecomponentof y orthogonalto u is

Ä

Ä

7

8

y yO D

6

4

D

D

Ä

Ä

8

4

1

2

Thesumofthesetwovectorsis y. Thatis,

Ä

Ä

Ä

7

8

1

D

C

6

4

2

"

y

"

yO

.y

"

yO /

Thisdecompositionof y isillustratedinFig. 3. Note: Ifthecalculationsaboveare

correct, then fOy; y yO g willbeanorthogonalset. Asacheck, compute

Ä

Ä

8

1

yO .y yO / D

D 8C8D0

4

2

SincethelinesegmentinFig. 3between y and yO isperpendicularto L, byconstructionof yO , thepointidentiﬁedwith yO istheclosestpointof L to y. (Thiscanbeproved

fromgeometry. Wewillassumethisfor R2 nowandproveitfor Rn inSection 6.3.)

6.2

x2

Orthogonal Sets

341

y

6

L = Span{u}

3

y – yˆ

u

1

8

x1

FIGURE 3 Theorthogonalprojectionof y ontoa

line L throughtheorigin.

EXAMPLE 4 FindthedistanceinFig. 3from y to L.

SOLUTION Thedistancefrom y to L isthelengthoftheperpendicularlinesegment

from y totheorthogonalprojection yO . Thislengthequalsthelengthof y yO . Thusthe

distanceis

p

p

ky yO k D . 1/2 C 22 D 5

A Geometric Interpretation of Theorem 5

Theformulafortheorthogonalprojection yO in(2)hasthesameappearanceaseachofthe

termsinTheorem 5. ThusTheorem 5decomposesavector y intoasumoforthogonal

projectionsontoone-dimensionalsubspaces.

Itiseasytovisualizethecaseinwhich W D R2 D Span fu1 ; u2 g, with u1 and u2

orthogonal. Any y in R2 canbewrittenintheform

y u1

y u2

yD

u1 C

u2

(3)

u 1 u1

u2 u2

Theﬁrsttermin(3)istheprojectionof y ontothesubspacespannedby u1 (theline

through u1 andtheorigin), andthesecondtermistheprojectionof y ontothesubspace

spannedby u2 . Thus(3)expresses y asthesumofitsprojectionsontothe(orthogonal)

axesdeterminedby u1 and u2 . SeeFig. 4.

u2

yˆ 2 = projection onto u2

y

0

yˆ 1 = projection onto u1

u1

FIGURE 4 A vectordecomposedinto

thesumoftwoprojections.

Theorem 5decomposeseach y in Span fu1 ; : : : ; up g intothesumof p projections

ontoone-dimensionalsubspacesthataremutuallyorthogonal.

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