Applied Multivariate Statistical Analysis 6th Edition Johnson Solutions Manual

2.1
26
Chapter 2
a)
I,
; i
I
... i : ; ,
,
----. . i I
: i . I
i
, ~
: :
. , I
oJ /l r ,
, .
... ------I
!
~
I
. ¡
, , , , I . ,
-A : t- ~ ..1'-, ø ':, . - ,i :=-: -j.1 3 d~j =
,:': ~ ~, ,,, . .,;' .,.. .: _ __ ~.; " ,1 '..,:/ ,1.. . . . /" 1-.~~ /. -~'"'-.A "'7' .! . _ ... i I .
: 'JO " : i i ; , , ,
../ :
7.
./
: '7K (-~ '~ =, -ii;;;;;.I~
l-' : I ,; I . ¡ ; ;
I I
i i :
~. . , I
. . : ,
! i I
i
I
,
: i
I ; ,
I i ,
I I , I
, , :
i !! I i : I
.
; '.,;! i/'i : g
'-' ,ILl / i-tI ! i i
,
, , 1
! I
I
. J :
.
I ,t
i I;
! : i I i : :, 'i; " I I
./
i ,,/_.
== i i I
b) i) Lx = RX = Iß = 5.9l'i
il) cas(e)
x.y 1
.051= ..= =
LxLy 19.621
- -
e = arc cos ( .051 ) ;, 870
; 1 i) proJection of L on x ; s
lt~i x = i
x
(1 1 31'is1is
3š = 7~35'35
c)
':
: i
I . ! I , :
. i
, I
= i i : l : i i
02-2: i.
!- .
i 'i
, i
I i
:i :3
~
~
. 1
I
;
I
.
;-- I 1
I i
. I
::ti I
i'
.1
i,
:i:-'.i
.1 ..
"T
~-:~~-';'-i-' .~._~-:.i" 1 .' :.. :. -"-- --_..-- ---
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Applied Multivariate Statistical Analysis 6th Edition Johnson Solutions Manual
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27
( -~
15)
r-6
-q2.2 a) SA = b) SA = - ~
20 1 a -6
(-1 :
-9
-: )
c) AIBI = d) C'B = (12, -7)
-1
e) No.
. (~
1). A i
2.3 a) AI so (A I) = A' = A
3 .
b) C'
.(: :l (C'f"l~ J)1 a 10
i2 4 J
(C' J' 'l- 1~
il). (t''-'-1 -ìa lõ,c =
3 1 iiÕ -iÕ ' 10 -Tõ
c)
(1:
7)'
U
8
': )
(AB) , = =
4 11 11
B'A' =
(i n (~ ~)
-
(~ ':)
= (AB) i
11
d) AB has (i ~j )th entry
k
a,. = a"b1, + a'2b2' +...+ a,,,b,,, = i aitb1j1 J 1 J 1 J 1 J R.=1
Consequently, (AB) i has (' ,)th .1 ~J' entry
k
c , , =
I ajR,b1i ,
Jl 1=1
Next ßI has .th row (b, i ,b2i ~'" IbkiJ and A' lias. jth1Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

28
column (aji,aj2"",ajk)1 so SIAl has ~i~j)th entry
k
bliaji+b2ibj2+...+bk1~jk = t~l ajtb1i = cji
51 nce i and j were arbi trary choices ~ (AB) i = B i A I .
2.4 a) I = II and AA-l = I = A-1A.
and 1= (A-1A)' = A1(A-l)l.
of Al or (AI r' = (A-l)'.
bl (S-lA-l)AS _ B-1
(f1A)B - B-1S' I so AS has inverse (AS)-1 ·
I
B-1 A- i. It was suff1 ci ent to check for a 1 eft inverse but we may
also verify AB(B-1A-l) =.A(~Bi~)A-i = AA-l = I ,
Thus I i = I = (AA - ~ ) I = (A-l)' A,I
Consequently, (A-l)1 is the inverse
2,6
¡s 12l r _121 r
1 :J .l:
~1
IT IT IT 13 = 1 69 = QIQ ,
QQI
=
-12 5 12 5
13 i3 IT i3 a 169
a) 5i nce A = AI, A' is symetric.
b) Since the quadratic form
x' Ax . (xi ,x2J ( 9
-:)(::1-
9xi - 4x1 X2 + 6X2
- - .. -2
2.5
~ (2Xi.x2)2 + 5(x;+xi) ~ 0 for tX,lx2) -l (O~O)
we conclude that A is positive definite.
2.7 a) Eigenvalues: Ål = 10, Å2 = 5 .
Nonnalized eigenvectors: ':1 = (2/15~ -1/15)= (,894~ -,447)
~2 = (1/15, 2/15) = (.447, .894)

29
b) A' V
-2 ) . 1 fIlS r2/1S. -1//5 + 5 (1/1S1 (1/IS, 2//5
-2 9-1/~ 2/~
c)
-1 1
(:
2) . (012 0041
A = 9(6)-( -2)( -2) 9 ,04 .18
d) Eigenval ues: ll = ,2, l2 = ,1
Normal;z~ eigenvectors: ;1 = (1/¡;~ 2/15J
;z =: (2/15~, -1I/5J
2.8 Ei genva1 ues: l1 = 2 ~ l2 = -3
Norma 1; zed e; genvectors: ;~ = (2/15 ~ l/~ J
A · (:
=~ = (1/15. -2/15 J
2) = 2 (2//5) (2/15, 1/15J _ 3( 1/1S)(1//s' -2/151 '
-2 1/15 -2/~
2.9 ) -1 1 (-2 -2) =i1131 11a A = 1(-2)-2(2) -2 1 - --
3 6
b) Eigenvalues: l1 = 1/2~ l2 = -1/3
Nonna1iz.ed eigenvectors: ;1 = (2/ß, l/I5J
cJ A-l =(t
;z = (i/ß~ -2/I5J
11 = 1 (2/15) (2/15, . 1//5J _ir 1/15) (1//5, -2/ß1
-1 2 1/15 3L-21 5

30
2.10
B-1 _ 1 r 4.002001 -44,0011- 4(4,D02001 )-(4,OOl)~ ~4,OOl .
( 4,OÒZOCl
= 333,333
-4 , 001
~.0011
1 ( 4.002-1
A = 4(4,002)~(4,OOl)~ -4,001
-: 00011
. ( 4.002
= -1,000,000
-4 , 001
-: 00011
Thus A-1 ~ (_3)B-1
2.11 With p=l~ laii =
aii- and with p=2,
aaii
a
= a11a2Z - 0(0) = aiia22
a22
Proceeding by induction~we assume the result holds for any
(p-i)x(p-l) diagonal matrix Aii' Then writing
aii a a
A = a
. Aii
(pxp)
.
.
a
we expand IAI according to Definition 2A.24 to find
IAI = aii I
Aii I + 0 + ,.. + o. S~nce IAnl =, a2Za33 ... ~pp
by the induction hypothesis~ IAI = al'(a2Za33.... app) =
al1a22a33 ,.. app'

31
2.12 By (2-20), A = PApl with ppi = pip = 1. From Result 2A.l1(e)
IAI = ¡pi IAI Ipil = ¡AI. Since A is a diagonal matrix wlth
diagonal elements Ài,À2~...,À , we can apply Exercise 2.11 to
p pget I A I = I A I = n À , .
'1 11=
2.14 Let À be ,an eigenvalue of A, Thus a = tA-U I. If Q ,is
orthogona 1, QQ i = I and I Q II Q i I = 1 by Exerci se 2.13. . Us; ng
Result 2A.11(e) we can then write
a = I Q I I A-U I I Q i I = I QAQ i -ÀI I
and it follows that À is also an eigenvalue of QAQ' if Q is
orthogona 1 .
2.16 (A i A) i = A i (A i ) I = A i A
Yl
Y = Y 2 = Ax.
show; ng A i A ; s symetric.
Then a s Y12+y22+ ,.. + y2 = yay = x'A1Axp _.. .. ..
yp
2.18
and AlA is non-negative definite by definition.
Write c2 = xlAx with A = r 4 -n1. Theeigenvalue..nonnalized
- - tl2 3
eigenvector pairs for A are:
Ài = 2 ~
'=1 = (.577 ~ ,816)
Å2 = 5, ':2 = (.81 6, -, 577)
'For c2 = 1, the hal f 1 engths of the major and minor axes of the
elllpse of constant distance are
~ = -i = ,707 and ~ =.. = .447
~1 12 ~ ~respectively, These axes 1 ie in the directions of the vectors ~1
and =2 r~spectively,

32
For c2 = 4~ th,e hal f lengths of the major and mlnor axes are
c 2 '- = - = 1.414 and
ñ:, .f
c _ 2 _-- - -- - .894 .
ñ:2 ' IS
As c2 increases the lengths of, the major and mi~or axes ; ncrease.
2.20 Using matrx A in Exercise 2.3, we determne
Ài = , ,382, :1 = (,8507, - .5257) i
À2 = 3.6'8~ :2 = (.5257., .8507)1
We know
__(' .376
A '/2 = Ifl :1:1 + 1r2 :2:2
,325
,325)
1. 701
A-1/2 = -i e el + -- e el _ ( ,7608
If, -1 -1 Ir -2 _2 ~ -,1453
- .1453 J
.6155
We check
Al/ A-1/2 =(:
~) . A-l/2 Al/2

2,21 (a)
33
A' A = r 1 _2 2 J r ~ -~ J = r 9 1 J
l1 22 l2 2 l19
0= IA'A-A I I = (9-A)2- 1 = (lu- A)(8-A) , so Ai = 10 and A2 = 8.
Next,
(b)
U;J ¡::J
¡ i ~ J ¡:~ J
10 ¡:~ J gives
- (W2J
- ei - . 1/.;
8 ¡:~J gives
¡ 1/.; J- e2 = -1/.;
AA'= ¡~-n U -; n = ¡n ~J
12-A 0 4o = /AA' - AI I - .1 0 8 - À 0
4 0 8-A
= (2 - A)(8 - A)2 - 42(8 - A) = (8 - A)(A -lO)A so Ai = 10, A2 = 8, and
A3 = O.
(~ ~ ~ J ¡ ~ J - 10 (~J
.gves
4e3 - 8ei
8e2 - lOe2
so ei= ~(~J
¡ ~
0
~ J ¡ :: J
8 (~J8 -
0
gives
4e3 - Gei
so e,= (!J
4ei - U
Also, e3 = 1-2/V5,O, 1/V5 J'

C)
34
u -~ J - Vi ( l, J ( J" J, 1 + VB (! J (to, - J, I
2,22 (a)
AA' = r 4 8 8 J
l 3 6 -9
r : ~ J = r 144 -12 J
l8 -9 L -12 126
o = IAA' - À I I = (144 - À)(126 - À) - (12)2 = (150 - À)(120 - À) , so
Ài = 150 and À2 =' 120. Next,
r 144 -12) r ei J = .150 r ei J
L -12 126 L e2 le2
. r 2/.; )gives ei = L -1/.; .
and À2 = 120 gives e2 = f1/v512/.;)'.
(b)
AI A = r: ~ J
l8 -9
r438 8J
l 6-9
- r ~~ i~~ i~ J
l 5 10 145
25 - À 5050= IA'A - ÀI 1= 50 100 - À 10 = (150 - A)(A - 120)A
5 10 145 - À
so Ai = 150, A2 = 120, and Ag = 0, Next,
¡ 25 50 5 J
50 100 10
5 10 145
r ei J' r ei J
l :: = 150 l::
gives
-120ei + 60e2 0 1 ( J
O or ei = 'W0521-25ei + 5eg VùU
( ~ i~~ i~ J5 lD 145 ( :~ J = 120 (:~ Jeg e2Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

35
gives -l~~~ ~ -2:~: ~ or., = ~ ( j J
Also, ea = (2/J5, -l/J5, 0)'.
(c)
(4 8 8J3 6 -9
= Ý150 ( _~ J (J. vk j, J + Ý120 ( ~ J (to ~ - to J
2.24
( 1
a
n
À1 = 4, =l=('~O,OJ'
;-1 = ~a)
1
b) À2 = 9 ~
=2 = (0,1,0)''9
a À3 = 1, =3 = (0,0,1)'
c) ~-l
For + : À1 = 1/4,
À2 = 1 /.9,
À = 1,3
':1 = (1 ,O,~) i
':2 = (0 ~ 1 ,0) ,
el = (OlO~l)1
-3

36
2.25
Vl/2 "(:
a OJ ( 1 -1/5 4fl5J
il
-.2 .26~
a) 2 o 'if.= ,-1/5 1 1/6= - 2 1 .1'67
" ~:i'67
a 3 4/15 1/6 1 ' 1 67 i
b) V 1/2 .e v 1/2 =
(:
0 0Jt 1 -1/5 4flJ (5 ° OJ i5 -1 4/3) (5 a
:J
2 a -1/5, 1 1/6 0 2 ° = -2/5 2 1/3 a 2
a 3 4/15 1/6 1 a a 3 4/5 1/2 3 a 0
= (~:
-2
4
1 n =f
2.26 a)
1/2 i /2
P13 = °13/°11 °22, = 4/13 ¡q = 4l15 = ,2£7
b) Write Xl = 1 'Xl + O'X2 + O-X3 = ~~~. with ~~ = (1 ~O~O)
1 1 i , i 1 12 x2 + 2 x3 = ~2 ~ W1 th ~2 = (0 i 2' 2" J
Then Var(Xi) =al1 = 25. By (2-43), ~
1X 1X ,+ 1 2 1 .19Var(2" 2 +2" 3) =':2 + ~2 =4 a22 + 4 a23 + '4 °33 = 1 + 2+ 4
15
= T = 3.75
By (2-45) ~ (see al so hi nt to Exerc,ise 2.28),
1 1 i 1 1Cov(X, ~ 2Xi + 2 Xi) = ~l r ~2 = "'0'12 +"2 °13 = -1 + 2 = 1
~o

2.27
37
1 1Corr(X1 ~ '2 Xl + '2 X2) =
, 1COy(X" "2X, + '2X2) 1
~r(Xi) har(~ Xl + ~ X2) =Sl3 :=
.103
a) iii - 2iiZ ~ aii + 4a22 - 4012
b) -lll + 3iZ ~ aii + 9a22 - 6a12
c) iii + 12 + 13' aii + a22 + a3i + 2a12 + 2a13 +2a23
d) ii, +~212 -. 13, aii' +~a22 + a33 + 402 - 2a,.3 - 4023
e) 3i1 - 4iiZ' 9a11 + 16022 since a12 = a .

2,31 (a)
38
E¡X(l)J = ¡,(l) = ¡ :i (b) A¡,(l) = ¡ 1 -'1 1 ¡ ~ J = 1
(c)
COV(X(l) ) = Eii = ¡ ~ ~ J
(d)
COV(AX(l) ) = AEiiA' = ¡i -1 i ¡ ~ n ¡ -iJ = 4
(e)
E(X(2)J = ¡,,2) = ¡ n tf) B¡,(2) (~ -iJ ¡ n = ¡ n
(g)
COV(X(2) ) = E22 = ¡ -; -: J
(h)
COV(BX(2)) = BE22B' = ¡ ~ -~ J (-; -: J (-~ ~ J - (~: -~ J
0)
COV(X(l), X(2)) = ¡ ~ ~ J
(j)
COV(AX(1),BX(2))=AE12B'=(1 -1) ¡~ ~J ¡ _~ n=(O 21

2,32 ~a)
39
EIX(l)j = ILll) = ¡ ~ J (b) AIL(l) = ¡ ~ -~ J ¡ ~ J = ¡ -~ J
(c)
Co(X(l) ) = En = l-i -~ J
td)
COV(AX(l)) = AEnA' = ¡ ~ -¡ J ¡ -i -~ J L ~~ ~ J - ¡ i ~ J
(e)
E(Xl2)j = IL(;) = ( -~ J (f) BIL(2) = ¡ ~ ; -~ J ( -~ i = ¡ -; J
(g)
( 6 1 -~1 i
COV(X(2) ) = ~22 = 1 4
-1 0
(h)
COV(BX(2) ) = BE22B' ,
= U i -~ J (j ~ -~ J U -n -
¡ 12 9 J
9 24
0)
CoV(X(1),X(2)) = ¡ l ::J ~ J
(j)
COV(AX(l) i BX(2)) = AE12B'

2,33 (a)
40
- U j J H =l n (¡ j J - l ~ ~ J
E(X(l)j = Li(l) = ( _~ J (b) Ati(l) = L î -~ ~ J ( _~ J - ¡ ~ J
(c)
( 4 i 6-i~J
Cov(X(l¡ ) = Eii = - ~ - ~
(d)
COV(AX(l) ) = Ai:iiA' ,
= (î -~ ~) (-¡ -~!J (-~ n -
¡234)
4 63
(e)
E(X(2)J = ti(2) = ¡ ~ ) (f) Bti(2) = ¡ ~ -î J ¡ ~ J = I ; )
(g)
Co( X(2) ) = E" = ¡ ¿ n
(h)
CoV(BX,2) ) = BE"B' = U - î ) L ¿ ~ J D - ~ J - I 1~ ~ J

(i)
41
( _1 0 J
COv(X(1),X(2))= -1 0
1 -1
ü)
COV(AX(l), BX~2)) = A:E12B1
= ¡ 2 -1 0 J (=!O J
1 1 3 i 01 -1 ¡ ~ - ~ J = ¡ -4,~ 4,~ J

42
2.34 bib = 4 + 1 + 16 + a = 21, did = 15 and bid = -2-3-8+0 = -13- -
(ÉI~)Z = 169 ~ 21 (15) = 315
2.35 bid = -4 + 3 = -1
- -
biBb = (-4, 3)
L: -:J (-~ J
= (-:14 23)
( -~ J · 125- -
( 5/6
2/6 ) il )
d I B-1 d = (1~1) 2/6
= 11/62/6 1
so 1 = (bld)Z s 125 (11/6)" = 229.17- - '
2.36 4x~ + 4x~ + 6xix, = x'Ax wher A = (: ~).
(4 - ).)2 - 32 = 0 gives ).1 = 7,).2 = 1. Hence the maximum is 7 and the minimum is 1.
2.37 From (2~51), max
x'x=l- -
X i Ax = max
~fQ
~ 'A!
~13
= À1
where À1 is the largest eigenvalue of A. For A given in
Exercise 2.6, we have from Exercise
2.7 ~ Ài = 10 and
el . (.894, -,447), Therefore max xlAx = 10 andth1s
-1 x I x Fl
maximum is attained for : = ~1.
2.38
Using computer, ).1 = 18, ).2 = 9, ).3 = 9, Hence the maximum is 18 and the minimum is 9,

2.41 (8) E(AX) = AE(X) = APX = m
(b) Cav(AX) = ACov(X)A' = ALXA' = (~
43
o OJ
18 0
o 36
(c) All pairs of linear combinations have zero covarances.
2.42 (8) E(AX)
= AE(X)= Apx =(i
(b) Cov(AX) = ACov(X)A' = ALxA' = ( ~
o OJ
12 0
o 24
(c) All pairs of linear combinations have zero covariances.
Applied Multivariate Statistical Analysis 6th Edition Johnson Solutions Manual
Full Download: http://alibabadownload.com/product/applied-multivariate-statistical-analysis-6th-edition-johnson-solutions-manual
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