This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: EXERCISE SET T.1 I] 12. The plane 1‘ + y — 42 = {l corresponds to oi Where a = [1, 1,—4}. Thus1 writing o as a column vector1
the standard matrix for the reﬂection oi R3 about the plane is 1 1
1 c c 2 1 1 —4 g —§ §
H=I—iaoT= 111 c —— 1 1 4 = —g g 3
Hire 13 4 4 15 i 1,
ﬂ '3 1 ' ‘ 3 t ‘t
nndthe reﬂection of the t'ectorh=l[l.ﬂ,l} about that pinneisgiren.incolmnn form, by
a 1 1 1
ﬁ ‘5 9 1 3
Hh= t t t[0 = t
1
1 ti 1 —3
loan 1—2—3—1 §1§§ﬁ
1.13+ H=;__2_MT= 91“ ﬂ 2 ‘2 4 '3‘ 2 = ﬁ % ‘E “1%
are I} [I I I] 15 —3 E El 3 g _g _% _§
0 {II [I l —1 2 3 l g _i _g 1.1
15 15 t T5 24. Referring to the construction in Exercise 13 {a}. the second entrg,r in the ﬁrst column of A can he
zeroed out: by muttiplying by the orthogonal matrix Q1 as indicated below: J§1§icclz1ﬁiﬂ¥—ﬁ Q.A=1¥_}2§jﬂ “134: "—3? 349‘
71— "IJ—Fi—"ﬁ ‘3' ‘1 '3 o —1 11
o o I 11 1 ‘3' '3 1 o o 1 From a similar construction, the third entry in the somnd column of QM can be zeroed out by
multiplying by the orthogonal matrix It“); as indicated below: 11___9_____0_1_11 e to j n m J: I I 2 2 2
_ﬂ _ﬁ
ogre.11: Di 3 3 in 0 J? 1‘? = ‘3' 2“ §
ﬂ:_—_£___l_§—3:‘j_ti o —1 o o c wﬁ
IT. 11 c l' 1 [I I] 1 11 c 1 From a third such construction, the fourth entry in the third colonin of {22122114 can be zeroed out by
multiplying by the orthogonal matrix Q; as indicated below: lolocﬁn§_ﬂ ﬁﬂ_142 2 2 “llﬂﬂﬂﬁ_ﬂi Glenn Q3Q2£11A= ""T """" " 2 2 = 2 2 =3
” “E4? % o 11 «Ed 11 c 2
c 11: g e o c 1 c c 11
Finally, netting t? = ol‘ogog‘ = ()1 (321123, we obtain the following QR—dooomposition of A:
1 2 1 ’32 “ti t4? .jﬁ e312 _J2ﬂ
A“132_”§?%’§ uo_e_qﬂ
31 3 u at a; e u a 2
[l
D [I i ii} I] I] [I 25. Since A 2 (JR. the system Ax = h is equivalent. to the upper triangular system R1: = QT]; whieh is: VIE 'Vﬁ 3:55 $1 $15 i 315 3 I?! 4 e a it; I: A? .ﬁ 713 e :1;
Solving this system by hack substitution yields 13 = 11:: = Let = 1.
D3. If s = iV'ﬁ. then w = v and the Hmlseholder reﬂection about {1; — wji maps 1‘ into w. DE. Let a=v—w={1.2,2}  {ﬂ.lJ,3]= [1,2.u1:l. Then thereﬂeotion ofﬂasbout all maps? into
mandtheplaneai mp0mlatua=+2y—z=ﬂurz=m+2y. P4. If A = QR is a QR—deenmpesltimr of A, then t? 2 AR“. From this it Inﬂows that the oolmnns of Q belong to the column space of A. In pertieuler1 if R“ = [3.1], then from Q 2 Ali—1 it follows
that CalQ} = chlﬂ_1l = a1.1i'31l4‘ll‘l' Sﬂjc‘llAll + ' ' ' + Sweet!” for eachj = 1, 2,. .. ,Jr. Finally. since dimfeoHAJJ = .l: and the vectors c1ﬁQ},cg{Q}, .. .1ekIfQ) are
linearle independent, it follows that they form a basis for eol{A] EXERCISE SET 7.11 12‘ {a} We ham [1: —2‘.'1 + V2 +2V3 = {%l—§1 —g} and V = 3v1 +ﬂvﬁ _2v3 = [—%l %lg]' Eb] II = Illuisll = tit2113 + {1}” +£21i= 3. IIVII = IIEVJsli = «[313 + {'1}E + {255 = m1 and u v = £qu  Evils = t—EJ[31+{1}{e]+{2}{2i= —1e. H..u= {%J“+I{§i“+{§F= ¥=suvu tft—%l=+f1—;’F+£t1*=\/li_i=~’ﬁr and uv= {tn—s} +{—§}E%}+ {—31:51 = e = —1u. 17' {all We have V: = 231 + 62 and V2 = —3e: + sea; thus 135...; =[{v1}3 [em] = [2 ‘3], I 4
_ l i .35.
(h) The row reduced echelon form. ol'IBiSJ = [1’ is '3; 111 14311.15 phi. =
I_ﬁ H i 1
...2L 1
II III
_ l
[c] Netsthur[PH,.Sg—1=[i 3] =i1_][_::=psw_
—1 (a: warm.va 42. thus [w13=[ has [sweeten = [f '3'] Li] = [3] [e] We have w = 3121  seq; thus [w]3 = Li] and [w]e = E’s—telwls = Li = 2a. {a} We have in = (ms rs, sin 29} and v: = [sin rs. —cns 29}; thus Pans = “j:
—'EUE
[b] If P = Par—.3 then. sinee F is orthogonal. as have PT a P‘1 = (PB—,SJ'1 = F’s—“s. Creme:—
rieally, this corresponds to the feet that reﬂection. about the line preserves length and thus is
an orthogonal transformation. .1:' £3?” a: I" "I 0—315 1'
31. WE have 1,." 1: _y"_5 1 1]. y and y" = {I I. I] y“ Thus
'1 15
1?" 1'35 0 ‘31s % I; '3 x Jig 3% “‘1? :e
_ _ 3
yjje'313—t—iéﬂvsitnv
e " s n n 1 I e e «s z D2. [3} {b} P2. The vectors VI’VEI _, Let B = {v1.v21va}. where v1 = [1,1,0]. v: = {1,0,2},va = {D.2,1} correspond to the col‘
umn motors of the matrix P. Then, from Theorem 7.11.3, F is the transition matrix from
B to the standard. bseis S = {31,eg,ea}. If P is the transition matrix from S 2 {eke3,33} to B = {whormwg}, then e1: W1+ W2.
B2 = w] + 2W3. and e; = Eur: + “'3. Solving these vector equations for W1, wg, and W3 in
terms ofel, eg, and on results in W1=EB1+%E:1 — 3e; = (E, %, —§ ,w: =§e1w§ea + geg
= {%,—%.%j. and w; = —§el + go: 4 gag = {—g, g. é]. Note that w1. w2, W3 correspond to the column vectors of the matrix P‘ . ..vk spmmﬁmﬁonlylfeveryveemrv mmembeexpressedeealjnesr oombiuatioo of them, i.e.. there exist scalars :11 c2, . . . . ck such that v = elvl + eye; +    + (2ka
banner {tn)5 = cﬁvljg + oﬂvﬂg +  u + ckivkjg and the coordinate mapping v —} {vja is onto it follows that the vectors v11v;:,..._.v;¢ span H“ Home] only if {v.3 B1{VE}B;J{Vk]3 span R“. CHAPTER 3
Diagonalization EXERCISE SET 8.1 3 —2 l I]
22. [a] [TV1]Ev= l ,, [TVglﬁv = ﬁ , [ﬁ3]Br= 2 r and [TV¢]3= 1 _
—3 (l T l
I] —? —ﬁ 11
[b}Tvl=3V;+1rE—3ﬁ=3 s + a —3 9 = 5,
B 1 l 22
{I —'i' —6 —42
TV2=2 3 Hi 3 +12} 9 = 32 ,
S l —lEI
U —T 6 —5E'r
Tv3=13 +2 3 +7 9 _ 31' ,and
El 1 1 1?
CI —7 —ﬁ —13
TV4=D 3 +1 3 +1 9 = 1? .
B l l 2 [c] For every vectur x in R”, we have
3‘ ={EI1 41:: — %H + lflxdl‘i'l 44351" 23: — %13 + gnh:
+ {—EII + gill:  £13 — émalva + {will + g3: + £533 — Emil“;
Thus, using the linearity of '1". it fullmrs 111m;
T1: = {gm — 4.2:; # éxa + %z¢}Tv1 +{Em1— 212 — h; + gmlﬂw
+ {—331 + 3x3  En — éanva + {—%:l + Ex: + ﬁrs, — gnﬂn which leads to the follmn'ng formula for T: $1 253. as 2H _ 29
I ‘ﬁFl + 132+ W13 E1314
2 .
TX=T I = I—FIl—mz—g—ﬁ33+%14
3 1," 55.11 — 501;  9315.3. + 911‘ {cl} Using the formula obtained in. part {c}, we have TEE}. Dull] = {—31,3T,12}_ 1 {II [I
24, If Tia the identity operator then [1"] = [1"]5 = [T] g. = In 1 :11. On the other hand1 we have TV] =
[I [I 1 3 ‘3 T 9 [I
19 _ _ 291: my 229
4 =% 3+ﬁ—5 —l—3éﬂ =§Vi+ﬁVI24ﬁVE~TV2—5 *mv;+Fvamvai
1 l 5 3 2
3 E E E3
= _ 1? r 119 a 115 . 35 1“
and TVs. H — 7331:, + Eva — EVE, thus [T1313 2 3;: % % 1
6 _E #229 .15
an m T 25+ L‘Et B = {V1 V2 .. . V1.1} 3.11.11 3" = {111 “2 u be bases fur Rn and Rm _
r r l I 1 _ _ _ J m 
1”» is the mm trmafﬂrmmiﬂn’ we have m} Iespe wcly Them Tﬁl=ﬂ=ﬂu1+ﬂu2+.._+ﬂum for each 1' = 1.2,. . .111. Thus [113:3 = [0[} lﬂ] 1's thezeru matrix. n —r u :53: 5'25 ‘3‘: 3]. We have T‘s": t] D 12 —ﬁ = —§5 =21“. Tv;== —ﬁa =—V3 +f3v31 and
'1' a e e 1 ‘3' '3’ T193: §E 2—ﬂv2—v3. Thue[T5= t] —1 —ﬁ =2 e —% 4? .Ft'cum thisweeeethet
73., 0 1’5 l 0 ti —t the effect of the operator T is tn retete teeters counterclockwise by an angle of 120 degrees about
the v1 axis {looking toward the origin From the tip et' in}, then stretch by a factor of 2. 134' {a} TWE we have [11(me = [Ttlﬂuﬂixiﬂ = [Tele'eixiﬂ = [Tzixlleri 1311113 Tlixi = Tzixia
{b} False. Fm' example, the zero nperater has the same matrix {the zero matrix} with Deepest to
any basis for R2.
[e] True. lfﬁ' ={v1,v2,....vn}and 1T13 = I. then The} = v]: for each in = 1,2,...111 and it
feilmes from this that. TIII} = x for all x.
{d} FhJee. Fer example, let B = {61,32}, 3’ = {133,131}, and Thu} 2 {1.3.3}. Then [Tram = I2,
but T is not the identity upereter. P3. If at is a vector in R“. then [T]3[x]a = [Tx]g = [I if and only if T): = 0. Thus, if T is oneto
nne. it fallews that [Thatch = D if end em];r if [x]3 = 6, Le. that [TIE is an invertible matrix.
Furthermnre, since [T_113!T}3 = [1“1 eﬂg = [NH = f, we haW[T_1]5 = [TBr1, ...
View
Full Document
 Spring '08
 anony
 Glenn, Jr., P‘1, rieally

Click to edit the document details