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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, ...
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