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Unformatted text preview: l l l _ 7 l U 0 21
16‘ We reduce l 2 3:1 to 0 1 0E —22 . ‘
l 3 6 3 0 U l 8 w r 21 '1
revealing, that 5: z 2117; — 22172 + 853. and [T15 = [—22J ,
8 1
2 Thenstquzllz 44142151014”
5—214321510—521” 125 o 50 50—25”0—5‘ b. Our commutative diagram: —2
1 ], and we ﬁnd the inverse .5"1 m be equal to g [ 1 2} 22 a.S=[
~21 :E:c, +L~2 T T(i)=Ai=c1A{;]+c2A[~12]
=c] +02 :50, “502 ms = [2;] 7’ {mm = [El
so a b c; ' 5c, d l 5 0
, C d Q ~ _5C2 ,au we quickly finde [0 "5]. :HélL n23 éllﬂLl
=H$L {1%Ll={3 .2} 30. Let‘s build B “columnvlwr'olulnu": 1'3 = HTW.)]u[T(ﬁz)Js{T(m)lsl ll? :2 75] HE [E :2 “all [Ella
{EL El. [E3 37. We want a basis 5 :2 (17“172) such that T070 = m". and T('Eg) z: (1172 for smm' scalars
u u and b. Then the B—matrix of T will he B = [[T('i7;)]5 [TUEHB] = 0 (I: ~ ll which is a diagonal matrix as required. Note that, T0!) = 17 r. 117 for vcctm‘s parallel to the
lino L (mm which we project, and T(1ﬁ) = 0 = 0147 for Vectors perpendicular to L.
Thus, we can pick a basis when: '17. is parallel to L and 172 is porpmlrlivulm'. for exmnplo. (Elm) AU, hum l‘lxurcisv 37, we see,
lmv, whilv the others must be perpemlk'ulm‘ the that we want one of our basis vectors to be parallel to the
line. We can easily ﬁnd such a hams: l I Wu will use the same approach as in Exerris 37 and 39, Any basis will) ’2 vemors in Llw
plnnv and one perpendicular to it will work nicely here! SO, lei, 171,172 be in the plane. 771 ml 0
run ln‘ 3 w and 17;; : ~2 (note that these must be independent). Then 173 should
0 l
lw perpendicular to the plane‘ We will use 173 2 1 ~tlie coefﬁcient vector. This is
2
3
perpendicular to the plane because all vectors perpendicular to 1 lie in the plane.
2
1 O 3
Sm Our basis is: 3 , —2 , 1
0 1 2
49 77 i 17 : ~2F. so that 'IF’ = 17.. 17‘ Le” [1 fit). 2L 0 :1? + 217, :50 that [575}5 =~ = 17+ 2117. 50 that = u (71‘? = 35+ 2m. See Figure 3.7. Figure 3.7: for Problem 3.450. (a If Lliv tip of '17 is a vertex, then so is the tip of 17+ 317 and éllﬁO the Lip of 17+ 3117 (draw
a sketclil). XV) know that. the tip P of 217+ (F is a vortex (see pay!» all Tliex’efnrv. Ll)“
tip 5 0163 : 1717+ 1313 = (221+ 15) + 5(317) + 40m) is a vertex as well. 56. Let S = {17, 172} where 171,172 is the desired basis. Then by Fan 3.4,], [1} = S and 2 5
3 w2. 32 13 1332”'12—
:5 . = u  '= ‘ ~ 7
l4] [3]” Sls [2 4 ' Hum” l2 4H5 :5] “[14 ~8l'
. .. i2 —7
Ti 1.: . v ‘
ie(esxredlmsism 14 , _8 . . , ‘ w 1‘ 1 .‘ ‘
6’2. We seek a basis ii, = ,172 = 31/ such that the matrix .5 z: [1); 1'2] 2 [ h satisﬁes r
the equation r— Solving the ensuing linear system We need to choose both 7. and t nonzero to make S invertible. For N wit: gives 3 : i: ‘ 1 ——1 _ 1 _ —1
wxainple. if we let 2 = 2 and t = 1, then b = [2 I], so that v, = [21.302 : [ 1]. Sean Ar. l l. Nul u Slilispm'vﬁinm‘ it (luvs not contain tliv neutralelement. that is, tlw fundinn f(t) = 0.
In] all I. 4. This sulisvt V is a subspace of P2: 1
t Th9 neutral claimant, f(1‘) = (l (for all f) is in V since / ()(lf 2 U.
u 1 i i l l
t If f and g are in V (50 tliat/ f =/ g = U) “1011 / (f +11) 2 / f + / g : (I. :m
u n n u . u that f + g is in V. A. chqud 1 x .1
o H ] 1.« in V (50 that/ f = 0) and I: is any constant, then / k] = k/ f = 0, so that
0 o 0 kfisinV. li‘f’t) 'bi+ct2t,he1/1f(t)di at+bfz+cf31 n+(’+C Uih b 6
=11 _ x = ﬂ, — = _ —: .=_____ ‘ + 0 2 3 0 2 3 2 3 ‘ , b c 2 1 2 1 '1he general elemem ofVm f(1‘): —§‘§ +bt+ct :b t—§ +c t ~§ ,so
1 ‘ 1 cl tt—.t2—'~b‘ fV. xa 2 315d 3h150 3X3
We. 36: \I <35} u?F.ur “\r’xcuxcbmkprk “KOM66 7% 0 Subspam cg: «Ex 3 Q 6
(km km: MMX [bng 13 x \]
60¢ ' \o. v; A: 6~ ‘5 kA: ta ma kn IN“ 14 . Yes 0 (0,0. 0% . . . .0, . , ) convvrges Lo 0,
~ IHim“.0L an. x O and mun»;C 31,, = 0, then 1i1n,...m(xt,. + y") : limuﬂat 1:" + linkwa y,. = 0, o If 13111,.“0L 17,. = 0 and k is any constant. then lim,._,,(k1',.) : Minn...“ 1:" = 0. .0 (l u 0 0 M 0 0
16. c d 2(100 +000 +c l U +(l (l 1 +r 0 0 +f 0 (l
U 0 U 0 1 0 O 1 The matrices O U , 0 O , 1 0 form a basis of
0 0 0 0 0 1 [Rizixzz9 so that dimming” : 6’ 1 0 (l 1 (i 0 (l (l U 0 0 O
(l , U
0 0 18‘ Any f in P,l can be written as a linear combination of 1,1312, , i l , t", by deﬁnition of P i
Also, 1, t, . V . ,1" are linearly independent: to see this consider a relaticin co + (511‘ + cur" = 0' since the polynomial co + (:11 + ‘ ‘ ~ + cut" has more than n zeros, we must have
co 2 c; 2 ~ n = c" z 0, as claimed. Thus, dim(P“) : n + 1‘ :52. We“whimringlkn‘thu“mm”.H H SW11 that i l u l; = a 11 2 ()
(( 11 ('{I (It, “0‘ (n‘ " ‘5 (i 11+ (I _ ‘24: (l . V
. 11+ ( [1+4 A «)(, U . ll n: rvqnn‘ml that u = C and b 2 “(L Thegenvmlclcninnl.is ['1 I, : J ‘l U l 1 U (l
u M” u l U +1: 0 *1 .Tlnis[1 “Jiu jIJisn basis. and tiln‘ (liinvnsinn is ‘J 48. if = «4 Jr I)! —l 112 44113 +m‘“. then f(—t) t: u — M + (1!2  (113+ ('l"1 and —j‘(——l) :
—n + bf — (#2 + (1'11" m“, a. f is even if f(~i) = fl!) for all t Cninpm‘ing mvﬁiuinnts wo ﬁnd that h = (I 1 (L so that
[(1) is of the {mm H!) = u + (12 + 0!". with basis 112.1% The (linu'nslon is 15‘ l). f is odd il' f(~t) : fU)‘ which is (,he (iﬂﬁl' if u z (' = t' 2 U, The ndrl pulyntnnials urn
()l' the form f(f) :2 I)! + {It}. with hush: 1.13 and dimension 2. 52‘ “"9 have in ﬁnd «unsuian u and l: snvh that tln‘ imn‘tinns 1"" and (1‘51 am solutinns
0f the (inferential equation f”([) —l uf’(;r) § hf(.r) : (l. Thus it is required Llialv C‘“ — 00*: + ba‘J : U, m 1 w a i (i : (l, and also Lhnl '25 v 50 + I) z 0. The mlntion of this system of twn equations in two nnknnwns is n 2 (3.?! z 5‘, so that the desired differential equation is fun) 4 (if/(1,) + 5/”) : (L s fniitwdimcnsional. 8 can be in nmsi n
ive n + l of infinite sequences 1 to Exercise. 57, they ‘
ntradiction: It IS easy to g 0 than the space V
n: solution But here is our CO 83511113 According to tl
its in V.
sequences, namely. .,(u,o,o,i..,oi1 66‘ Argue indirectly and with diniﬂ’) = n.
linearly independent eleinex linearly independent inﬁnite
.i),(0,1,o,0,...)‘(0,o,1,0,...), ..
he (n + l)i.h place. basis of V by omitting the re
ﬁnitedimensional, and, in fact, n.9,,“ .0, i i .); in the last salucncc (1,0,n,n,i
the l is in t we list 9! V V ‘ ' ~ gm, dant elements from ‘ ¥ h
dun ur basrs is a ‘ blll)’ 57‘ We can consiirnct a dim(v) S my Since 0 It. follows that V is '
list” of the original list ghl ...
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