This preview shows pages 1–4. Sign up to view the full content.
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: 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 Thenst_,AS=112 44142151014”
5—214321510—521” 125 o 50 50—25”0—5‘ b. On: 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 ,an we quickly finde [0 "5]. :HélL n23 éllﬂLl
=H$L {1%Ll={3 .2} 30. Let‘s build B “colunnkbyr'olulnn": 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 (funk) such that T070 = m". and T('Eg) z: (1172 for smno scalars
u u and b. Then the B—matrix of T will he B = [[T('ﬁ;)]5 [TM(3)15] = 0 (I: ~ ll which 17 r. 11'? for \'(K:t.<)l‘s parallel to the is a diagonal matrix as required. Note that, T('U)
F for vectors perpendicular to L. lino L (mm which we project, and T(1ﬁ) : 0 = 01
Thus, we can pick a basis when: '17. is parallel to L and 172 is porpmlrlivnlm'. for exmnplo. MW) 4”, hum l‘lxurcisv 37, we see, the).
lxnv, wlnlv the others must be perpendicular the t 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 with ’2 vemors in Lho
plnnv and one perpendicular to it will work nicely here! SO, let 171,172 be in the plane. 771 ml 0
mm 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 ~the coefﬁcient vector. This is
2
3
perpendicular to the plane because all vectors perpendicular to 1 lie in the plane.
2 3
Smunr basisis: 3 , —2 , 1
2 49 174 17: ~21". so that u" = _».7_ 17‘ Le“ [rags : [‘1] rm. 2L 0 :1? + 217, :50 that [575}5 =~ = 17+ 2117. 50 that = ‘2 n (71‘? = 35+ 2m. See Figure 3.7. Figure 3.7: for Problem 3.450. (a If th tip of '17 is a vertex, then so is the tip of17+ 317 and éllﬁO the Lip of 17+ 3117 (draw
a sketchl). XV) know that. the tip P of 217+ (F is a vortex (sec pan» all Therefurv. tho
tip 5 0163 : 1717+ 1313 = (221+ 23) + 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 w 2 . 3 2 1 3 1 '5 3 2 ”'
=5 2 ,. '  h 12 —7
[4] s[5 [2 4.Hg11<.eb—[2 HS 3] [M *8].
 . . 12 7
Tl 1.9 . l v ‘
Ieumred mmsxs 14 , '8 ,
. .H 2: y .__~__1‘y‘.‘ﬁ
()2. We seekabasxs v, =  ,m: 1 suchtha: thematnxb—lv; 1V21~ Z t bat/lb es
. r 0 , . l I. , ‘
the equation r— _1]. Solving the ensmng hneax system M «an: gives 3 : [ We need to choose both 7. and t nonzero to make S invertible. For ‘ 1 ——1 _ 1 _ —1
wxample. if we let 2 = 2 and t = 1, then b = [2 I], so that v, = [2] .31»; z [ I]. ﬂea/am Ar. l l. Nul u snlispm'vnincv it (luvs not contain thv neutralelement. that is, tlw fnnﬂjmx f(t) = 0.
In] all I. 4. This subsvt V is a subspace of P2: 1
t Th9 neutral element f(1‘) = (l (for all f) is in V since / ()(lf 2 U.
u 1 1 3 l l
t If f and g are in V (50 that/ f =/ g = 0) when / (f +11) 2 / f + / g : (I. :m
u n n u . u that + 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
(iv—m Rm NMX [offal ‘3 \ \I
604: ' \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 (i 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 0 0
0 1 , 0 0 r U U formn basis of
0 1 The matrices O U , 0 O , 1
0 0 0 0 0 [Rizixzz9 so that dimming” : 6’ 18‘ Any f in P,l can be written as a linear combination of 1,1312, , i l , t", by deﬁnition of P ‘
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, dimU’“) : n + it :52. Wemplunlcinglkn‘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: rvqim‘ml that u = C and b 2 “(L Thegenvi’nlcitintuitis ['1 I, : 1 ll U l 1 U (l
u M” u l U +1: 0 *1 .Tlius[1 “Jiu jIJisn basis. and tln‘ (liinvnsinn in ‘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 Cumpm‘ing muﬁiuinnts wo ﬁnd that h = (I 1 (L so that
[(1) is of the {mm H!) = u + (12 + 0!". with basis 112.1% The (liim'nslon is 15‘ l). f is odd il' f(~tl : f(i)i which is the (iﬂﬁl' if u z (' = t' 2 (l, The ndrl pulvnmnials urn
(if the form f(f) :2 I)! + {It}. with hush: 1.13 and dimension 2. i , . A , ,r.. .
52‘ We have in ﬁnd vonstnnts u and l: sm‘h that thv lUIlK’lJUIlS 1' ‘ and (1 "‘ arc SOlllLlDuS
1 0f the (inferential equation f”(ir) —l uf’(;r) § hill) : 0. Thus it is required that c‘ —
00”" + be" : (l, m 1 w a l l: : (l, and also that '25 v Fm + I) = 0. The salutiou of this system of twn equations in two nnknnwns is n 2 (3.?! z 5‘, so that the desired differential e‘lua‘ion is f "m 4 Wm + 5m) : n. finitwdimcns’ional. can be at mosi n
e n + l of infinite sequences 15 Exercise 57, there ‘
radiction: It is easy to giv ,
at the space V
10 solution to
But here is our cont assume th According to tl
s in V.
sequences, namely. .,(u,o,o,i..,oi1 66‘ Argue indirectly and with diniﬂ’) = n.
linearly independent element linearly independent inﬁnite
.i),((),1,0,0,.. .),(0,0,1,0,. ..), . , he (n + l)l.h place.
idant elements from the basis of V by omitting the redui . v < m Since our
ﬁnitedimensional, and, in fact, dnn( ) w , n.9,,“ .0, t ‘ .); in the last smucncc (1,0,0, n,‘
the l isint listghnﬂgmv
*‘ ‘ ‘ ' “sub 57‘ We can construtttn bass b a
It follows that V is list” of the original list ghi ...
View
Full
Document
This note was uploaded on 09/09/2008 for the course MATH 2940 taught by Professor Hui during the Fall '05 term at Cornell University (Engineering School).
 Fall '05
 HUI
 Linear Algebra, Algebra

Click to edit the document details