Midterm 1 15-16 Spring.pdf

# Midterm 1 15-16 Spring.pdf - :99 Q4(12 p Let S = cfw_1,2,3...

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Unformatted text preview: :99“ Q4 (12 p.) Let S = {1,2,3} and U = {f 6 Fun (S) : f(1) — 2f(2) +3f(3) = 0} be a W subspace in Fun (S). Find a basis for U in terms of the character functions thg. {3. uzf Houseman |m_2{m+3rra=o} 5m W n to (s) cm be klaﬂHﬁacj with, at +h'lolA (ﬁ’lljdzlejfUl) with respect +0 HA9- deiS ‘PXMKX’Lq/Gjl =e, U}, ‘ { (thaLBﬂSlj-ﬂllﬁtal) irate. Ba ={(1;110)I('3I014\Z .‘s a basis ; m1(2,1,o)+)u~3.0,¢):(QOp) =9 iffdho 2 We ‘ 94 H3”) é/Uz )we have. [2W f:(.§lf(oﬂ may )1 “(3 ¥:1(2J1,O)+/A(v3,0[4) zb 2fi21_3;(s}=11 )AJR / ﬁé qupunf Bu) \/ ”\$931), =°[ (QI4IO)I(’3,O/4)}={L%ﬁ+%zj-3%i+%8[f 1'5 51 has ' Q5 (10 p.) Consider the vector space EL 2 {:23 > 0} with the operations a: + y = my (vector addition) and Ar = :13" (scalar multiplication). For the vector 2 E 1R+, ﬁnd the subspace Span {2} in R+. Justify your answer. ISpuan} :QERAQ/ Ilégz} =o[ 21 “will? 1 x31.2_:1_ or Seine AER: METU Northern Cyprus Campus Math 260 Linear Algebra m1 Exam ‘ . 27.03.2016 “I! L. Dept/Sec: Name 2 Time I 13: 40 Signature Student No Duration 1 110 minutes 5 QUESTIONS ON 4 PAGES TOTAL 100 POINTS null-III-l- Ql (6+6+6=18 p.) Test whether the following subsets are vector subspaces or not. Justify your answer. Last Name (i) U = {f E Fun(S) : f(a) = O} in the vector space Fun (8) of all functions on a set S where a E S is a ﬁxed element; 4’: onto. +rcbixtb+rom eu, Vftbl)f(c)elR.=g u Hp f4) V435 e U :({+3)(a)=f(ﬁl+3(a):0+0= Oa==} {+3 G U cavjteu, MIR: (rf)(a)enf(a]:r.0=0=§ rf e U ;%U ,4; Fum(5l (it) U = {y E D (—30, +00) : y’ + 5y 2 0} in the vector space ’D (—00, +00) of all contin- uously differentiable functions on the real line; «4:0 Cline 28W function) is a. SoluHon for -H«z. clig'ctnliql ecln. : 5:O=¥>3':O O+EO=O\$ U¢¢ J Ct) Vuhgz e U : (Ly-+3311- 5(31r91):5333531+532=(H1'+5.5434595335sz i =9 34rﬂléu_ ‘ l l l (iii) U = {(awa w) e R4 : m + w = —1} in the vector space n4. tr imam, (0,qu 6% =1» rum HOW/”U -‘ (4) Vu:(xq,3,ﬁ1,w1l, buggy—124%) 6U, : ’tH-V: (Xvi-x1,31+91J—214-allw14-wz) ¢ '1), because.) (Xitxz) +[W1+W2.)=(x,+wr)+(xz+wz)_—_vi 4-0-1) ;—2, 39-1. #W ‘15 (10+- 0. \JEC‘lD-r 5U195PSICL 0F we? Q2 (10+10+10=30 p.) Determine whether the following sets of vectors are linearly Q3 (10+10+10=30 p ) Consider the vector space de endent or inde endent. In the case of linear de endence; rovide a nontrivial linear p p p ,p ( ) 733 (R): {P493 J=ao+a1x+azx2+a3r3 : a0,al,a2,a3€R} relation. Justify your answer. of all polynomials of degree at most 3 and its subspaces U = {P(\$) E P3 (R) I {11— — 0.0 + 012} and U2: {10(1') E 733(1R ) 1 (1-2 = CLO + a1}. (4') {ems 63”” ”34“} in C(—x,+oc-); X _eX+1: 6 ex ==t> EL +1+ (“€48 =3 O (4') Find bases for U1 and U2,respectiv'.e1} What are the dimensions d1m(U1) and dirn( (U2)? ,1, 3x +1 0 U4 =4a°+ (““85“ 8141+ 034‘s?“ Hols/(0°) 01112, 2011le 4 a. 9R 4C R X X. , , , _ . ,. ‘ n0ﬂ+hma4 4.le [Elam/1, s» 42.6 + Oe +1449 __ 15a ,1 b {444,o,,,,,,04(04104 400,,0144M1111140r 44,: 44884 88888884“ ‘- ‘8 u1*\ Wt regime»: 0:0 4 111,41 141:0; 0,, 44,4; 594 =>Tlnts SPA—is i (11) {o—\$x+l(\$—- lat-)(-2),x3—1}111733(R);O 146 6—21 4fjk4X+44+4f(xM~3x+Q—4+9CX3 14: (“51 441-1-931— ——94+(~X-+j4—3‘64X +KX + 9x3 7 (14b3,, 0104513,,314‘43430‘0v4'l' 01:49."? 6434‘)?» 5111110145 1\44+sc, 111111, >4 4 u [I e ‘ — (1 +61 xttao+a1lxﬁ=4®0 a 040401 a 364 4;} (heal-U WWW/4:14 3 3““- UQ,--{ <3 4 +643X J 1J 1’3 lgm‘gdlﬁugl :3 080“” *0 XZ @3442," 4440,404, (0,1,4,,o) 40,00,444= 4.4+le 4+“) Y 3%” b88188 8M4 ,1 1r 0 0 ~11,- m 0 111,111 8 8 1311111;er 4815;112:1153241: “/ @m “‘8’ 1i)": O - (1,1414), :{a,=ao+a,, anal aizaomﬂzfjo. =0, a1=a\$2 , “Mug :40111-014 X1+ 03" K3?“ "4C6 (14,641,094 4 0,, (1345414 \$11,: 4:9: 0 @TWS 5141‘s Lineerlxa INDEPENDENT éguinuL34404 4,404, (OWOO ﬂju or 4012315,?“ Uw’uz: (ﬁe, {x+x, 1)::44 41\’Lv,+]tv,,:0 =80}: 1,0“ 411a,x+a,x 240434 “01199341643 '4/ ' —Spm{5u,,w,, V1 V2— (111) “102”’10)(111)(21~10)}i8R3 4 51.99- 4111. @4244 Remaﬂ: 31116.2 dWlR34= 3 ,‘-\:\ms wt 04- L, (£14m 1‘s or (iii) Find a basis for U1 + U2 Is it true that U1 + U2 2 733 (R)? Justify your answer. Erma 4511,4114; +0 a basis ,10, 41,441,, . X(,,40c4+,ulo. 1' 04414044441902.4404 ZWQO) B. «(44 0,04 {D4,,,404 (0,00,)4 41444101410 5 “11'“12 VIP @1— W80 4”“qu ALF/IA 1 basis 48" U1 4, - pp . —-2950 ‘45 ,- ﬁ? 1+2} _)A_Q_R+~§3:—l" O r inﬁanﬂﬂ. 1:2 1 W 95%K #:1- 1):: f1 bf: ”in e: 3 “4,4/164\$ aﬂtoioﬁ 0]: EUR) 0 430,545 40!“ U1+KUL 9 ' L V(X15.41M4ew- OWN—1leUWNHJALO/w844*750188444-9010H104 I. I " ' _ 1— =t>4l+e4 9441+ 4—( §_ .4. - @244, 401414- 44 CD, 4,0\+(- L14 ”(*4 4 ,(4443 ("119/4 O4 “(O/0404 #119:X.?\.+}l_13, /UL 9: '2‘: 74: W 9.91:4 €54,425; x 3 [U u _ 1+ is , ~. 4 DEPENDENT: 5 Lmeﬁfg 6994—5- (x-is—ELJU: 3——-[ [X+EJ“B4 45-.— UU 7b ?% (4R4 :SPan(BU4+U\” {sanon’rrmr14 l/MQB’ Fla/404mm Oh WIS 364’ ~:}>le/\156€44' ...
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