MATH 218 Final with Solution, Fall 2010

0 0 -2 1. Let A= 12 BS : 10 3 (a) Find the eigenvalues of A. (9 points) (b) Find a basis for each of the eigenspaces. (10 points) (c) Deduce that A is diagonalizable and write a formula calculating As, (6 points) Aad I —r» o -2 dot 1 a> | = se © , ro» 3-Aa ° > 2 4 ») Ah padi a (a-a(3-*% F i oA Ave WiLery <b a (a ->) (- 3 oe _ ted I 5 = - x Gpb)rce’s »=\ \ ALES J bEIR { Laat le _2u 5 \ ve (7) =! a hie 2 oO |g for A ‘ S \ 3) 6 Q e4(8) ER CoC oo mm Ee ~™ =—hK 2. Consider the following system: 21+ 223 =2 22, +423 = 6 @2+273 =4 (a) Find the least squares solution of the above system Ax = b. (12 points) (b) Find the projection of b onto the Column Space of A. (8 points) Ee 3. wwf] Fem ieerse mea} w (a) Find a basis for W. (7 points) (b) Find a basis for W+, (8 points) (c) Use parts (a) and (6) to find an orthogonal basis for R* with respect to the Euclidean inner product. (10 points) @) Wr RF Y-ABZaAO "2 = -K- am _ ° ie { ° o) (3 as % (e) \ oO > ~ z = K co] ~— 4s fe) te > —x- Yr Ze -\ x 4 ND ; WW: ’ Lf We ae we Loa, VS y A ») Dasis Foc WwW . Sr (% WwW nyt, =O ~~ a-d=0 oo a © = Ca oo b-Aeo & je po 9.73 =O —_—> elm SS on nullspu U4 Beg, Wumuen. aad=b am, base, ORWLOSOW KLLE E va 1 Vay SF) V2) Uy By wing Grose Shut 4. Let T: P3 — Pz be the linear transformation defined by: T(p(z)) = p'(z) — p” (x) + p(0). (a) Let B= {1, 2,27, 23}, find [T]s. (6 points) (b) Find a basis for the nullspace of T. (8 points) (c) Find a basis for the range of T. (6 points) (d) Let BY = {1,2 —1,22 — 22, 2%}, use (a) to find [T]e-. 1 (10 points) cj) TD = o_o +14 =. =(Ti» = fi ) °o ae RE TH = 4-o0r0 =4 =>), = (2) ° (6 pe ~ TMs 2% ~2 +,0= ~-2+2~x TO) = (#) = 32x*_€x+0 = O — 6% 13X™ _, (Cat), = Cs) oOo } + LO =axtrbxrrextd/TCp@)) =o} =Lp ofl pO — pOorpe) - 0} Ae 9p alta eed / 3BAXxX*>4+2bxXFC ~Gaxrab)rd =o} ip \o ue? =1 PO = a8 Fb xerox +d/ 3ax2+H2b-6au)x + emit bad ty Nt am) 2aG=o ‘ . ee _ =ox?¢ bxt+extd/ ab cao. | $e ened =} -p@ ane bx iY c-2b+id=0 ={ p@qaxrr bx roxrd/ a=o «£4 oud. d= —C} = hp p eee / pe = Cx —c ) CER} ' mA POo/ parm exe, ce} eae. f xi} S>) RO) = wpow | 10), Tx) Ow), TC) } sale aeataal | 4, 4,-2+2x,-6x ae = Pe | 4, - 242%, —6x 43205 ~22%, xf “cry, og Cr, ar / Cr] 14, X-1, % rc i=) Pye = B oe. OVA os eo ips Uae Oo” NnvO ~0990 J 00 peas o0Org 0900+ 100 oord ° +900 kts Sy \ > te = e0°- oA Oo-0 6 ~ 9-00 oo ~O090 Pe) 000+ : o\O ov Wo ro) TVoo © ~000 | O\o NS _ w ss ~e all ef Tres = 2,029 Yoo gde> No oi" dd-o i 1800, x7 00 NO9 9g . ——~_ @¢ 0 ee Q°e°or Estero Goo 10 5. Answer the folowing INDEPENDENT questions. (10 points each) (a) e For which values of t is < Pg >= SP p(z)q(x) dx an inner product on V = Po? © For which of the values of t obtained above would the two polynomials P(x) = x+1 and g(x) = 2 be orthogonal? = axtt+ hae. <PP>= Gen) *che Meu. APpipe= Os. pre = { GoYdeno <s a *rhatc) * cx =0 &>. a=b=e=0 Cc a [ego b's arc" + 2abats 2be x sme] = p=0 =O ok bas Bb 2c “rd2abr ube +l Bac. a - ta - £6" = bet aab gt 5 Oe Uwe Can Ane Kat PSO 5 <pip>>e Piao. f G0) dx >O alvays whine be. © fartdr- 0 wren] s ‘aa © ‘ (= G+4 ~€*_2b=0 (©) ~—(t*+24—-9) = G& +4)(b-2A=0 ‘fea —4 baz {i (6) Let W be a subspace of V = M2 10 8 4 I= 0} J enda= 241]: LettT:wWiw given by T(M) = AM. e is cal IT]e(6 ) ‘a, e [T]z to-coinpute T(3r — a). /y 4) x2, with basis B = {7, A}, where ir? (2) @ T@)= AL.= » A =oTt+1A-© Cer), - (3) @ TA)= A An AZ (2 NE = (SF 2) @ > (0 a) C= 36 =s]C, Jon = 94 a) -e(5 24 (2%) C.+C, = = See, =>. THA)=0T+9A/- =5 CH); = (J oe ; ACE DY) ome) for. A)~ LT], @r-a),. <n “(PGI (2) @ = (FG@r Ve fo = T(3I-A) = 0r.-64- al =-6A NP TG4r-A)=- a@® ‘ee 12 (c) For which value(s) of k would the vector v = [ 4 ] k ce) i = 7 the matrix A [ 1-1 ] ? be an eigenvector of Av= = Aw Lf ) (FDU, WD CS) =" Bh A= -a Ke A=-—aQ