11th - EXERCISE SET 7.? 2. Using Formula is] we have projlx...

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Unformatted text preview: EXERCISE SET 7.? 2. Using Formula is] we have projlx = fin = {$192, 5} = (-fi. $1. On the other handr since sinti= 3% and oosfl: —fi, the standard matrix for the projection is as i a it elm-[e _E as E ‘21} H 29 1 o o g 2d. The reduced row echelon from of the matrix A is R = o 1 D o ', thus the general solution of c o 1 —§ ‘i —3 o the system Ax = II] can be expressed as x = t 1 , or [to avoid fractions} 3: = s 3 . In other words. r 2 1 the solution space of Ax = O is equal to span{a} where a = {-3, 111 1.2}. Thus. from Theorem 7.12. the orthogonal projection of v = {1,1123} on the solution space is given by . lit-v 11min": “flush hi-3,fl,11ti=i—%,fl.fi.fii 3|]. In Exercise 20 we found that the standard matrix for the orthogonal projection of R3 onto the plane W with equation 2:1: — y + 32 = fl is 1 ll] 2 -E- P = H 2 13 3- —ti 3 5- Thus the standard matrix for the orthogonal projection onto ii“ {the line perpendicular to W} is 1 II] I] 1 1|] 2 —fi 1 4 -? 5 I—Pzero—E 2 1:1 3:1: —2 1—3 0 c1 —s s s o -3 a D5. {a} True. Since projwu belongs to W and 1310ij11 belongs to W*, the two vectors are orthog— coal. {1}} False. For example, the matrix P = [E satisfies ii"2 = P but is not symmetric and there- fore does not correspond to an orthogonal projection. {e} True. See the proof of Theorem TIT. [d] 'I‘me. Since P3=P, wealsohatre [I—P}2=I—2P+Pa=I—P: thusI-Fisidempo tent. {3} False. In fact. since projmlmjb belongs to oolIfA}, the system Ax = projmlmyh is always consistent. P3. Let P be a symmetric n x in matrix that is idsmpotmt and has rank k. Then W = eoliP'} is a k-dimnansioual subspace of R“. We will show.r that P is the standard matrix for the orthogonal projection oi R" onto W, i.e., that P1: = proij for all x in H". To this said, we first note that beelongstoWaudthat x=Px+{x—Px}=Px+U-Plx To show that F): = proij it sufiees {from Theorem 114] to show that {I — Pix belongs to WJ'. and sinoo W = doc-HA} = until”), this is equivalent to showing that P} u {I # Pi): = [I for all Y in R“. Finally1 sinoe PT = P = P" {P is sgmunetrio and idempotent}, we have P“ — P} = P—P9=F—P=Gandso Py - {I — Pix = {PfiTfl — Plot 2 yTPTEI — m = yTPU — P]: = 3T0}..- o for every x and 3r in E“. This completes the proof. EXERCISE SET "LB 2. The solitutus of A are lineme indepeuoeut anti so A has full column rank. Thus the system An: = b has a unique [wet squares solution given by -1 i 2 x=EATA}'1ATs=[” D] [2 1 J #1 =i 9 Cl ii —il l l. 1 21 -14 The least squares error vectoris 2 2 —2 —4 b—Ax= —l — l 1 lirz—L ---IE 1 3 1 _ 21 and it is my to check that this vector is orthogonal to each of the oolumns of A. 1 1 12. The quadratic least squares model for the given data is Mir 2 y where M = i I: and y = _l . 1 4 1 o l 2 The least squares solution is obtained by solving the normal system MTMV = MTy which is 4 *1 6 s1 1 4 Ii 10 trg = fi 5. ll} 13 1:3 14 Sinoe the matrix on the Ieft is nonsiognlar, this system has a unique solution given by I 3 I it. 4 4 ti _ l 1 —§ g i —1 1J2 = 4 l3 ll] 6 = —% E *2 fi = —2 e; e is is 14 t -2 1 14 3 Thus the least squares quadratic lit to the given data is y = —1 — £1 + £33. 1 e. in 1 :1 1 12 ill T _ 1 1 l 1 m: E El: 15. .- M— II. at: -" in] '[EII' and 1 In H.“ 1. In 9'1 my = 1 1 1] = [ El" Thus the normal system can be written as I1 1:: "' In ' Eh!“ E; 3%] [El = [555;] A In . DE. We have 0 IF] = Il'and only if Ax + r 2 b and if} = 0. Note that Arr = [I if and only if r is orthogonal to eoltirl]. It follows that h - Ax belongs to tic-HAP and so. from Theorem 13.4. x is a least squares solution of Ax = b and r = b — Ax is the least squares error vector. P2. [E l: is orthogonal to the column space of A then projmlwb = I]. Thus1 since the columns of A are linearly independent. we have Ax = projcufl 411: = 0 if and oroi}r if x = fl. EXERCISE SET T3 6- in) ijwx =lx'V1}VI +ix- wit: = [like +i2lvz = {is i1 i1 il+ l1, i1-l.-1}| =E E E _l _1} 22‘ 2‘ 2 lib} preiwx = ix Univ. + no vain + {swam = {1m + {23% + mm, = 11;, gr _%, _% 24. The dimension of the range is equal to me} = e + £5 + is = 1. 39. If w = {mom}. then the vector W n. b c u=1|w||=(m'xm’€a2+bg+3) is an orthonormal basis for the 1-dimensiens1 subspaoe W spanned by w. Thus, using Female [6}, the standard matrix for the orthogonal projection of Ra onto W is 1 a, l n2 oi!) no _ T ._____ =—. t? be P—uu—G2+b,+cz 5' [r1 h “l a2+bz+c2 “b c 04: bl: (:2 D5. (21) oo1{M} = coliP] {11] Find an orthonormal basis for eol[P} and: use these vectors as the columns of the matrix M. {c} No. Any orthonormal heels for ool{P} can he used to form the columns of M. PS. We must prme that my E SPEH{W11W3 . . . ,wf} for eachj = 1,2..... The proof is by induction on J. Stop 1. Since v1: W11 we have in E spon£w1}; thus the statement is true for j = 1. Step 2 {induction step}. Suppose the statement is true for integers .t: which are less than or equal toj1 i.e., for it =1,2,....j. Then Wm ' “1 “EH “"2 W'+l ‘ Vii V'+1=W+I——'—"—"Vl——V --~~'-3—v- ’ I [I‘Ir'ill2 ilw |2 3 ll‘ef2 J and since In E s1mn{ur1}r v: E Spflfl{W11W2}1..., and v,- E span{w1,w1i,.. . . wj}. it follows that v3+1E Bpon{w1,wg1... ,w,,w_,+1}. Thus if the statement is. true for each of the integers in = l,2‘,...,j then it isaloo truefork=j+L These two steps complete the proof by induction. ...
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11th - EXERCISE SET 7.? 2. Using Formula is] we have projlx...

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