MATH 218 Exam 2 with Solution, Spring 2007

a AMERICAN UNIVERSITY OF BEIRUT Mathematics Department Math 218 - Quiz II Spring 2006-2007 = es We oe es Section: 9 (@ 9:30) 2 (@ 11:00) 1 (@ 12:30) Time: 60 min Unjustified answers will not receive any credit I- Give a precise definition of the following expressions. (a) (4 points) Vector Space ¥ | A vector space is a nonemply set wit Be operctions ; addition con scolar mublhali cahon Rat cabs hy the following axioms: fer ald UY, w EV and scalars mr. WS usw EV (2)a4 = Hu (2) (ktm)u= kutmu os Se (8\ k(utrr)= kutkv 4 thane N (9) e(mu) = Clem) uw thak O,sucu foralQueV (s) soe ae —wewv for all ue V such tat u+(-u) = Oy (6) ku EV (b) (4 points) Span {vy, va, ..., un} Spon 3%)--- %% | a5 thx. | of epee, > aseh Spon $e, a w= Shop ky +- 2 k >», ks-- kane scalars} = ‘ 1 set of aff \meay Combinations (c) (4 points) Linear Transformation \ A \inear tyansfarmahon is a futon TR ae +Rox em be represented oy a malt , +Rot ae THX) = Ax» Gr al x s or Tt ub & funchon icles: 1 TR >» R”™ sahstyi tha Following ‘3 TMu+¥)=Tl) 4T 4») nod. Tie) ee Ta) for 000 c EIR, u, VER” (d) (4 points) Column Space of a matrix A "The column Space of a metyix% A (s me Secs of he, calumn ve chors of AR. II- Consider the vector space Mo2 with standard addition and scalar multiplication. Let W be the subspace of all matrices of the form Gy 38y 2-2 6 12 —-—30 such that (z,,y,z) € NullSpaceg where B= |—5 -—10 25 |. 2 4 —-10 (a) (9 points) Find a basis for W. 3 : = ] We oi|6 suet cpa emutng > ay" ofS | 2 i ee he al Oe = E ae le z m \ 2 =6 ° CO =5 ! @ 6 \e 2 2 fe) aa > K=-2t+5s ie yt? at+5s ¢43 |= i l4— Ste | 2 [zy a re at-3s B = ie - 4 ae «ee k+5k=0 =yk=k=o St=p3. | p Step2: yf SJrkls flr" 2 Byes JONS Pea YP epee \ kj +k, = 0 = in|] ph ae 4 [3 nye (b) (2 eee) Find dim(w)* is a basis fre (c) (4 points) Find a set of 4 vectors that spans W. O) Teds. any Aircon combinahon of ce!) ~ | Seo, #¢ can tale een eee aul III- Part A: (10 points) Given the vectors v1, v2,...,U, in a vector space V. Show that Span{vj, v2, ...,Un} is a subspace of V. Spon ty vy} =} kaye kas [ «.1- --k ove scalars\ O Since W,-- UW ave in V +Pon ke. 4- -- le ws, EV So Spar $y j-— % 4 < V = }® Tf wwe tom (eS, 4 \ ee Me waves (kg )ye-- Cleyt 5% ee i aie ® Se Seu 4, -- vad SO TF ueV wat bheR. ws krw4-— ko, kus kk aw+-— kk,» ESpen}-u{ Thempore, Span$ry,---v,} 2 2 subspau of V Part B: Bpoints far { Let v; = (1,3, 1, -3), ve = (2, 7,3, -6), v3 = (—4, -15, -7, 12), the yusa vg = (5, 17,7, —15). (a) (12 points) Find a basis for the space spanned by vj, v2, v3, v4 that is a subset of {vj, v2, v3, va}. feo hs ak Sileay ase —fl 2 4 d Mel 3 4 5 4 So 7 =~ & et) =). < =R b bee a Bo ae ee o.5 BaS —3 =6 \2 -15 i) | -3 3 eo eS Ener) a 72 7 : {wv 253 cas basis for Sper dm, wrx a6 , b 6 (b) (10 points) Write each vector that is not in the basis as a linear combination of elements in the basis. a ye) 4 “ta =) ie bok ee co ° : ° Bp LP) Ca ae 6 . ° 3 Wand mypham 5 Oo Lz powls fo. Re Cn et CMAP Pe G d aN IV- (18 points) Consider two linear transformations T; and T2 where the standard matrices representing them are A; and Ag respectively. Given that the standard basis of R° is {e1, es, e3}, we know the following: To(e2) = T; (3e1) + (2, T»(2e2) = 57, (e1) = (5 Find all possible one-to-one linear transformations T such that if A is the standard matrix representing T then the first column of A — A, and the second column of A — Ag are all zeros. (Note: To determine a group of linear transformations it is enough to give the general form of their standard matriz.) ie oo r b=(2,h45) be (5,5,3,10) 2 [el =% . 1 * =) Kotte =Sl --ghaSs, K-By = [2 8 = (10,5, -5)-25) + (15,2/5,32) So fea | | 4 @® -2 -5 » res 5 fin T tp be tree bone Aveo shapalld tae oly @ triad cS a f &@ ajayayt 45 @ * )- Le . i + aa 2 == Lb \o Oo 5 Aaho a. il oe ee o © ~Sathctl] 9 | ; < \e o- i= © o | 3 6 2 Qath-d\o os Ss a) 0 6 do Oo 0 aarbd fo So we med —Sexbetd and saalliaal not bath yw : 3 V- Answer the following questions (a) (8 points) If the row vectors of a matrix A are linearly independent, and the column vectors are also linearly independent, explain why A is a square matrix. Soy A Win. Mm xn aS eal) { | m| ma (b) (10 points) W is a space spanned by the elements {v1, v2, ..., Un} of a vector space V. We know that if we want to check the linear independence of {v1, v2, ..., Un} we have to solve a system of 4 equations. Moreover, if we put vj, V2, ...,Un as rows of a matrix A, then the row echelon form of A has exactly 4 non-zero rows. Determine W precisely. Justify your conclusion. Ta wows Ancor independance. we nocd to i ae ee w,-—- VY Rave Y ie J Note Hak w.- — % are vecbors x a ea Sine wre devallad nde Taw. on AoOws a, molyix A. mm REF of A fa, ex rae eet evoKy Y Kea ding 5 le vw oe R Wigs a y Therefore, _ — v, 3 = IR