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Test3(Key) - TEST 3(Math 250 A l(a Define what it means...

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Unformatted text preview: TEST 3 (Math 250 A) \ l. (a) Define what it means for a linear map T : V —> W to be an isomorphism. (30 pts) A iimtzéav Wka “M’Q 11Vw-MVV l5 ccxlied «av; tramwghwm 1 «ya gr Qwefiaa-vevw omé owl'o. // § .9» if}? (b) suppose that dimV= dimW. Then show that a linear map T : V —> W is an isomorphism if and only if it is one-to—one or onto. (Hint: Use the Dimension Formula) “z" a i, 1.5“ 4' P: \ l3“ Mia, Qimemiw iiiwwéa we have '. alswV 11’ Jamil/mi} “M QEMMMWMMWE if”? _, ‘ N a m JUMV: «ii miN Now; in? l LS ‘OVUiO 31:? lefiEMiTg 133' ditmw fig> glitz/av :7 airwavl "i' GilWi/Vr :73 1 t ‘ _, i Jimmfi : O a”? T J§ l‘l =53) (Zita/aifiéx/T if: {.3 iii) dzmh/ Z: diMilwsiTi} L, if _ . . _ . M‘ffit‘iwm WW :«ahmgaaml T M. (c) Show that the followmg linear map 1s an 1somorph1srn: l , ‘5 is “in @mi‘fla T : Mm (R) —> R4 MM a b T =(a+b+c,b+c,c,d) c d (Hint: Use part (b) above) Synge. div-M (ix/lbw, {W l : Ur : citwllgu' / WE Omi; Weed “i0 WOW “HNCUM i U l“ l in be aw memo , I New; “M if} «5:? may? l' a" T 3:} : {0;Q,®,0} fix} {raw-m sitariéi :{amwl k (L a}. ' <;> .1" G‘K-er-lfifl-Q ) 0i: Q ‘ o 3 ~, ___, Ma n: e z 3 ‘Cl Kai/T : {(9);}? ‘ SO. 2 $ 55 3-; {QWVID 3;: V I j ”4 g~ & at?) d] z 0 2. (a) True or False: If the linear mapT : V —> W is an isomorphism, then T (30 pts) i ' sends a basis of Vto a basis of W. {I 0 so) 0 e!) TV we , ”3,644 (b) Let T : R2 a R2 be linear, defined by: T(x, y) = (x, y — x) . Explain why T has an inverse T ‘1. Find a formula for T “1. (Hint: It will be useful to see Where T sends the natural basis of R2) 'T it“ QM? WWW” EEC“ "i“ a; rim fireman. Mama, T Cavygfifi 41,2 9364'” §(no),(on>}\ in: ii}; harm S(:,—5l,(e,§}3 “if ,5 a»; News!" , , Tar w; a a a W’ : 01‘. NOW; T"5 )5 §xflifivmamaei 197 ' l Li’ 3} {U93 QM M {(3,5) {j} (c) (i) Does the linear mapT : R4 —> R4, defined by: T(x,y,z,t) =(x+y, y+z, z+t, t+x) have an inverse? (Hint: Look at the Ker(T) ) 52 /; ~l; 31"333 #{ofagmo} {5 jig/{VT N0; 3356C. T :5 5/5035 i~3 ‘ indeed; 3 55 ‘- 3 (ii) For the linear map T : R2 —> R2, defined by T(x,y) = (ky—x, y), k e R, gm Kai/7:339} show that T “3 = T . 7;, (Hint: It would be easier if you computed T 2) Tzixrfl =5 T3T€>mfi> :: T3375 xpj) " (:57: (54»; my??? ‘3» {“3533 in, “:3: 33:5 :5) TT: .‘W Z) T: ~ 5” 3. FindthematrixMT ofT: R2—9R4, definedby T(x y): (2x— y,3x+2y,4y,x), relative to the bases B = {(1,1), (1,0)} and D — “natural basis of R4”. Use the matrix to compute T07) , Where L? = (1,2). (10 pts) (Hint: Make sure that L7 is expressed with respect to the appropriate basis before you pass it through the matrix M r . The L7 given above is with respect to the natural basis of R2) l f. 4-“ a 31 Mil,“ 3.. T13} i) L: (2W3 j 331;”;3) :3 3; 333395333313 2: 3(gi359109'fi” Eiflflawflfifi “33'" ”3353555133 E x \ J y a?» 3? @3513; 4333,} T3bo) : 32~D,34~O,03§} 32,3,0; 13 : 2010/0/93“3330);,0/ogv} C(00) )3 i l )0 3 2‘1 .31. ; W I: - 369 o ,) SO} M3 £303} » 5’ IQ}; l 3 , , , "33 » ‘ x. 3M 2333 New”; "Di veiai‘wt’ 5c 53 ,3 5555555555 A: 235 :5 {"5535 5535 v— "M3253 ("”3 egi‘av O’ 3 q M ’34 T: 3E3 23): :33 3w :0 \3 M33 T L3; ”‘45 5.5} 3“” (0)3335”; i 1 3233‘ 33-; (Nséiic time} Mr“ (“M wai’ B) " 33333333} 33,11,53353333) 4. Let T. R2 —>R2 be linear defined by T(x, y): (—x, y), and consider the (30 pts) bases B = “natural basis of R2” and D = {(1,1), (-l,0)}. Find: (a) The matrix MT of T relative to the basis B. w . 7 3 .3 r” l {(130)1333 ‘” 333(3‘0 O3 3 33333333 , 351317“ W (3 33/3 m. , ”“555 110,1};{0/1‘): 03(3,o}3+l(o;§> (b) The transition matrix P from basis B to basis D. (1)0320 03;); )3 3433"} 3503 f) 5’0 E35 _W\ "313w “’13? x 3.3 m Hair; so]; (0,1); l3lil)i 565203 3 3 ' 3] (c) Use P above to find the matrix N T of T relative to the basis D. “ “#35533 ~ 05‘540‘5 “3%.” g”; 3 MT” 33333 “(vii/33021533: 93* {2—5} ...
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