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# Linear Algebra MTH 309

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This 14 page Class Notes was uploaded by Donny Graham on Saturday September 19, 2015. The Class Notes belongs to MTH 309 at Michigan State University taught by Staff in Fall. Since its upload, it has received 18 views. For similar materials see /class/207312/mth-309-michigan-state-university in Mathematics (M) at Michigan State University.

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Date Created: 09/19/15

MATH 309 HOMEWORK SOLUTIONS 11 2b WehavexERUSUT ltgt sERUSoerT ltgt zERoerSorET ltgt xERoerSoerT ltgt zERorQESorzET ltgt xERorzESUT ltgt zeRuSuT 21 2d We have x72y4z 1 y72z 71 172 4 1 0 172 71 R12R2gtR1 1 0 0 71 01 72 71 39 In terms of matrices Letting z r we obtain z 71 y 7 2r 71 Solving for Ly we get the solution set S 452 71727quot7177 l r E R yz r02171710 l 7 ER 23 5c We are given 3 4 1 21 2 23 0 2 44 4 So using row operations 0 0 1 1 1 R1lt gtR2 2 1 2 0 0 12R1R1 1 12 1 0 0 2 1 2 0 0 0 0 1 1 1 0 0 1 1 1 0107474 01077 0107474 1 12 1 0 0 R1712RgeRl7 RliRg Rl 1 0 0 R2lt gtR3 0 1 0 74 74 0 1 0 74 74 0 0 1 1 1 0 0 1 Letting x4 r we have 1 r i 1 x2 7 4r 74 x3 r i 1 Solving in terms of 7 gives the solution set S xl23x4 1 7r7744r717rr l r E R 1727374 T 174717117 47170l7 E R which is a line in R4 Theorem 15h Prove TV TV TV Theorem 15d associative law for scalar multiplication 1 commutative law for R 17 V associative law for scalar multiplication Theorem 15d 15 12 Axiom 6 T 8V TV 8V for R T 8V T s11112 7117 de nition of R T s11T s12 7 T svn de nition of scalar multiplication in R Ti1 5111 T02 502 7T1 5 distributive law in R T01 T02 7Ten 5111 5112 7511 de nition of addition in R T11112 71 501702 71 de nition of scalar multiplication in R TV 8V de nition of R 18 6 N Considerthe matrix Z Thena0bandb2c0200 So this matrix is inSandSy C1 Consider the matricesA Z 7A 3 2 E S Soaba b b2c07 and 7 7 aa bb b2ciONowAAi 60 dd bb 2cc b20b 20 0200SoAA ES Ta b i TC rd Multiplying a i b by T give Ta 7 Tb7 also Adding a b and a b gives a a b H7 also C2 Keeping the notation of C17 we have TA Tb 7 2Tc Tb 7 2c T0 0 So TA E S 19 2 a The line through 00 and 01112 is 8U17U2 l s E R The line through 00 and T017T02 is tT11T112 l t E R tT11112 l t E R 51112 l s E R where the last equality is gotten by letting s tT and noting that as It runs through R then so does 5 tT b The distance from 07 0 to 01112 is d My 11 The distance from 00 to T017T02 is x T111 T112 1T21 11 lTl 11 11 Td Review pp 58760 2 If S Q R is contains more than one element7 it must contain an element other than 07 say a E S with a 3A 0 Since S is closed under multiplication7 1 1 a2 E S And since a7 0wehavea2gt0 4 a 1 V associativity of scalar multiplication 1 V multiplicative inverse in R V multiplicative identity in the vector space 6 V W 7 x V W X de nition of subtraction V W EX associative law for addition V W 7 x de nition of subtraction 8 Calculate both sides of the associative law for addition with p ax b7 q 0x d7 and 5x 6f Then pz qm 695m bdmac ezf acfzbde while 1095 Q9 695195 WHJ df95 C 5C69 adf We can make these different by letting p 7 qx 2s and rx 3s so that qz 5 3s 3 but p 69 69 5x 1 So these operations do not make R1 into a vector space 10 Let Q denote the set of quadratic polynomials a No We have 2 isz E Q since 171 7E 0 But 2 if 0 E Q since the coe icient of 2 is now 0 b No We have x2 E Q but 0x2 0 E Q as in part a c No7 because it is not closed under either addition or scalar rnultiplication 12 Since S is a subspace of V7 it is a subset which is a vector space in its own right So it will still be a vector space when considered as a subset of T 16 a N We have 0V 0 is in S since it is of the form TV with r 0 E R So S 3A 0 011fu7w E S then u TV and W 8V where r75 ER So u W TV 8V de nition of S 7quot 8V distributive law in V E S since 7 s E R by additive closure of reals 02 If u E S then u TV where r E R and consider 5 E R So su 5rv de nition of S TSV associative law for scalar multiplication in V E S since rs E R by rnultiplicative closure of reals b If 0 E L then 0 rev 1 X for some r0 E R by de nition of L Solving for x gives x TOV So Lrvx l rERrvEr0V l TERTT0V l rER5V l sER where7 as 7 runs through all the reals7 s r 7 r0 does so too c Suppose rst that L is a line which is a subspace of V Then L must contain the additive identity 0 and so go through 0 Now suppose that L is a line passing through 0 By part b7 we know that L TV 1 r E R for some V E V And by part 217 any line of this form is a subspace Review pp 87789 1 a If one rnultiplies by 0 then one can change the solution set of the corresponding equations For exarnple7 consider the system of equations z Ly 2 This has as solution a single point Ly 17 2 But if we multiply the second equation by 0 then we get the system z 1 which has as solution set a whole line7 narnely 17 10 r01 l r E R b If one replaces Ri with Ri ORJ then Ri is unchanged So there is no reason to prevent this from happening7 although it is also not very useful 4 a If a leading 1 is in the rightmost column7 then that row must have the rest of its entries 0 and so correspond to the equation 0 1 So the solution set is S b In this case we can solve for the leading variables in terms of the free variables and so S 3A 0 6 We have 71 1 72 0 72 71 1 72 0 72 1 71 1 1 1 R2 R1 a R2 0 0 71 1 71 2 72 1 3 1 R3 2R1 a Rs 0 0 E3 3 E3 7R2 gt R2 0 0 1 1 1 0 0 1 1 1 0 0 3 3 3 R3 3R2 gt R3 0 0 0 0 0 This corresponds to the system iliig 3741 Let 2 r and x4 5 Solving for 1 and x3 give 961962963964 2 247271 47M 7 257721 87 5 So the solution set is S 00 1 0 r1 1 0 0 572011 l r s E R a plane in R4 b We have 0 1 2 1 3 R1 lt gt R2 2 0 4 i6 6 2 0 4 i6 6 0 1 2 1 3 5 3 16 712 6 5 3 16 712 6 12RleRL 1 0 2 i3 3 S o 1 0 2 i3 3 0 1 2 1 3 0 1 2 1 3 5 3 16 712 R375R1R37R373RZR3 0 0 0 0 718 The last row corresponds to the equation 0 718 and so the solution set is S 0 11 a A homogeneous system always has the solution 0 and so can never have an empty solution set b Take any equation of the system Since 0 is a solution plugging in every variable equal to 0 on the left hand side must give a true equation So that equation must be 0 0 and so the right hand side must be 0 making the equation homogeneous 12 a Yes if the reduced echelon form of the augmented matrix has a row with a pivot one in the rightmost column See the answer to 421 b No there are other possibilities such as e and g c No since there are 6 unknowns to have a single point as a solution there would have to be 6 pivot ones in the reduced matrix But there are only 5 equations and so only 5 rows for pivot ones 1 No since it is not possible to have this solution by c e Yes if each row has a pivot 1 for one of the variables then there are 5 pivot ones and so exactly one of the variables is free This will give us a line f No there are other possibilities such as a and g g Yes if 4 of the rows have pivot ones for variables and the 5th row is all zero then there will be exactly two free variables This will give us a plane h No there are other possibilities such as a and e 31 4e We have x 12 x3 2x2 x We want to nd abc E R with x32 xa1bz2cx32cx3bcx2abxa Equating coef cients gives 0 1 b c 2 a b 1 a 0 Solving by row reduction or otherwise gives 10 b 1 01 So z12 2 3x2 33 4 To show 2 2 6 0 0710 1 1 233 1 713 72 linearly independent we consider a226 0 b0710 1 c1233 d1713 72 0 000 1 The corresponding system of equations has matrix 2 0 1 1 0 1 0 0 1 0 2 71 2 71 0 l 0 1 0 1 0 6 0 3 0 which row reduces to 0 0 1 71 0 0 1 3 i2 0 0 0 0 0 0 Since d is a free variable7 the vectors are dependent Letting d 1 gives a 71 b 71 and c 1 and so equation 1 becomes 72727670 7 07101 1233 17137 72 00007 so we can solve for 17 713 72 and get 17 71737 2 272767 0 07 1707 1 i 17 27373 To show 22607 071017 1233 are independent we consider 12727670 b07717071 C1727373 007 07 0 The corresponding system of equations has matrix 2 1 0 1 0 0 0 2 71 2 0 l 0 1 0 0 6 3 0 which row reduces to 0 0 1 0 0 1 3 0 0 0 0 0 The only solution to this system is a b c 0 so the vectors are independent 34 6 Consider 11 10 10 11 i abcd ad alool li0 01l1dl11l l bd cd39 Row reducing the coef cient matrix gives 1 1 1 1 1 0 0 0 1 0 0 1 0 1 0 0 0 1 0 1 0 0 1 0 0 0 1 1 0 0 0 1 Since this is the identity matrix7 we have a basis 35 20 Compare S x 372 x to the basis 17Lz27x3 adding basis elements one at a time to S as long as S remains linearly independent First we check if 1 E spanx 372 So we wish to check if 1a3 bx2 bx2 ab3a Equating coef cients gives b 0a b 031 1 which has no solution since the rst two equations force a 0 and the last gives a 13 So 1 E spanx 372 x and 1x 372 x is linearly independent Now we check if z E span17 z 372 So za1bx3c2z cx2bcxa3b This has solution a 737 10 0 So m E span1 312 x and we can not add m Now we check if x2 E span1 37 2 So zza1b3c2xC2bca3b This has solution a 3b 710 1 So x2 E span1x 372 x and we can not add 2 Now we check if x3 E span1x 37 2 So zsa1b3c2xC2bca3b This has not solution since there is no 3 term on the right So x3 E span17x 372 x and 1x 37 2 7 3 is a basis 36 4c Writing V7 15 ia10 b00601 7597 00 01 10 1 we see7 by inspection7 that a 1b 97 c 5 So the coordinate vector is MB 9 5 52 6 To prove the Theorem it suf ces to show that the matrix B IA l is the inverse of the matrix AB This is equivalent to showing that ABB 1A 1 I and B lA 1AB I For the rst equation ABB 1A 1 ABB 1A 1 associative law AIA l B and B 1 are inverses AAA I is the identity matrix I A and A 1 are inverses For the second B lA 1AB B 1A 1AB associative law BIB l A and A 1 are inverses BB 1L I is the identity matrix I B and B 1 are inverses Review pp 13274 1 21 Consider 117271 b215 017 747 7 07 07 0 which corresponds to 1 2 1 0 1 0 73 0 2 1 74 0 which row reduces to 0 1 2 0 1 5 7 0 0 0 0 0 Since the reduced matrix has a column corresponding to a free variable7 the vectors are dependent b Since the rst two columns of the reduced matrix contain the leading ones7 the two corresponding vectors are independent and are a basis for the span of the set So the geometric gure is plane with equation a1271 b215 l ab E R 2 21 Consider azg b3 2 0amp3 2 s d3 2 z 1 0 Then d 0 since p4 is the only polynomial with a constant term This forces 0 0 since p3 is the only polynomial left with a rst power of By the same argument with 2 we get I 07 and now only p1 is left so a 0 So this equation only has the trivial solution and the polynomials are independent b Since dimng 4 and we have 4 linearly independent vectors7 they must also span and so be a basis c Suppose n3 bz3 x2 Cxs 2 m d3 2 z 1 5x3 4m 7 2 Equating the constant coef cients gives d 72 Now equating the z coef cients gives 4 d c 72 c so 0 6 Using x2 gives 0 bcd b672 so I 74 Finally7 using 3 we gave 5 abcd a74672 so 15 So 5z34x72 5x3743x2 6x32x 723x21 d From part c7 we have 1 5 l 1le 2 1721 7 a We are given spanVW V We want to show spanv W7V 7 W V spanvw By Section 327 problem 137 it suf ces to show V WV 7 W E spanvw and ii VW E spanV W7V 7 W Part follows since V W 1V 1W and V 7W 1V 71W For part ii7 to get V aVW bV7 W a bv a7 bw we need 1 b 1 and a 7 b 0 which has solution a b 12 To get V CVWdv7w c7dVc7dW we need 07d0 and 07d1 which has solution c 127d 712 Part ii is now proved b We must show that the equation aV W bV 7 W 0 has only the trivial solution This can be written as a bV a 7 bw 0 Since V7W are independent7 the only solution of this last equation is ab 0 and a7 b 0 Adding the two equations gives 2a 0 so a 0 Now plugging back in gives 0 a b 0 b so b 0 Thus the only solution is a b 0 as desired 9 a No7 dirn M23 6 and 5 lt 6 so 5 vectors can not span b No7 by the same reasoning as part a c Yes7 removing an element of a basis would give a linearly independent set of 5 vectors 1 No7 we could have 5 multiples of the same vector e Yes7 adding a vector to any basis would give 7 vectors which span f No7 using the same example as d g No7 dirn M23 6 and 7 gt 6 so 7 vectors can not be independent h No7 for the same reason as g i Yes7 any basis would have 6 vectors and both span and be independent j Yes7 if 6 vectors span they must be independent If they were not7 they could be contracted to a smaller independent set which still spans7 contradicting the fact that every basis of M23 must have 6 elements k Yes7 for the same reason as i l Yes7 if 6 vectors are independent7 they must span If they did not7 they could be expanded to a larger independent which would span7 contradicting the fact that every basis of M23 must have 6 elements 10 Consider 117 71 07 2 b71102c1110d02172 0000 In terms of matrices 1 171 0710 1 1 0 00 0 0 2 0 10 00110 l2 00 1 0 1 2 1 1 0 72 0 Since the 4th variable is free7 0 2 1 72 can be solved for in terms of the other three vectors Since the 0 0 i 0 which row reduces to o 1 other vectors 17102711021110 correspond to leading variables7 they form a basis for the span of the original set 15 a To show S U T Q spanV1 7Vm7W17 7Wm we must take any x E S U T and show that x is in the span If x E S U T then x E S or x E T If x E S then7 since S spanV1 7Vm7 we have scalars such that x a1V1 7 amvm So we can write x alvl 7 1me 0W1 7 Own which shows x E SpaHV1 Vm7W1 7Wm If x E T then7 since T spanw1 7Wn we have scalars such that x blwl 7 bnwn So we can write x 0V1 7 0Vm blwl 7 Jran which shows that7 again7 x E SpaHV1 Vm7 W1 7Wm So in either case7 x E spanV1 7Vm7W1 7Wm7 and we are done b We must show that if x E spanV1 Vm7W1 7Wm then x E W From the hypothesis7 there are scalars with x alvl 1me blwl 7 bnwn Let y alvl 1me and z blwl bnwn so that y E spanV1Vm S7 2 E spanw1wn T and yz x Since y E S we have y E W since S Q S U T Q W Similarly7 since 2 E T we have 2 E W since T Q S U T Q W Thus7 since W is a subspace7 x y z E W which completes the proof pp 2087210 1 a The product will be in M54 b 2771 213 8 2 a We have AB2ABABAABBABA2ABBAB2 b If AB BA then we can subsitute into the answer for part a and get AB2A2ABBAB2A2ABABB2 A22ABB2 c If A B2 A2 2AB B2 then we can equate this expression with the one for A B2 in part a to get A2 AB BA B2 A2 2AB B2 Cancelling the squares gives AB BA 2AB And subtracting AB from both sides gives BA AB as desired 3 a Let the rows interchanged be the 2th and jth In any product BC7 the kth row of B is used to construct the kth row of BC Since the kth row of E is the same as the kth row of I for k 3A Lj such a row in EA will be the same as the row in IA which is the same as the row in A The 2th row of E is the same as the jth row of I So the 2th row of EA will be the same as the jth row of IA which is the same as the jth row of A Similarly7 the jth row of E is the same as the 2th row of I So the jth row of EA will be the same as the 2th row of IA which is the same as the 2th row of A This shows that EA is A with the 2th and jth rows interchanged b Say the 2th row of E is multiplied by the nonzero constant c As in part a7 all the rows of EA are unchanged except the 2th The 2th row of I is the standard basis vector el for R So the 2th row of E is cei If C is the jth column of A then the zj entry of EA is cei Cj Cei 07 calj Since this is true for all j the 2th row of EA is 0 times the 2th row of A c Suppose we add 0 times the 2th row to the kth row Again7 all rows of EA are unchanged except the kth Letting 07 be as in part b7 we have that the khj entry of EA is co ek Cj Cei 07 ek 07 calj akj Since this is true for all j the kth row of EA is the kth row of A plus 0 times the 2th row of A 1 Let R be an elementary row operation and let E be the corresponding elementary matrix Letting A be the result of applying R to A Then by parts aic7 A EA It follows that A B EAB EAB Using parts aic again7 this last matrix is the result of applying R to AB as desired 7 a We consider 1 2 0 1 0 0 1 0 0 1 0 2 1 0 3 7 1 0 1 0 reduces to 0 1 0 0 0 1 so the inverse is 0 0 1 0 71 0 0 0 1 0 0 1 i3 1 1 i3 1 1 b We consider 1 1 0 1 0 0 1 0 0 752 2 3 752 2 0 2 4 0 1 0 reduces to 0 1 0 72 52 72 so the inverse is 72 52 72 i1 1 5 0 0 1 0 0 1 1 71 1 1 i1 1 9 a Row reducing the given matrix gives 1 0 1 0 1 1 so the rank is 2 0 0 0 b Row reducing the given matrix gives so the rank is 3 OOH OHQ HOG c Row reducing the given matrix gives 1 0 so the rank is 2 0 OHO D 00 1 Row reducing the given matrix gives 1 0 i1 72 l 0 1 2 3 i 0 0 0 0 so the rank is 2 l 0 0 0 0 10 a If A7B E Mn7n and rankA rankB n then A and B are invertible by Theorem 512 But then AB is invertible by Theorem 59 Using Theorem 512 again7 we see that rank AB n which is what we wished to prove b We rst show that we must have rankA n by contradiction So suppose rankA 7 71 Since A E Mn7n we have rankA 7 71 implies rankA lt 71 Let A be the row reduced echelon form of A So the last row of A will be zero since rankA lt n It follows that the last row of AB is zero and so rankA B lt 71 Using Theorem 55 repeatedly7 we see that AB is the result of applying the same row operations to AB But doing row operations does not change the rank since such operations do not change the nal row reduced echelon form So we should have rank A B rank AB n This contradicts the previous inequality for rank AB and so rankA 3A 71 can not happen We have shown that we must have rankA n and so A L exists by Theorem 512 By Theorm 587 this implies that A 1 is invertible So7 using Theorem 512 again7 rank A L n We now have rank A l rank AB 71 So using part a7 we have n rank A 1AB rankA 1AB rank InB rankB which is the second condition we wished to prove 61 8 We must show that T preserves addition and scalar multiplication Tz 2H 1Tif 21 ziin 2H 3 Tli ZlTlZ ill TH 21gtTltli 21gtl 2111 2M1 21 63 8b Applying Dto each element of B gives D1 0 Dz 1 1 Dz 1 2z 1 2x 27Dz 71f 3z 71 3x2 7 6x 3 Expressing each of these as a linear combination of elements of B 17 717 x22x17 37323 7 1 gives 0 010x710x22x10z373z23717 1 110z710z22z1037323x717 2z2 412x710x22x10z373z23717 3276x3 7121712z713z2210x37323z71 Using these coef cient coordinates as the columns of our matrix gives 0 1 4 712 0 0 2 712 AT 0 0 0 3 0 0 0 0 64 1 a We have T1 x Tx 2 Tz2 3 and using the coordinates in 177273 gives 0 0 0 1 0 0 0 1 0 l 0 0 1 2 b We have D1 07 D 17 Dx2 x Dx3 3x2 and using the coordinates in Lnxz gives 0 1 0 0 A 0 0 2 0 0 0 0 3 c We have D o T1 17 D oT 2x7 D o Tz2 3x2 and using the coordinates in Lnxz gives 1 0 0 0 2 0 0 0 3 d We have 0 1 0 0 T 8 8 1 0 0 AA 0 0 2 0 0 1 0 0 2 0 0 0 0 3 0 0 1 0 0 3 66 3 Since idVV V for all V E V keridVVE V l idVV 0V E V l V0 0 To compute irnidV we need to nd all W E V such that there is V E V with idVV W But for every W E V we have idVW W and so imidV V 67 221 HT R5 a R4 is any linear rnap then rankT g dirn R4 4 So nullityT dirnVErankT dim R5 E rankT Z 5 E 4 1 But then T can not be one to one since such a map has nullityT 0 71 1 For the base case7 if n 1 then the left side is 1 DH k1 while the right is 1 12 1 So both sides agree when n 1 For the induction step we assume 71kn71n711 n71nn27n39 2 2 2 So iknilkn n27nn n27n2n n2n nn139 k 2 2 2 2 71 28 We are given that SnO is true Now S 704 and letting n 1 in this equation gives Si Sno So we have that Si is true We are also given that S true irnplies Sn1 true for n 2 no Replacing 71 by nn0 71 in this statrnent gives Snn011 true irnplies Snn0 true for n no 7 1 2 no Translating in terms of S and simplifying the inequality gives S true irnplies Sg true for n 2 1 H w H H So we have shown that Si is true and S true implies ST L1 true for n 2 1 But by the Principle of Mathematical lnduction7 we can conclude that S is true for all n 2 1 Replacing n by n 7 no 1 in this statement gives S72 01 is true for all n 7 no 1 Z 1 Translating this back in terms of S and simplifying the inequality gives Sn is true for all n 2 no which is what we wished to show 72 9 Since the matrices ar 3 gtlt 3 we can use diagonals to compute a b c det d 6 f aez39bfgcdh7ceg7afh7bdi g h 239 On the other hand a d det b e ae dhcgbf7gec7ahf7dbi 0 f eta Comparing the 6 terms in each expression7 we see that the match up and so the determinants are equal 81 8 g We wish to proved that an upper triangular matrix A with diagonal entries all am has these diagonal entries as its eigenvalues Since A is upper triangular7 so is AI 7 A We proved in Exercise 72 79 that the determinant of a diagonal matrix is just the product of its diagonal entries So detI7A A7a117ann 0 gives the eigenvalues A 111 am as desired 82 4 a We know that if A and A are matrices for the same linear transformation in two different ordered bases B and B 7 then A is similar to A So let T have matrix Hag in the ordered basis B V7W Then TV 2V and TW 3W Now consider the matrix of T in the basis B W7V Since TW 3W amd TV 2V7 the matrix for T in B is 7 30 Ailml Review pp 2627264 6 a To show T preserves addition7 let pq E 1 as desired Tp Q 96 1p 9W 96 1p 962 11962 96 1p 9 29 0911962 Tp TM To show T preserves scalar multiplication7 let p E P2 and r E R T079 96 DUN962 96 1Tp 962 Wt DID962 TWP b We compute the following coordinates7 using the fact that 1 E B is the only element with a constant term and x2 z E B is the only one with an x2 term to nd the coef cients by inspection T1 x100030x3x02x0x1 Tz1z11x10x30x3z0xzx1z2L Tz2z x12z2 1 2z3 2z2x1 4963 2 95 1 x 2z2z 1x 1 So 0 0 0 0 72 AT 0 0 2 01 1 7 Wecompute 1 1 1 1 2 71 0 71 1 3 1 3 0 0 1 1 1 1 1 2 71 0 2 1 0 1 0 0 0 1 1 1 1 1 0 1 0 0 1 0 1 0 0 0 1 So 71711 P 71 2 0 3 0 0 9 a To show T preserves addition Tax2 bxc dx2 exf Tadx2 bex 0 f ad2cfbe a2cbd2fe Tax2bxcTdx2exf To sho T preserves scalar multiplication Trax2 bx 0 T7 ax2 rbx TC Ta 7 2rc Tb ra 20 b 7 Tax2 bx c b To show T onto take pq E R2 and nd axz bx c E P2 with Tax2 bx c pq So by de nition of T we want a 2cb pq So choosing a pb 10 0 gives the desired equality c Since T is onto we have imT R2 and so rankT dimRZ 2 Furthermore we have nullityT dimlP g 7 rankT 3 7 2 1 dWehave T1021020210001 2 0 1 Tx020101010101 gt A0 1 0 Tx212001011001 e Row reducing A just multiplies the rst row by 12 which does not change the number of nonzero rows So rankT 2 rank A 11 a If T is onto then rankT dimlll lo 11 So nullityT dimM43 7 rankT 12 7111 b If T were one to one then we would have nullityT 0 So rankT dim M43 7 nullityT 12 7 0 12 But this is impossible since imT Q P10 so rankT dimlll lo 11 13 a We are given TV AV E R So m is the number of rows of AV which is the same as the number of rows of A which is 4 b We are given V E R In order for the multiplication AV to be de ned we must have that n which is the number of rows of V equals the number of columns of A which is 5 c Since row operations do not change the row space the non zero vectors of the reduced form are abasis 1720100 011000 0 0 1 7777 777 d A basis for the column space is obtained by taking the columns of A corresponding to leading ones in the reduced form 1 0 0 1 E1 E1 72 7 1 7 3 3 0 e Since imT is spanned by the columns of T the answer is the same as in part d f To nd kerT we set the two free variables equal to r and 5 Solving for the variables of the leading ones gives kerT2rE sriss0 1735 E Rr21000s71071101735 E R So a basis is 21000 710E11O g We have rankT nullityT 3 2 5 dim R5 19 a De ne T R a R by Tz z and T R a R by T z Ex Then it is easy to verify that T and T are isomorphisms But T T x Tz T z z Ex 0 for all z E R This is not an isomorphism since it is not one to one every element of R gets mapped to 0 b To show CT is an isomorphism we must verify the three properties of an isomorphism To show CT is linear check it preserves addition CT V W CTV W de nition of CT CTV TW T preserves addition CTV CTW distributive law for scalar multiplication CT V CT W de nition of CT and scalar multiplication CT TV C TTV de nition of CT T preserves scalar multiplication commutivity and associativity for scalar multiplication r CTV de nition of CT To show CT is onto pick V E V and nd V E V with CT V V Since T is onto there is u with Tu V Let V 1Cu which is well de ned since C 71 0 Then CTV CT1Cu C1CTu 1 V V To show CT is one to one note that since T is an isomorphism we have dimV dim V And since we have just shown CT is onto we have nullity CT dimV E rankT dimV E dim V 0 So CT is one to one Review p 300 2 A nonnegative integer 71 means 71 Z 0 So the base case is n 0 The left side equals kiioi7 02 2 while the right is 201 7 120 11 1 So both sides agree when n 0 For the induction step we assume 172ni1 7 21 k0 So 1 1 quot 1 1 2 71 1 22ni11 2n171 2n71 2n 7 2n 4 a Expanding about the rst row and then using the diagonals gives 2 T 33 1 2 3 det 0 7 4 3 5det 7 4 3 5430637607971470 6 5 3 1 5 3 1 b This matrix is obtained from the matrix in the rst part by multiplying the 2nd row by 111 So7 by Theorem 74c7 the new determinant is 111 70 7770 c This matrix is obtained from the matrix in the rst part by adding 100 times the 2nd row to the 1st row So7 by Theorem 74e7 the determinant stays at 70 Review pp 344345 2 a We have A710 3 4 det 12 A73 76 A375A26AAA72A730 720 6 A8 So A 07 27 3 b Row reducing the matrix when A 0 gives 1 0 i1 1 0 1 72 letting z 1 gives z Ly 2 and eigenvector u 2 0 0 0 1 Row reducing the matrix when A 2 gives 1 0 712 1 0 1 0 letting z 2 gives m Ly 0 and eigenvector V 0 0 0 0 2 1 0 712 3 0 1 16 letting z 6 gives m 3y 71 and eigenvector W 1 0 0 6 c Letting B e17 e27e3 the standard basis and B uvw using the vectors from part b7 we have P lAP D where D is the matrix of eigenvalues 0 0 0 D 0 2 0 7 0 0 3 and P is the change of basis matrix from B to B which has columns ulg7 MB wlg Since B is the standard basis7 xlg x for any vector x So 1 1 3 P 2 0 71 1 2 4 Since V is an eigenvector for A with eigenvalue A we have AV AV So ATIVAVTIVVTVTV

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All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.