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# Elem Linear Algebra MATH 2300

WMU

GPA 3.59

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This 10 page Class Notes was uploaded by Ms. Deborah Gaylord on Wednesday September 30, 2015. The Class Notes belongs to MATH 2300 at Western Michigan University taught by Staff in Fall. Since its upload, it has received 12 views. For similar materials see /class/216822/math-2300-western-michigan-university in Mathematics (M) at Western Michigan University.

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Date Created: 09/30/15

Mar 2006 Selected Problems7 Chapter 3 Math 230Mackey Revised Feb 2008 Let C be the set of complex numbers De ne addition in C by abi 0di ac bdi for all abcd E R 1 and de ne scalar multiplication by aa bi aa abi for all scalars 04 E R for all ab 6 R 2 Show that C is a vector space with these operations Note The Student Study Guide7 p397 has an outline of the solution We present here a model of a complete solution It is important to study this example thoroughly7 and read the discussion with care Discussion Recall that C is de ned to be the set of all numbers of the form 1 bi where 17 b E R In symbols7 the set of complex numbers is written as C a bi ab 6 R The number a is called the real part and the number b is called the imaginary part of the complex number a bi Note that the plus sign is to be treated as a formal symbol7 in the sense that a bi cannot be simpli ed any further We should examine the de nition of addition and scalar multiplication in C carefully Let7s rewrite 1 to bring out an important point not all the plus signs that appear in 1 mean the same thing Let us write C to denote addition of two quantities in C7 R to denote addition of two quantities in R to denote the formal plus between the real and imaginary parts of a complex number Here is 1 rewritten to bring out the difference between the meaning of the plus symbols that appear in it a bi 0 0 di a R c b R di7 for all abcd E R 3 This equation tells us how to add two complex numbers left hand side to get a new complex number right hand side It says we have to add the real parts which we know how to do7 since these are two real numbers7 and add the imaginary parts which we also can do7 since the imaginary parts are also real numbers For example7 1 says that the sum of the complex numbers 2 7 i and 77 Si is the complex number 75 7 4i7 because 2717751 2777151 by1 75741 Similarly7 2 contains different kinds of multiplication on the left hand side and on the right hand side Let us write 41 to denote multiplication of a complex number by a real number7 R to denote multiplication between two real numbers Then 2 becomes 04 41 a bi 04 12 a 04 12 bi for all scalars 04 E R and for all ab E R 4 which says that multiplication of a complex number by a scalar is to be done by multiplying both the real part and the complex part by the scalar So for example 72 7 7 7277221 4 714 2 31 Solution of Problem 3 To show that C is a vector space we must show that all eight axioms are satis ed A1 We must show that addition as de ned by 1 is commutative Let x 1 z2i and y yl y2i be an arbitrary pair of complex numbers Here zlx2y1y2 are arbitrary real numbers We need to show z y y x We have y 1 21 111 ygi substituting for z y 7 1 M 2 y2i by 1 which de nes addition in C 91 1 yg x2i since zlx2y1y2 E R and addition of real numbers is commutative M yzi 1 zgi by 1 which de nes addition in C y z substituting for s y Since this equality holds for all s y E C we have established that addition as de ned by 1 is commutative A D We must show that addition as de ned by 1 is associative Let x 1 mi y yl ygi and z 21 zzi be arbitrary complex numbers We need to show y 2 zy 2 We have x y z z y1 ygi 21 22i substituting for y z z y1 21 yz 22 i by 1 which de nes addition in C zL zgi y1 21 112 22 i substituting for z zL yL 21 2 yz 22i by 1 which de nes addition in C zL y1 21 x2 yg 22i since zlz2y1y22122 E R and addition of real numbers is associative On the other hand x y z zL xgiy1 ygi z substituting for s y zL yl 2 yg i 2 by 1 which de nes addition in C zL yl 2 yg i 21 22i substituting for z zL yl 21 x2 yg 22i by 1 which de nes addition in C Thus we have shown that z y z m y z for all yz E C Thus addition as de ned by 1 is associative A3 A4 A5 A6 We have to show that there exists an element in C7 to be called 07 such that z 0 z for every x E C Consider the element 0 0i Clearly this element belongs to C since it is of the form a bi here a and b have both been chosen to be 0 E R Then x00i z1z21001 1 0 952 0i 1 igl x substituting for x by de nition of addition in C7 see 1 since 1 2 ER and a0afor alla ER So7 if we designate 0 as the element 0 0i in C7 we see that axiorn A3 is satis ed We have to show that for each x 6 C7 there exists another elernent7 designated by 7x such that z is 0 Note that 0 is the element identi ed in the proof of the previous axiorn So we start with z 1 zgi7 an arbitrary element in C Consider the element in C de ned by izl 7x2i We will represent this element by 7x We claim that z ix 0 We have x is 1 mi 721 7Zi 961 19 2 WWW 1 1 2 2i substituting for z and is by de nition of addition in C7 see 1 properties of addition in R 0 0i properties of addition in R 0 de nition of 0 This establishes our clairn7 and proves that A4 holds in C Let 04 E R be an arbitrary scalar7 and z 1 mi y yl ygi be arbitrary elements in C We have to show that az y 04x ay We have ay a121ylyzi 041 11 2 y2i 04 11 Ot2 y2i Ml avg1 axz ay2i On the other hand7 04x1 zgi ay1 ygi owL azgi04y1 Oxygi mm avg1 meg ay2i substituting for L y by de nition of addtion in C azay substituting for z and y by de nition of addition in C Thus we have shown that az y 04x 04y as desired This proves that axiorn A5 holds in C Let 0 B be arbitrary real nurnbers7 and z 1 zgi be an arbitrary complex number We need to prove that 04 z 04x x Consider 04 5W 04 5 1 21 a 6m a 2621 Owl 1042 51 On the other hand7 04 alt 120 lt120 0amp1 a21 1 20 Owl 5 04 51 Thus we have shown that A6 holds in C substituting for z rnultiplication distributes over addition in R i substituting for z de nition of scalar multiplication in C7 see 2 by de nition of addition in C by de nition of scalar multiplication in C7 see 2 since rnultiplication distributes over addition in R by de nition of scalar multiplication in C7 see 2 by de nition of scalar multiplication in C7 see 2 A7 Let 0 B be arbitrary scalars and z x1 zgi be an arbitrary complex number We have to show that a 04 x Notice that on the left hand side we have only one multiplication between a scalar and a complex number while on the right hand side there are two such operations Now a a x1 zgi substituting for z a z1 a gi de nition of scalar multiplication in C see 2 a zl a mgi since 046 12 E R and multiplication of real numbers is associative On the other hand 04 x oz z1 zgi substituting for z 04 z1 xgi de nition of scalar multiplication in C see 2 a xl a mgi de nition of scalar multiplication in C see 2 This establishes that A7 holds in C A8 We have to show that 1 x z for all z 1 mi in C Note that the left hand side of the equation denotes the operation of scalar multiplication between the scalar 1 and the complex number x We have 1 x 1 zL mi substituting for z 1 1 1 xgi de nition of scalar multiplication in C see 2 1 1 mi since 1 2 6 R and 1 is the multiplicative identity in R s Thus the last axiom A8 holds in C We have proved that all eight axioms hold in C Hence the set C with addition and scalar multi plication de ned by 1 and 2 respectively is a vector space 31 p 122 10 Let S be the set of all ordered pairs of real numbers De ne scalar multiplication and addition on S by ax1z2 0410 2 5 17M EB 1792 1 917 0 6 Show that S with the ordinary scalar multiplication and addition operation ED is not a vector space Which of the eight axioms fail to hold Solution We can safely conclude that axioms A7 and A8 are satis ed since these axioms involve just scalar multiplication which has been de ned in the ordinary way Only axioms A1 through A6 need to be checked since these involve vector addition which has been de ned in a non standard way You should verify that 69 satis es Axioms A1 to A4 When investigating A4 you might notice that z 12 E S has many additive inverses for example 7x1 722 h g 7x1 0 l T are all additive inverses for x lndeed every x E S has in nitely many different additive inverses in S But A4 does not require uniqueness of the additive inverse so the fact that additive inverses are not unique is irrelevant and A4 does hold To check A5 we have to check if az 69y 04x 69 ay for all vectors s y E S and all scalars oz 7 So let x 12 and y y1y2 be two vectors in S and 04 be a scalar We work on each side of 7 separately starting with the LHS 04x 69 y 04x1 y10 by de nition of EB given in 6 04x1 yl 040 by de nition of scalar rnultiplication given in 5 Gal ay10 by distributivity of multiplication over addition in R Now we work with the RHS of 04x 69 ay az1 x2 69 ay1y2 substituting for x y 04x1 0amp2 EB ay1ay2 by de nition of scalar rnultiplication given in 5 owl ay10 by de nition of EB given in Thus we see that the LHS and RHS of 7 are indeed equal and so A5 is satis ed Next we next check if A6 is true This means we have to determine if 04 z 04x 69 s for any vector x 1 x2 6 S and for all scalars 046 8 We work on each side of 8 separately starting with the LHS 04 z 04 6 1 2 substituting for z 04 x1 04 z2 by de nition of scalar rnultiplication given in 5 04x1 6 am 6 by distributivity of multiplication over addition in R Turning now to the RHS of 8 04 69 s 04x1 0amp2 EB sh xg by de nition of scalar rnultiplication given in 5 owl 6310 by de nition of EB given in Thus in order for A6 to be true we need 04x1 zhowg 2 04x1 z10 for all choices of oz 6 and z 12 Clearly this is not true For example if zhxg 01 and Oz B 1 then LHS 02 while RHS 00 and hence LHS 31 RHS Thus A6 does not hold and S with scalar multiplication and addition de ned by 5 and 6 fails to be a vector space Remark Observe that quite a radical change was made to the de nition of vector addition yet only one axiorn A6 failed to hold 31 p 122 11 Let V be the set of all ordered pairs of real numbers with addition de ned by 17 2 21792 1 117 2 92 9 and scalar rnultiplication de ned by 04 o 12 0412 10 5 ls V a vector space with these operations Justify your answer Solution Since vector addition is de ned in the ordinary way axioms A1 7 A4 which involved just addition alone will be satis ed You should check as was done in the previous problem that A5 is also satis ed We turn now to A6 This means we have to determine if 04 B o z 04 o z B o z for any vector x 12 E S and for all scalars 046 11 We work on each side of 11 separately starting with the LHS 04 B o z 04 B 0 2122 substituting for z 04 z1x2 by de nition of 0 given in 10 owl 621 2 since addition distibutes over multiplication in R Turning now to the RHS of 11 04 o z B o z a o 12 B o 12 substituting for z 0412 xhzg by de nition of 0 given in 10 04x1 6 2x2 by de nition of vector addition given in Thus in order for A6 to be true we need owl 12 04x1 z12z2 for all choices of z 12 E S If x2 31 0 clearly this equation will not be true Thus A6 does not hold and S with addition and scalar rnultiplication de ned by 9 and 10 fails to be a vector space 32 p 132 5 Determine whether the following are subspaces of P4 First we recall that P4 consists of polynomials of degree less than 4 We can specify the set P4 as follows P4 a0 alz 1222 1323 a1a2a3a4 E R a Solution Let S be the set of polynomials in P4 of even degree Let p1x x2 2x 1 and p2x 7z2 3x 1 Since p1z and p2z are of degree 2 they are both in S However p1z p2x 5x 2 has odd degree and thus is not in S We conclude that S is not closed under and so is not a subspace of P4 c Solution Let S be the set of polynomials 102 in P4 such that p0 0 That is the subset S consists of those polynomials in P4 that evaluate to zero when x 0 You may notice that this means S 01 1ng 1ng a1a2a3a4 E R We have to determine if S is a subspace of P4 a ls S non ernpty Consider for example the polynomial z 2 This is a polynomial of degree less than 4 and takes on the value 0 when x 0 so z 2 is in S and S is non ernpty b Next we have to check if S is closed under addition Let p1 and p2 be polynomials in S Does p1 p2 belong to S We need to check the value of p1 p2 at the input 0 So we consider p1 102 0 1010 1020 by de nition of polynomial addition 0 since 101102 6 S so 1010 0p20 0 0 Since p1 102 is zero when the input is 0101 102 6 S and so S is closed under c Next we check if S is closed under scalar multiplication So let 04 be a scalar and let p E S To check if ap E S we have to evaluate ap at the input 0 ap0 04100 by de nition of scalar multiplication in P4 040 since p E S so 100 0 0 Since ap evaluates to zero when the input is 0 ap E S and S is closed under Since S non empty and closed under both operations S is a subspace of P4 32 p 133 17 Let A be an n gtlt 71 matrix Prove that the following statements are equivalent a NM 0 b A is nonsingular A c For each b E R the system Ax b has a unique solution Solution Our strategy is to show that a implies b that b implies c and that c implies a Once this chain of implications is established we can prove that any one statements implies the other two For example b gt a will be true because we have b gt c gt a The proof required consists thus of three parts For each part we must keep in mind what is assumed and what is to be proven 1 a gt b Proof Assumption NA Recall that the nullspace of A is the set of all solutions to the homogenous equation Ax 0 NA 0 means the system Ax 0 has only the trivial solution Hence when A is reduced to echelon form there are no zero rows Otherwise we would have a free variable and hence a non trivial solution to Ax 0 Since A is n gtlt n and its echelon form has no zero rows A is row equivalent to I Hence A is invertible nonsingular D 2 b gt 0 Proof Assumption A is nonsingular Hence there exists an n gtlt 71 matrix A 1 with the property that AA l A lA In We use A 1 to solve Ax b as follows Ax b gt A 1Ax A 1b multiplying on the left by A 1 gt A lAx A lb by associativity of matrix multiplication gt Ix A lb since A 1A I gt x A lb by a property of I Since A 1 is unique A lb is unique and the system Ax b has a unique solution namely x A lb Note no assumption was needed on b so this argument is valid for every b E R D 3 c gt a Proof Assumption For each b E R the system Ax b has a unique solution In particular we note that the zero vector is a vector in R so by assumption the system Ax 0 has a unique solution We know that x 0 is always one solution of Ax 0 Our assumption tells us that this is the only solution This means NA 1 33 p 145 13 Prove that any nonempty subset of a linearly independent set of vectors is also linearly independent Proof Let S be a linearly independent set of vectors Let T be a nonempty subset of S We must prove that T is also a linearly independent set of vectors We will prove this result by contradiction Suppose on the contrary that T is a linearly dependent set If we denote the vectors in T by V1V2 Vk then linear dependency implies that there exist scalars 041042 ozk not all zero such that a1V1a2V2akvk0 12 If we let W1 W2 Wn denote the remaining vectors in S remember T was contained in S then we can silently sneak these Wi7s into 12 by rewriting this equation as alvl042V2akvk0W10W20Wk0 13 Since not all the ads are zero we have a non trivial linear combination of all the vectors in S equalling the zero vector This says that S is a linearly dependent set But this contradicts what we were given So our assumption that T is a linearly dependent set if false Hence T is a linearly independent set Since T was an arbitrary subset of S we conclude that every subset of a linearly independent set is also linearly independent 1 33 p 145 16 Let o1o2 on be a spanning set for the vector space V and let 1 be any other vector in V Show that oo1 1 are linearly dependent Solution Since o1 on span V and o E V we can express 1 as a linear combination of 111 on ie o 041 agog anon for some scalars 041042 Ozn gt o 7 o 7o 041111 042112 anon adding 71 to both sides gt07oa1o1o 2o2o non Now we have a non trivial linear combination since 1 is scaled by 71 ofthe vectors o 111 112 on equal to the zero vector Hence these vectors are linearly dependent 8 337 p 1457 17 Let Ulvg 1 be linearly independent vectors in a vector space V Show that 112 1 cannot span V Solution Assume the contrary7 that spanv27 vn V Then 01 is expressible as a linear combination of 112 711m ie7 vl 04202 04303 any for some scalars 041042 04 gt 01 7 01 711 041111 04202 any adding 7111 to both sides gt011CY21204313OML But now we have a non trivial linear combination since 01 is scaled by 71 ofthe vectors 111 112 1 equal to the zero vector Hence 111112 1 are not linearly independent7 a contradiction Thus 112 1 cannot span V 347 p 1517 12 In Exercise 3 of Section 2 p 1327 some of the sets forrned subspaces of R2 In each of these cases7 nd a basis for the subspace and determine its dirnension Solution a Let D2 denote the set of all 2 gtlt 2 diagonal rnatrices7 ie7 WW glwm a0 10 00 Observethat0 b7a0 0b01Thus KHESHSH is a spanning set for D2 Next7 observe that B is clearly a linearly independent set Since 8 is a linearly independent set that spans D2 8 is a basis for D2 Since 8 contains 2 rnatrices7 the subspace D2 has dimension 2 e The set of all 2 gtlt 2 matrices whose 11 entry is zero can be represented by S0 aabc R b 0 0a 01 00 00 Observethatb Ca0 0b10c01Thus mg m 8H8 fl is a basis for 87 since 8 is clearly a linearly independent set that spans 8 Since 8 contains 3 rnatrices7 S is a 3 dirnensional subspace of the vector space of all 2 gtlt 2 matrices 367 p 1697 11 Let A be a 5 gtlt 8 matrix with rank equal to 5 and let b be any vector in R5 Explain why the system Ax b must have in nitely many solutions Solution Since A has rank 57 there are 5 non zero pivots reduced to 1 gtlt gtlt gtlt 0 1 gtlt gtlt 0 0 1 gtlt 0 0 0 1 0 0 0 0 Hgtltgtltgtltgtlt gtltgtltgtltgtlt gtlt gtltgtltgtltgtltgtlt X X X X X The augmented matrix Alb can be gtltgtltgtltgtltgtlt Since the echelon form of A has no zero rows7 this system will always be consistent7 regardless of the choice of vector on the right hand side Since there are 3 free variables7 this system has in nitely many solutions

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