### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Linear Algebra II MATH 790

KU

GPA 3.6

### View Full Document

## 20

## 0

## Popular in Course

## Popular in Mathematics (M)

This 27 page Class Notes was uploaded by Edgar Jacobi on Monday September 7, 2015. The Class Notes belongs to MATH 790 at Kansas taught by Staff in Fall. Since its upload, it has received 20 views. For similar materials see /class/182387/math-790-kansas in Mathematics (M) at Kansas.

## Reviews for Linear Algebra II

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/07/15

Annihilators Linear Algebra Notes Satya Mandal September 217 2005 Let F be a led and V be vector space over F With dimV n lt 00 As usual V Will denote the dual space of V and VM V 1 De nition 01 For a subset S Q V de ne the annihilator annS f E V fu 0 for allu E S We also use the notation SO annS We may also temporarily use the notation annV S annS to underscore the fact that annV S is a subspeee of V 2 First note that for S Q V7 the annihilator annS Q V is a subspace of the dual space 3 Also note for S Q V7 if W SpanS then annS annW 4 Now let S Q V Then according to above de nition7 annihi lator of S is annS annV S is a subspace of V In this case7 there is another natural annihilator of S as a subspace of V as follows annVSuEVfv0f0r all fES 5quot G 1 It is possible to mix up two annihilator of S The rst one is a subspace of the double dual VW and the second one is the subspace of V We Will justify that these two annihilators are same Via the natural identi cation of V and V Let L V VW be the natural isomorphism Let S Q V Then La7mV 8 annV 8 Proof Let 1 E annV S Then fv 0 for all f E 8 Hence Lvf fv 0 for all f E S So7 Lv E annV 8 Therefore La7mV 8 Q annV 8 Now let G E annV 8 Since L is an isornorphisrn7 we have G Lv for some 1 E V We need to show that v E annV 8 But for f E S we have 0 Cf Lvf fv Therefore 1 E annV S So7 annV S Q LannV Hence the proof is complete Now7 for asubspace W Q V7 the annihilator WO annV W has two annihilators They are annV W0 and annV W0 It follows from above that LannV W0 annV W0 Since7 L is a linear isornorphisrn7 we also have dirna7mV W0 dirnamzV W0 8 Theorem 01 For a subspace W Q V we have W annV a7mV7 also written as W W00 Proof Write U annV annV It is easy to see that W Q U Therefore7 it is enough to show that dimW dimU We have dimW dimannV dimV Also7 by the same theorem7 dimannV Wdima7mV a7mV7 dimV dimV It follows that dimW dimannV a7mV7 dimannV annV dimU So the proof is complete Lemma 01 Suppose V is vector space of nite dimension dirnV it over F Let fg E V be two linear funetionals let Nf be the null space off and N9 be the null space ofg Then Nf Q N9 if and only ifg of for some 0 E F Proof Obvious i If g 0 then 9 of with c 0 So we assume that g a 0 So dirnNg 7271 Since Nf Q N9 we have f a 0 and dirnNf 7271 Therefore Nf N9 N3ay Now pick 6 N Since dirnV n it follows that V N Fe Also fe a 0 and 96 a 0 Write c gefe Claim that 90 First note 96 cfe Now for n E V we have n y A6 for some y E N and A E F Therefore 9 921 we W6 Ac e Cfy Ae cffv So the proof is complete Following is Theorem 207 page 110 Theorem 02 Suppose V is vector space af nite dimension dim V n7 over F Let g7f17 fT E V be linear functionals Let N be the null space ofg and Ni be the null space of f2 Then N1 0 N2 0 0 NT Q N if and only ifg ELI cifi for some Ci 6 F Proof Obvious i We use induction on r to prove this part Case 7 17 is the above Lemma 01 Now7 we assume the validity of the thoerem for r 7 1 functionals and prove it for 7 Write V NT Let 9 GlVHfi f1lV77 471 fT1lV7 be the restrictions of the respective functionals to V lnduction ap plies for these functionals and it follows that 9 Ti Cifi 21 for some Ci 6 F This means7 for all m E V NT7 we have gm This means7 for all m E V NT7 we have gm Write 171 h 9 i Cifi Then NT V Q Null 7 Spaceh By the case r 1 or by Lemma 01 It follows that h chT for some CT 6 F Hence T71 7 9 Zcifi h Zcifi i1 i1 and the proof is complete Polynomial Rings Linear Algebra Notes Satya Mandal September 277 2005 1 Section 1 Basics De nition 11 A nonempty set R is said to be a ring if the follow ing are satis ed 1 R has two binary operations called addition and multipli cation 2 R has an abelian group structure with respect to addition 5 The additive identity is called zero and denoted by U 4 Distributiuity For 7112 6 R we have ay2 aya2 and y 2k ya 2a 5 We assume that there is a multiplicative identity denoted by 1 a 0 Note the multiplication need not be commutative So7 it is possible that my a ya Also note that not all non zero elements have an inverse For example Let R MmlF be the set of all n X n matrices n 2 2 Then R is a ring but multiplication is not commutative Following are few more de nitions De nition 12 Let R be a ring 1 We say R is commutative if my ya for all my 6 R 2 A commutative ring R is said to be an integral domain if xy0gta0 or y0 3 Let A be another ring A map f R a A is said to be a ring homomorphism if for all my 6 R we have f y f fy7fxy ffy and f11 4 For a ring R7 an Rialgebra is a ring A with together with a ring homomorphism f R a A Remark 11 Let l be a eld and A be an Fialgebra The textbook calls such an algebra as Linear Algebra Note that A has a natural vector space structure Exercise 11 Let l be a eld and f F a A be ring homomor phism Then f is 1 1 This means that ifA is an Fialgebm then lF Q A Proof It is enough to show that if f 0 then m 0 Are you sure that it is enough Assume m a 0 and f 0 We have f113071 m l ff 1 307 1 i 0 2 Polynomials We do not look at polynomials as functions Polynomilas are formal expressions and in algebra they are manipulated formally De nition 21 Let l be a eld and N 07 1727 be the set of non negative integers 1 Let f denote the set of all functions f N a F So7 f a07a17a27 12 E F is the set of all in nite sequences in F De ne addition and multiplication on f naturally see the book f is called the power series ring 2 Let X 07170707 6 f Then any element f a2 6 f can be written as 00 f ZWXZ 2 0 with apprpriate meaning of in nite sum attached OJ Notation Usual notation for the power series ring is EX f Elements in EX are called power series over F 4 Let EX f E EX f a0a1XanX7L7a E F Note that EX is a subring of We say that EX is the polynomial ring over F U Importantly two polynomials f7 9 are equal if and only if coe icients of Xi are same for both f and 9 Theorem 21 Let EX be the polynomial ring over over a eld F 1 Suppose f79791792 E EX and f is non zero Then f9 0 5 9 0 and U91 f92 5 91 92 2 f E EX has an inverse in EX if and only if f is a nonzero scalar 3 Section 4 Division and Ideals Theorem 31 Division Algorithm Let F is a eld and EX be a polynomial ring over F Let d a 0 be a polynomial and degD n Then for any f E EX7 there are polynomials q7r E EX such that fqdr r0 degr ltn In fact q7r are UNIQUE for a given f 3 Proof Write a proof Corollary 31 Let l is a eld and EX be a polynomial ping over F Let f be a nonzero polynomial and c E F Then fc 0 if and only if X 7 c divides f in EX Further a polynomial f with degf n has atmost n roots in lF Proof Obvious i By division algorithm7 we have f X 7 CQ R Where R762 6 EX and either R 0 or degR 0 We have 0 fc R0 R Therefore f X 7 CQ For the proof of the last assertion7 use induction on n 31 GCD De nition 31 Let l be a led and EX be the polynomial ring Let f177f7 E EX be polynomials7 not all zero An element d E EX is said to be a Greatest common divisor gcd if 1 dm v ll7r7 2 If there is an elment d E EX such that am v i1r then d d Lemma 31 Let l be a led and EX be the polynomial ping Let f177f7 E EX be polynomials not all zero Suppose d1 and d2 are two GCDs off177fT Then d1 udz for some unit it E F Further if we assume that both d17d2 are monie then d1 d2 That means monie GOD off17fT E EX is UNIQUE Proof By property 2 of the de nition7 d1 udg and d2 vdl for some u7v E EX Hence d1 uvdl Since d1 a 07 we have uv 12397 so u is an unit Now7 if d17d2 are monic then comparing the coe icients of the top degree terms in the equation d1 udg it follows that u 1 and hence d1 12 This completes the proof Remarks 1 Note that Z has only two unit7 1 and 1 When you computed GCD of integers7 de nition assumes that the GOD is positive That is why GCD of integers is unique De nition 32 Let R be a commutative ring A nonempty subset I of R is said to be an ideal of R if 1 my 6 I i y E I 2 ER7 yEI yEI Example 31 Let R be a commutative ring Let f17 7f 6 R Let I ZER Z91f1H39ngT f0 9 ER Then I is an ideal of R This ideal is sometime denoted by f17 7 Also IRf1RfT Theorem 32 Let l be a led and lle be the polynomial ring Let I be a non zero ideal 0leX Then 1 Hde for some d E EX ln fact7 for any non zero d E I with degd least7 we have I FXd Proof Let k mlndegf f 6 I7 f a 0 Pick d E I such that d a 0 and degd k Question Why such a d exists Now clairn IEXM Clearly7 I Q EXd Now7 let f E I By division f qd r with r00rdegrltkNoterf7quIWeprover0 lfr7 07 then degr lt k would contradicts the rninirnality of k So7 r 0 and f qd E EXd This completes the proof Theorem 33 Let l be a led and EX be the polynomial ring Let f7 fT E EX be polynomials not all zero 1 Then f177fT has a GOD In fact a GOD d off17fT is given by dq1f1quot39qrfr for some qi E EX 2 Two GODs dz er by a unit multiple 5 A mom39c GOD is unique Proof Write 1FWUHHM By above thorern7 I EXd for some d E EX We calirn that d is a GCD of fh 7 First note7 dq1f1quot39qrfr for some qi E EX Since fi 6 I7 what have df Now let 1 E I be such that d f 7 for l 17 77 We need to prove that d d This follows from the above equation This completes that proof that GCD exist We have alrady seen 2 and 3 before 4 Prime Factorization De nition 41 Let EX be a the polynomial ring over a eld F 1 An element f E EX is said to be a an reducible over l if f gh for some non unit g h E EX equivalently7 degg gt 0 and degh gt 0 2 f E EX is said to be irredubible over l if it is not reducible 3 A non scalar irreducible element f E EX over l is called a prime in EX Lemma 41 Let R be an integral domain For non zero g 6 R7 Rf Ry if and only if f ug for some unit in R Proof Easy Lemma 42 Let R EX be the polynomial ring over a eld F Let p E R be a prime element and f E R Then Rf Rp R ltgt p does not divide f Proof i We prove by contradiction Assume that p f Then f dp for some d E R Hence RfRp deRp Rp a R So7 this part of the proof is complete Assume p does not divide f By Theorem 327 we have Rf Rp Rd for some d E R Therefore f ud and p ed for some u7v E R Claim d is a unit If not7 since p is prime7 U is an unit Hence f ud uv lp That means7 p f This Will be a contradiction Therefore the claim is proved and d is a unit Hence Rf Rp Rd R The proof is complete Theorem 41 Let R EX be the polynomial ring over a eld F Let p E R be a prime element and y 6 R Then pfg either pf or py Proof Assume p 1 f9 and p does not divide f We will prove that p l 9 We have f9 pm for some it E R Also by above lemma 42 Rf Rp R Therefore 1 pf yp for some p11 6 R Hence 9 zfg yp mop yp This completes the proof Corollary 41 Let R 1le be the polynomial ring over a eld 1 Let p E R be a prime element and f1f2 f E R Then Plf1f2quot39fr gt Plfi forsomei1r Proof Use induction and the above thoerern Theorem 42 Unique Factorization Let R 1le be the poly nomial ring over a eld 1 Let f E R be a nonzero element Then fuP1P2quot39pk where u E 1 is a unit and p1 pk are monie prime elements In fact this factorizton is unique except for order Proof First we prove that factorization as above of f is possible Let degf n we will use induction on 72 Case n 0 If n 0 then f is an unit and we are done Case n 1 lnthis case f uXv withuv E Fandu a 0 Write p X vu The p is prime and f up Case n gt 1 If f is prime then write f uX an1X 1 a1X 10 with up E F and u a 0 Write p fu The p is rnonic prime and f up Now if f is not a prime then f gh with 1699 lt n and degh lt 72 By induction 9 and h have factorization as desired The product of these two factorizations will give a desired factorization off So the proof of existance of the factorization is complete Now we will prove the uniqueness of the factorization Suppose fup1p2pkvq1q2qm where uv are units and p qj are monic primes Assume degf 71 By comparing coef cients of X we get u 1 Therefore we have 9P1P2quot39PkqiltI2quot39qm where g fu is monic Now p1 l q1q2qm By Corollary 41 p1 l qj for some j we may assume j 1 and p1 l ql Since both p1q1 are monic primes we have p1 ql Hence it follows p2pkq2qm Therefore by induction k m and pi q upto order This com pletes the proof lnner Product Spaces Linear Algebra Notes Satya Mandal November 217 2005 1 Introduction In this chapter we study the additional structures that a vector space over eld of reals or complex vector spaces have So7 in this chapter7 R Will denote the eld of reals7 C Will denote the eld of complex numbers7 and l Will deonte one of them In my View7 this is Where algebra drifts out to analysis 11 De nition Let l be the eld reals or complex numbers and V be a vector space over F An inner product on V is a function 7 V X V 6 l such that 1 am by72 1m72by727 for ab E l and 7112 6 V 2 my W for my 6 V 3 am gt 0 for all non zero m E V 4 Also de ne m H This is called norm of m Comments Real Case Assume l lR Then 1 Item means Ly y 2 Also I and 2 means that the inner product is bilinear Comments Complex Case Assume l C Then 1 Items 1 and 2 means that the Loy d2 any E2 12 Example On R we have the standard inner product de ned by my 221 myi Where m 177n E R and y y17yn E R 13 Example On C we have the standard inner product de ned by my 21 E7 Where m 177n E C and y y17yn E C 14 Example Let lF R OR C and V For A myLB b2 6 V de ne inner product A73 Zahar 2 De ne conjugate transpose B P Then A7 B traceBA 15 Example Integration Let l R OR C and V be the vector space of all Rivalued continuous functions on 07 1 For f7 9 E V de ne l m 0 fydt This is an inner product on V In some distant future this Will be called L2 inner product space This can be done in any 77space77 Where you have an idea of integration and it Will come under Measure Theory 16 Matrix of Inner Product Let F R OR C Suppose V is a vector space over l With an inner product Let 617 7en be a basis of V Let pm ehej and P p27 6 Then for v 161 mnen E V and Wy1e1ynen E Vwe have 171 1710 Emifjpij 17 7TLP E H N 03 Hgt 5quot G This matrix P is called the matrix of the inner product With respect to the basis 617 76 De nition A matrix B is called hermitian if B B So7 matrix P in 1 is a hermitian matrix Since my gt 07 for all non zero v 6 V7 we have XPYtgt0 for all nonizero XElF It also follows that P is non singular Otherwise XP 0 for some non zero X Conversely7 if P is a n X n hermitian matrix satisfying such that X137t gt 0 for all X e r then X Y X137t for X e r de nes an inner product on lF 2 Inner Product Spaces We will do calculus of inner produce 21 De nition Let l R OR C A vector space V over l with an inner product gtk7 is said to an inner product space 1 An inner product space V over R is also called a Euclidean space 2 An inner product space V over C is also called a unitary space 22 Basic Facts Let l R OR C and V be an inner product over F For v10 E V and c E l we have 1 llcvlll0lllvll7 2 HUHgt0 if va O7 3 v10 l S v w H Equility holds if and only if 10 We It is called the CauchySwartz inequality 4 vw lt v w It is called the triangular inequality Proof Part 1 and 2 is obvious from the de nition To prove the Part 37 we can assume that v a 0 Write WMU HvH Then 21 0 and wvv wvv 2 wvvxvvw 0 SH 2 H2 Mi v 27w wi vvw H w H H v H2 H v H2 H v H2 This establishes We will use Part 3 to prove Part 47 as follows H v 10 H2H v H2 v7w wiv H w H2H v H2 2R lv7wl H w H23 H v H2 2 l WU l H w HZSH v H2 2 H v NH 10 H H 10 H2 H v H H w W This establishes Part 4 23 Application of CauchySchwartz inequality Application of 3 of Facts 22 gives the following 1 Example 12 gives 1 1 scalHiya 21 21 21 for my 6 1R 2 Example 13 gives 1 13 Z l M WW2 l 11239 W 21 21 21 for my 6 C 3 Example 14 gives traceAB 1 traceAAlZtmceBB12 for AB E 4 Example 15 gives 1 forme lft12dt12 mt 2 ml2 for any two continuous Civalued functions f g on 01 21 Orthogonality 24 De nition Let l be R or C Let V be an inner product space over F H N 03 Hgt 5quot G 5 00 Supppose v10 E V We say that v and w are mutually orthogonal if the inner product 1710 0 OR equivalently if 1071 0 We use variations of the expression 7mntnally orthogonal and sometime we do not mention the word 7mntnally 7 For v10 E V we write 1 l w to mean i and w are mutually orthogonal A subset S Q V is said to be an orthogonal set if vlw for all 1510687 1 a it An othrhogonal set S is said to be an orthnorrnal set if H v H 1 for all UES Comment Note the zero vector is otrhogonal to all elements of V Comment Geometrically7 i l it means i is perpendicular to in Example Let V R or V C Then the standard basis E e1L7 7en is an orthonormal set Example In R2 or C27 we have the ordered pairs 1 my and w 157 are oththogonal Caution Notation Ly here 25 Example Cosider the example 15 over R Here V is the inner prod uct space of all real valued continuous functions on 07 1 Let Then fnt cos27rnt gnt sin27rnt 17f17917f27927m is an orthonormal set Now consider the inner product space W in the same example 15 over C Let fniyn hm ex 27rmt W p Then hmn0717172772777 is an orthonormal set 26 Theorem Let F R or F C and let V be an inner product space over F Let S be an orthogonal set of non zero vectors Then S is linearly independent Therefore cardinalityS S dimV Proof Let v115 71 E S and 1010202Cnvn0 Where Ci 6 F We Will prove Ci 0 For example7 apply inner product gtk7 m to the above equation We get 0101701 0202701 39 Cnvn7v1 0701 0 Since v17v17 0 and v27v1 v37v1 vmvl 07 we get 01 0 Similarly7 Ci 0 for all 239 17 772 27 Theorem Let F R or F C and let V be an inner product space over F Assume 617627 7eT is a set of non zero orthogonal vectors Let 1 E V and UCi 102 2CT T Where Ci 6 F Then v7 ei T H 6239 H2 Ci forz391r Proof For example7 apply inner product 761 to the above and get U761 0161761 Cl ll 61 HZ So7 01 is as asserted and7 similarly7 so is Ci for all 239 7 28 Theorem Let 1 R or 1 C and let V be an inner product space over 1 Let v1v2 1 be a set of linearly independent set Then we can construct elements 61 62 6 E V such that 1 61 62 6 is an orthonormal set 2 6k 6 Spanv1 vk Proof The proof is known as GramSchmidt orthogonalization process Note v1 a 0 First let vi 61 H v1 H Then 61 H 1 Now lat 02 127 6161 62 ll 02 02761 1 ll Note that the denominator is non zero 61 1 62 and 62 H 1 Now we use the induction Suppose we already constructed 61 ek1 that satis es 1 and 2 and k S 7 Let we 211 U167 629 6k 1671 ll vk Zi1vk7 i i Note that the denominator is non zero 6k H 1 and 6k 1 e for 239 1 k 7 1 From construction we also have Spanv1 vk Span61 ek So the proof is complete 29 Corollary Let 1 R or 1 C and let V be a nite dimensional inner product space over 1 Then V has an orthonormal basis Proof The proof is immediate from the above theorem 28 Examples Read examples 12 and 13 page 282 for some numerical com putations 210 De nition Let l R or l C and let V be a inner product space over F Let W be a subspace of V and v E V An element 100 E W is said to be a best approximation to v or nearest to v in W7 if Hv7w0H S Hv7wH for all wEW Try to think what it means in R2 or C when W is line or a plane through the origin Remark I like the expression 77nearest77 and the textbook uses 77best ap proximation I will try to be consistent to the textbook 211 Theorem Let l R or l C and let V be a inner product space over F Let W be a subspace of V and v E V Then 1 An element 100 E W is best approximation to v if and only if v 7100 T w for all it E W 2 A best approximation 100 E W to v if exists7 is unique 3 Suppose W is nite dimensional and 6176277 n is an orthonormal basis of W Then n 100 Zvekek 191 is the best approximation to v in W The textbook mixes up orthogonal and orthonormal and have a condition the looks complex We assume orthonormal and so ei H 1 Proof To prove 1 let 100 E W be such that v 7 100 T w for all it E W Then7 since it 7 100 6 W7 we have H viw H2H viwowow H2H viwo H2 H woiw H2le viwo H2 Therefore7 100 is nearest to v Conversely7 assume that 100 E W is nearest to i We will prove that the inner product i 7 100710 0 for all it E W For convenience7 we write v0 v 7 100 So7 we have Hvo Hz llviw H2 Eqn71 9 for allw E W Writev7w v7w0w07w v0w07w Since any element in 10 van be written as 100 7 w for some 10 E W Eqn l can be rewritten as H v0 HZSH v0 10 H2 for all 10 E W So we have H vo HZSH v0 10 H2H v0 H2 2113 lv07wlHwH2 and hence 0 S 2Rev0w w H2 Eqn 7 H for all 10 E W Fix w E W with w a 100 and write v0w0 7 w 739 w0 7 w 7 H we 7 w H2 Since 739 E W we can substitute 739 for w in Eqn ll and get l 007100 w l2 0lt7 Hwo wHZ Therefore v0w0 7 w 0 for all 10 E W with 100 7 w a 0 Again since any non zero element in W can be written as 100 7 w it follows that v0w 0 for all 10 E W So the proof of 1 is complete To prove Part 2 let 100101 be nearest to 1 Then by orthogonality 1 we have H100 7 101 H2 wo 7 101100 7 101 wo 7 w1w0 7 v v 7 101 0 So Part 2 is established Let 100 be given as in Part 3 Now to prove Part 3 we will prove 10071 l e for 239 1 n So for example wo 7 v61 10061 7 v61 v616161 7 v61 0 So Part 3 is established 212 De nition Let l R or l C and let V be an inner product space over F Let S be a subset of V The otrhogonal complement SL of S is the set of all elements in V that are orthogonal to each element of S So SL UEVZ UJ LU for all 1063 It is easy to check that 1 SL is a subspace of V 2 0L V and 3 VL 0 213 De nition and Facts Let l R or l C and let V be an inner product space over F Let W be a subspace of V Suppose v E V we say that w E W is the orthogonal projection of v to W if 10 is nearest to v 1 There is no guarantee that orthogonal projection exists But by Part 2 of theorem 211 when it exists orthogonal projection is unique 2 Also if dimW is nite then by Part 3 of theorem 211 orthogonal projection always exists 3 Assume dimW is nite De ne the map 7139W V a V where 7rv is the orthogonal projection of v in W The map 7139W is a linear operator and is called the orthogonal projection of V to W Clearly 7 WW So 7139W is indeed a projection 4 For 1 E V let Ev v 7 7139Wv Then E is the othrogonal projection of V to Wt Proof By de nition of 7fw we have Ev v 77rWv E WL Now given 1 E V and 10 E WL we have 1 7 Evw WWvw 0 So by theorem 211 E is the projection to WL Example Read Example 14 page 286 214 Theorem Let l R or l C and let V be an inner product space over F Let W be a nite dimensional subspace of V Then 7139W is a projection and WL is the null space of 7139W Therefore7 V W 619 WL Proof Obvious 215 Theorem Bessel s inequality Let l R or l C and let V be an inner product space over F Suppose U17U27 7vn is an orthogonal set of non zero vector Then7 for any element 1 E V we have H1le n l Who l2 2 s H v H2 k 1 Also7 the equality holds if and only if n v Z Ulvkivk H vk H Proof We write7 1 6k H we H and prove that H0760 l2 lt H v H2 k1 Where 617627 7671 is an orthonormal set Write W Spm 17 27 7en By theorern 2117 n 100 2v7ekek 191 is nearest to v in W So7 v 7 100100 0 Therefore7 H v H2H v 7100 100 H2H v 100 H2 H we H2 So TL H v H2 2 H wo H2 2 l m V k1 12

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "Selling my MCAT study guides and notes has been a great source of side revenue while I'm in school. Some months I'm making over $500! Plus, it makes me happy knowing that I'm helping future med students with their MCAT."

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

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.