### Create a StudySoup account

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

Already have a StudySoup account? Login here

# Applied Linear Algebra I MTH 261

PCC

GPA 3.98

### View Full Document

## 57

## 0

## Popular in Course

## Popular in Math

This 14 page Class Notes was uploaded by August Feeney on Monday October 19, 2015. The Class Notes belongs to MTH 261 at Portland Community College taught by Staff in Fall. Since its upload, it has received 57 views. For similar materials see /class/224650/mth-261-portland-community-college in Math at Portland Community College.

## Reviews for Applied Linear Algebra I

### 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: 10/19/15

MTH 261 Mr Simonds class a graphical discussion of spanning vectors and translation vectors 2x6y4 The augmented matrix representation of this 4 x 12 y 8 Consider the system of equations 2 4 1 3 2 system is 4 which is equivalent to the reduced row echelon form matrix 0 J 8 0 0 H x3y2 So the original system of equation is eqUIvalent to the system of equations 0 that is the two systems have the same solution set Since the first equation in the reduced system has two remaining variables we need to designate one variable as free as the other as dependent Sticking with orthodoxy let39s make y the free variable ie y is free to be any number I The value of x is now dependent upon the value we choose for y If y t then x3y2 3 x2 3y 3 x2 3t 2 3 t 2 3 So solutions to both systems of equations have form t 0 t 1 2 3 This makes our translation vector 0 and our spanning vector 1 J Figure 1 shows a graphical representation of the spanning vector with its tail drawn at the origin Since there is only one vector in the spanning set linear combinations of the vectors in the spanning set consist entirely of multiples of this one vector If we were to draw every possible multiple of 11 onto Figure 1 we would end up with the line shown in Figure 2 Figure 1 Figure 2 The spanning set consists of this one vector Every multiple of the spanning vector lies along this line Page 1 of 2 MTH 261 Mr Simonds class a graphical discussion of spanning vectors and translation vectors 2x 6 Take a few poinTs of The line in Figure 2 and plug Them inTo The expression 4 my Unless you x y 0 goof up you will geT 0 every Time ThaT is every poinT on The line in Figure 2 saTisfies The sysTem 2x6y0 The seT of all soluTions To This sysTem is called re null ne of The 4x 12y 0 of equaTions 2 6 3 maTrix 4 12 So a spanning seT for The null space of This maTrix is 1 J 2 2 The TranslaTion vecTor for our original sysTem is In Figure 3 I39ve drawn The vecTor 0 originaTing from several poinTs ThaT are in The null space If you connecT all of These vecTor heads you geT a new line ThaT does noT pass Through The origin 2x 6 y If you do your Take a few poinTs from This new line and plug Them inTo The expression 4x 12y 4 ariThmeTic correchy you geT 8 every Time ThaT is every poinT on This TranslaTed line is a soluTion 2x6y4 To The sysTem 4x 12y 8 In summary linear combinaTions of The spanning vecTors resulT in soluTions To The sysTem 2x 6y 0 4x12y039 Adding The TranslaTion vecTor To any linear combinaTion of The spanning vecTors resulTs in a soluTion h 2x 6 y 4 To T e Sysfem 4x 12 839 The TranslaTed null space gives The soluTions To The y sysTem The linear combinaTions of The spanning seT solve The sysTem I 2x 6y 4 39 4x 12y 8 2x 6y 0 4x 12y 0 I 76 5 r4 73 72 71 N 17 s This is called The null space of 7 2 6 J 6 4 12 I Figure 3 l Page 2 of 2 MTH 261 Mr Simonds class Vector Spaces A vector space is de ned by a nonempty set of objects V together with two operations called vector addition EB and scalar multiplication These entities de ne a vector space if and only if each ofthe following properties are true V 11 e V V e V w e V and V k 6 1R 1 6 1R These 10 properties are called the axioms ofa vector space Closure axioms 11 The set V is closed over vector addition 11 EB V e V 12 The set V is closed over scalar multiplication k 911 e V 2 Identity axioms 21 The set V contains a zero vector additive identity El 0 e V 3 11 ED 0 u 22 1 is the scalar multiplicative identity 1 u u 3 Additive Inverse axiom 31 Every element in V has an additive inverse V 11 e V El u e V 3 11 EB u 0 4 Commutative property axiom 41 Vector addition is commutative 11 EB V V EB 11 5 Associative property axioms 51 Vector addition is associative 11 EB V EB w 11 EB V EB w 52 Multiplication is associative k 9 1 11 kl 11 O Distributive property axioms 61 Scalar multiplication distributes over vector addition k 9 11 EB V k 311 ED k 9 V 62 Scalar multiplication distributes over scalar addition k 1 11 k 9 11 ED 1 11 The most commonly encountered vector spaces are the sets of all n X 1 column vectors 1Rquot coupled with traditional vector addition and scalar multiplication these type vector spaces are called Eucia ean anes However several other sets and operations satisfy the axioms of a vector space so to avoid repetitiveness mathematicians keep the definition of a vector space as general as possible so that the proof of additional properties covers all such structures Vector spaces are but one of many algebraic structures studied by mathematicians Some other common algebraic structures are groups rings and fields Each of these structures begins with a set of objects and one or more operation Each of them requires closure but the other defining properties vary from structure to structure Page 1 of 8 MTH 261 Mr Simonds class LeT39s go ahead and prove someThing ThaT is True for all vecTor spaces WhaTever The specifics of a given vecTor space The addiTive idenTiTy for ThaT space is unique We39re going To prove This using The principal of great bx confraachoI In This Type of proof we assume The opposiTe of whaT we are Trying To prove and show ThaT This assumpTion leads eiTher To an irreconcilable cg radicTion or To a sTaTemenT ThaT is ob39ecTivel false 39 39 1 Al L K J xLJ y We 9 Luquot 39lv OA 44 0 VA F Avian ll n o Ii 6 JJii w 1l7 rm A 5 x V 0 0 o a 5 01 c M L s v A irx u quotL q l l 13147 Skio 0 5 7 J g I 14M VLqu 05 0 0 05 ujo J 5 05 0 This lasT facT we proved is one of several elemenTary properTies of vecTor spaces While These properTies are fundamenTal To The naTure of vecTor spaces we don39T call Them axioms because They all follow direchy from The axioms of a vecTor space ThaT is each of These elemenTary properTies can be proven To be True based upon The 10 properTies sTaTed in The definiTion of a vecTor space Some elementary properties common to all vector spaces k u0 2 k0andoru0 71 u7uVueV 1 The zero vector of any given vector space is unique 2 A vector s additive inverse is unique 3 0 11 0 V 11 e V 4 k 0 0 V k e 5 6 Once we have proven a properTy abouT vecTor spaces we can add ThaT facT To our arsenal when consTrucTing oTher proofs abouT vecTor spaces For example elemenTary properTy 3 is proven on page 11 of The Penney TexT LeT39s use ThaT properTy To help us prove elemenTary properTy 6 What dowe need to establish L ED 1 33 l O a NJ J has 3 A 9 A 39quot 303 an I 0 OF 7 A OEI Z Page 2 of 8 lew L1 if izir 57 MTH 261 Mr Simonds class To prove ThaT a given sTrucTure is a vecTor space we need To show ThaT all 10 of The axioms of a vecTor space are True for ThaT sTrucTure SomeTirnes The proof of a properTy is noThing more Than a sTaTemenT of iTs validiTy someTimes There is more work involved 2 5t LeT39s define V exit te g LeT39s de ve Toape H iTVeLV 7 3 39 7 25t 25t 25t5t 2 5t 2 Sin 1 2 1 2 and k 371 1 371 2 3713713 371 37kt LeT39s prove ThaT This sTrucTure is a vecTor space CUJL l 1 5439 511 2 YltIL i 3 EL 3 4 SH 2Kk 2 l KJC l 316ch J KH 1 felvix iw 1 54 2 24 H 14quot 2 3 3 j4 3 5 Z yl proI 3 4 quot x 2 5 a 3 E I CH 10 1S D i 30a T 11 TM Page 3 of 8 MTH 261 Mr Simonds class A J J1 JK id crl I 2 L I NK J 1115f 49 3 39 3 7 99531 3quot V 145Ljev 7 1 391 Y 1 9 S fJ39w 3 Jlt 3gt 3 Jc lt gt Lquot 31644 L quotgt Z I 1 4 1 quot v x 3 Jr s kH LL 1 n 4 1 CH2 H 1 yL 1 chI 511 36315 3quot 1 3 39Ll 2 2 in 1451 1 1rJ39l 3 392 x1 KRAM39C VD P Ofer ki 5H psi VP in 5 nltx LPUJCI Page 4 of 8 MTH 261 Mr Simonds class The associafive properfies 25t1 25t2 25t3 ea ea 3 5 3 5 3 5 51 25t15t2 e 25t3 3 t1 t2 3 t3 ea 25t3 3 t3 5 t1t25t3 3 t1t2 t3 25t15 t2t3 5 55 2 3 t1t2 2 3 t1 t2t3 3 1 25t1 25t2t3 3 t2t3 t 251 25t25t3 3 t1 3 t2 t3 ea 25t1 25t2 e 25t3 3 t1 3 t2 3 t3 25t 2Slt k l k 3 t 3 lt 2 5lt fizslgi 3 Y 2 3 t 52 k 15 25t kl 3 t Page 1 of 2 MTH 261 Mr Simonds class The dis rribu rive properfies waitife tifDk Frzii k iiiifif 25kgg z warm 312353 2322 32239 2 iiij iZZ 255 255 k k 2 klt 3 3 3 5 5 kl 5 61 62 kt t5lt ltkrgt Ir i39gtet1 25kt 251t EB kl 235tt 2 3 3 kt 3 lt 25t 25t k l 3 t 3 t Page 2 of 2 MTH 261 Mr Simonds class To prove ThaT a given sTrucTure is Imfa vecTor space you only need To show ThaT any one of The 10 axioms of a vecTor space is false for ThaT sTrucTure 3ft addiTion and scalar mulTiplicaTion in The TradiTional manner 2 5t tell does 107 creaTe a vecTor space if we define vecTor LeT39s prove ThaT The seT V Olequ 35 owl L waving 3 2V M i moi 05 1 2 51 A c a 3 5 I Ill239lgl6rc all 7CLle LeT39s prove ThaT The seT of all 3 x1 vecTors does Imf creaTe a vecTor space if we define scalar mulTiplicaTion in The TradiTional way buT define vecTor addiTion To be The cross producT Crux FMJJd u W CW 4 4 l 9 Ul lramp 1 Laqu L LeT39s prove ThaT The seT of all 3 x1 vecTors does Imf creaTe a vecTor space if we define scalar mulTiplicaTion in The TradiTional way buT define vecTor addiTion as u v 0V u Elm v 63 zr l H lJIKJg IJc K l39y For a 43 km M vatHr 4 1 lt1quot 75 l tr l 6 U J IS Page 5 of 8 MTH 261 Mr Simonds class Subspaces Suppose that V is a vector space and that WC V Then we say that W is a subsguc of V if W is closed over and gt9 Theorem The null space of an m x n matrix is a subspace oflk Proof We can establish both closure properties if we can show that kuveWVkel and VueWveW For simpliciTy39s sake we dispense wi rh The formalized symbols for vec ror addi rion and scalar mul riplica rion when working wi rh Euclidian spaces Ana L iLa 5 IzAii393 A KULi39ir l K b L V r lt V Page 6 of 8 MTH 261 Mr Simonds class Theorem The column space of an m x n matrix is a subspace of Proof LeT39s define M2X2 To be The vecTor space wiTh defined as TradiTionaI maTriX addiTion and 9 defined as TradiTionaI scalar mulTiplicaTion LeT39s furTher define To To be The subseT of all 2 x 2 maTrices wiTh a Trace of O LeT39s show ThaT To forms a subspace of MHZ Page 7 of 8 MTH 261 Mr Simonds class LeT39s define T1 To be The subseT of all 2 x 2 maTrices wiTh a Trace of 1 and show ThaT T1 does m form a subspace of MHZ Define fzw To be The subseT of all real valued funcTions ThaT map 2 To 0 and H2 The subseT of all real valued funcTions ThaT map 0 To 2 If we define f To be The vecTor space of all real valued funcTions wiTh vecTor addiTion and scalar mulTiplicaTion defined as expecTed Then one of our subseTs forms a subspace of f and one does noT LeT39s prove This Pracfice HW AnTon secTion 51 pp226 228 1 17 odd 19a AnTon secTion 52 pp238 239 1 9 odd Page 8 of 8 I N A vJ xigf w 114 VJ f L l m4 3 10 A Q 0 S 3r 0 D 39uf ye a 0 f 9 S L L I II L T 2 474 Fug f J quot39J Oj I 7 P 7gt rwm5 01 3391 K D0 139 O I n gt a e u I 11f 0 I o int fl y an I n IIy Fry039ll I I lt o I 0 l l 1 I 1 W 39 r l f v TLVULVI h 0 fa Arf J39 1 1 1 f l 39I JJ by I amp Wafffr WA cf 1 H lf v I 0 9 7 o fp q n t C r77 39fr J j 3 qay f v I MIA mom j M w 7 o o fbf o 31 j 0j tf Q 31VV J on jalll a fvlr39 I C Cvr A I

### 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

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "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!"

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### 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.