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Applied Linear Algebra I

by: August Feeney

Applied Linear Algebra I MTH 261

Marketplace > Portland Community College > Math > MTH 261 > Applied Linear Algebra I
August Feeney
GPA 3.98


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This 8 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 35 views. For similar materials see /class/224650/mth-261-portland-community-college in Math at Portland Community College.


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Date Created: 10/19/15
MTH 261 Mr Simonds class a numerical discussion of spanning vectors and translation vectors 2x5y4z 5wc1 x 4y4z 5wc2 Consider the system of equations x5y 8210wc3 3x7y82 10wc4 12 15 4 5 The spanning vectors for any system of this form are and 0 Let39s make sure that we understand what this means 12 15 4 5 These vectors span the set k 1 1 0 where k and l are both real numbers We can 4 5 fully communicate that with the notation k 1 1 0 k E R l E R Let39s replace k and l with a couple pairs of specific numbers and see what vectors get generated I39m truly just randomly picking two pairs of Makes sure that you do the math yourself to verify these vectors If we let k 2 and l 9 the vector that is generated is If we let k 7 and 11 the vector that is generated is 11 Let39s plug the entries from each of these vectors into our system and see what the constants turn out to be Once again make it real to yourself by actually doing the calculations Page 1 of 4 MTH 261 Mr Simonds cass a numerical discussion of spanning vectors and translation vectors Using 159 53 2 9 the results are Using 249 83 7 11 the results are Hmm seems like a mighty big coincidence that both of these vectors generated by the spanning vectors create nothing but constant terms of 0 Let39s see if this is always the case 12 15 12k151 4 5 4k 51 k l 1 0 k 0 1 I When we substitute these expressions into our system the constants do in fact always turn out to be 0 The reader should definitely verify the algebra 2 12k 151 5 12k 151 4 12k 151 5 3 12k 151 7 4k 514k 510 4k 514k 510 4k 51 8k1010 4k 518k 1010 AA AA 2 5 4 5 0 1 0 12 15 0 1 4 4 5 0 0 1 4 5 0 If we find the RREF of the matrix the result is 1 5 8 10 0 0 0 0 0 0 3 7 8 10 0 0 0 0 0 0 x122 15w 5 This gives us the system of equations Designating z and w as our free y 42 5 w 2 variables and specifically assigning Z s and w 1 gives us Page 2 of 4 MTH 261 Mr Simonds class a numerical discussion of spanning vectors and translation vectors x122 15w0 3 x 12215w 3 x 12315t y 4z5w0 3 y4z 5w 3 y4s 5t So every soluTion To The sysTem in which all four consTanT Terms is O has The form 12s15t 12 15 4s 5t 4 5 3 t s 1 t 0 1 To recap The spanning vecTors span The seT of all soluTions To The sysTem of equaTions where each of The consTanT Terms in The sysTem is zero This seT of squTions when wriTTen as column vecTors is called fIe null ace of re caef cenf mafrxv Now IeT39s add The vecTor 10 3 00 To each of our spanning vecTors don39T worry where This vecTor came from I did some irrelevanT advance work To deTermine iT 10 12 2 10 15 25 3 4 1 3 5 8 and 0 1 1 0 0 0 0 0 0 LeT39s plug The enTries of each of These vecTors inTo our sysTem and see whaT consTanTs geT generaTed Again you should verify These calcuIaTions yourself 2 25141 505 2255 840 515 2 4141 502 25 4 840 512 251 81100 5 and 255 8 80101 5 3 27181 1009 3257 880 1019 Page 3 of 4 MTH 261 Mr Simonds class a numerical discussion of spanning vectors and translation vectors MeThinks The vecTor39 10 3 00 has Taken The null space of The coefficienT maTr ix and 2x5y4z 5w5 x 4y4z 5w2 TransIaTed iT To The soluTion seT To The sysTem x5y 8210w 5 3x7y82 10w9 10 12 15 3 4 5 ThaT is a general form for The soluTion seT To ThaT sysTem is S 1 t 0 1 2 5 4 5 5 1 0 12 15 10 1 4 4 5 2 0 1 4 5 3 You should verify ThaT The RREF for is 1 5 8 10 5 0 0 0 0 0 3 7 8 10 9 0 0 0 0 0 2 5 4 5 0 1 0 12 15 0 1 4 4 5 0 0 1 4 5 0 Compare ThaT To The RREF for which is 8 10 0 0 0 0 0 0 3 7 8 10 0 0 0 0 0 0 See The connecTion Page 4 of 4 MTH 261 Mr Simonds class the Google search algorithm part 2 and a bit of bonus info Suppose There are only 4 pages on The WWW and Their link sTrucTure is as illusTraTed below The link dangling random eXiT and Google maTrices for This web are lisTed 0 050 03 0 0 quotEai39 E39 5 0 5 0 0x 0 0 L D 5 0 0 1 0x 0 0 0 0 0 0 0x 0 0 lt 25 25 25 25 R 25 25 25 25 25 25 25 25 03 75 25 4625 03 75 4625 25 4625 03 75 4625 25 0375 8875 I 0375 25 0375 0375 G85L85D15R The sTandard Markov chain iTeraTion algoriThrn from one sTaTe vecTor To The neXT is QHUG 0 25 25 If we leT Km 25 iT is easy To calculaTe The neXT 2 sTaTe vecTors 25 196875 2758984375 X0 me 303125 and Km GXm 3595703125 409375 2626171875 090625 1019140625 We can inTerpreT These sTaTe vecTors Thusly GusTaf is placed in from of a compuTer and opens The web browser One of The four pages opens aT random wiTh each page having The same probabiliTy of being The one ThaT opens ThaT is The meaning of sTaTe vecTor zero xl0 Page 1 of 4 MTH 261 Mr Simonds class the Google search algorithm part 2 and a bit of bonus info State vector 1 Km are the probabilities for the second page Gustaf finds himself on There is roughly a 20 chance that the second page he visits is Page 1 a 30 chance that the second page he visits is Page 2 a 41 chance that the second page he visits is Page 3 and a 9 chance that the second page he visits is Page 4 Remember that we are considering the first page he visits to be the one that is randomly loaded when the browser opens Remember too that he can visit the same page in consecutive states State vector 2 xlzl are the probabilities associated with the third page visited by Gustaf These state vectors are fofa z degenden upon the initial state vector Let39s now suppose that the web browser always opens on Page 1 Let39s calculate and write down Km x0 and xlz Please follow fire a39I39I39ecfMs I39ll fire box Store the Google Matrix as x Type the vector 1 0 0 0 T onto your entry line and press ENTER Type xans l onto your entry line and press ENTER the result is xll Press ENTER the result is Km 1 1 thpr If we continue to press enter we will get successive state vectors Because the Google matrix is primitive we know that the Markov chain will converge to the steady state vector regardless of the initial state vector Continue pressing ENTER to determine the Google steady state vector for this web 3 1 v 393 N 339 These calculations all happened quickly because we are working with very small matrices and vectors The calculations would take significantly longer if our Google matrix were a trillion by a trillion That39s the motivationfor using the link matrix in the calculation lots of zero entries mean lots less calculation time Page 2 of 4 MTH 261 Mr Simonds class the Google search algorithm part 2 and a bit of bonus info Let39s go ahead and figure out the iteration algorithm in terms of L gt K x l 39X 639 x J 13 IL Y quot a V Z 2 439 g l I YYLXK IYS quota II 6 a Y I a v1 0 1 n Lquot l YSLXW 4 25 Let39s assume again that Kw Let39s calculate X0 and Km using our new iteration 25 algorithm and compare the results with the values stated on page 1 of this document Pease foIow fire redans I39ll fire box 1 Store the Link Matrix as y 2 Store the vector L 111 T as z 3 Type the vector 252525 25T onto your entry line and press ENTER 4 Type 85yans l854ans121z154z onto your entry line Press ENTER the result is XU Compare the result to the value stated on page 1 5 Press ENTER the result is Km Compare the result to the value stated on page 1 Continue pressing ENTER and see that state vectors converge to the steady state vector relatively quickly Page 3 of 4 MTH 261 Mr Simonds class the Google search algorithm part 2 and a bit of bonus info U turnll 2 73 74 73 17 73 4 9 73 12 6 7 7 3 2 79 Let A and B 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Find the dot products of the row vectors from A with the row vectors from B What do these dot products tell us about the null spaces and row spaces of me respecti ve matrices 3 u o IrL 1 a a o MM A 639 l39 391 I b J k D 1151 D aw PHI 39 R mu m 7 MI 13 Do these properties always go hand in hand J 39QOLJ A NuIIUS lt Row A J Ju9 Mum mum J 5 Jim mum Q J 4 V Howmightthis beuseful L J JIIP C L 5 L l r a I Hl A M J r x A U G J 0 u 3 Ll g 5 d 6 F l udquot a JJc39bw B Ukug nusr ILvl39Il JLquot 0 SrLt uo n Page 4 of 4


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