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# VECTOR CALCULUS I MTH 254

OSU

GPA 3.6

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This 11 page Study Guide was uploaded by Miss Johan Jacobson on Monday October 19, 2015. The Study Guide belongs to MTH 254 at Oregon State University taught by B. Peterson in Fall. Since its upload, it has received 28 views. For similar materials see /class/224438/mth-254-oregon-state-university in Mathematics (M) at Oregon State University.

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Date Created: 10/19/15

Ei envalues Ei envectors and uadratic Forms Exam les Mth 254 Mar 12 2001 Bent E Petersen Filename 254w2001eigenvectsexampmws Linear Algebra supplement for Vector Calculus 1 In this worlsheet I give a few examples of eigenvalues and eigenvectors of matrices in the fust part In the second part I provide the answeres as given by Maple to the problems in the Mth 254 Study Guide Parks 200001 revision Ifyou try a few of these examples by hand you will develop a deep appreciation for Maple or a similar tool gt restart gt withlinalg Warning new definition for norm Warning new definition for trace Part 1 Examples Example 1 gt A1matrix33 122222 3 6 6 l 2 2 A1 2 2 2 3 6 6 gt eigenvals A1 032 I Al has distinct eigenvalues Thus Al is diagonizable Let39s diagonalize it explicitly gt v1eigenvectsA1 V1i311010101121210 gt sl for k from 1 to 3 do slsl union op3opk v1 od unassign k 51 101011210 gt S H matrix30 1 S arrayl 3 l 0 augmentS1opsl Page I 0 12 S110 1 1 1 0 39gt inversesl amp A1 amp Sl D1evalm 0 0 0 D1 0 3 0 0 0 2 Alternate Calculation for Example 1 There is an easier way to do these calculations Let39s start over Fiist we compute the eigenvaules of A1 and put them in a diagonal mattiX We can do this as follows gt D1bdiageigenvals A1 0 0 0 le 0 3 0 0 0 2 Now if A1 is diagonizable the it will be 39similar to le so we ask Maple to check gt issimilar A1 D1b P1b true The retum value indicates A1 is diagonizable In this case Plb has been assigned a mattix which plays the role of the inveise of 81 above let39s check it gt evalmP1b S1binverseP1b 1 1 1 6 3 6 1 2 2 3 3 3 1 1 1 0 3 2 SM 6 0 1 6 3 0 It doesn39t look the same but let39s check that it works gt inverseslb amp A1 amp Slb evalm Page 2 That worked as expected Example 2 39gt A2matrix33 74 44 8 1 4 1 8 7 4 4 A2 4 8 1 4 1 8 gt eigenvals A2 l 9 9 9 This time we have an eigenvalue 9 with algebraic multiplicity 2 We do not yet know if A2 is diagonizable or not Let39s compute the eigenvect01s 39 gt v2 eigenvects A2 v2914119201115450 We see that 9 has geomettic multiplicity 2 Thus A is diagonizable Let39s diagonalize it explicitly N gt s op3op1v21 union op3op2v21 s2011140411 N gt S matrix30 S2array1310 gt S2augmentS2op52 4 1 0 S2 1 4 1 1 0 1 39gt inverse52 amp A2 amp 2 D2evalm 9 0 0 132 0 9 0 0 0 9 Again as expected Example 3 gt A3matrix33 11 1 13 1 120 Page 3 1 1 1 A3 1 3 1 1 2 gt eigenvals A3 2 1 1 Again we have an eigenvalue 1 with algebraic multiplicity 2 We do not yet know if A3 is diagonizable or not Let39s compute the eigenvectors lgt 53zeigenvects A3 s31211121011 Oops We see the eigenvalue 1 has geometric multiplicity 1 only one eigenvector for 1 is listed by Maple Thus A3 is not diagonizable The quotclosestquot we can get it the Jordan canonical form gt jordan A3 CON Ot IO D Il O You probably suspect we could have used the jordan lnction above to verify the diagonizability of A1 and A2 Ifso you are correct but it would have been less ln 39 gt jordan A1 2 0 0 0 3 0 0 0 0 39 gt jordan A2 9 0 0 9 0 0 9 Example 4 This time consider a symmetric matrix We know that a symmetric matrix has real eigenvalues and is diagonizable whether it has repeated eigenvalues or not gt A4matrix44 01 1211o 1 1o 112 1101 Page 4 0112 1 1 0 1 Ali1 0 1 1 2 1 1 0 Ifwe compute the exact eigenvalues we get a nearly incomprehensible expression so let s compute approximate eigenvalues instead gt evalf eigenvals A4 16 2 1681330643604977 3323404276086478 2 103915 1 357926367518500 2 103915 I Those imaginary parts should not be there They are due to roundolT as you can see by doing the calculation at several precisions For example I gt evalf eigenvals A4 22 2 1681330643604977375475 1 10211 3323404276086477625772 1 10211 357926367518499749704 1 103921 I That A4 is diagonizable we can verify by computing the Jordan canonical form gt evalf jordan A4 2 0 0 0 0 1681330643 3 10399 I 0 0 0 0 3323404277 0 0 0 0 357926367 Example 5 Here is another symmetric example 39gtA5matrix440123101221013210 0 1 2 3 l 2 0 l l 0 gt jordan A5 Page 5 2J2 0 0 0 0 2 2 0 0 0 0 2q10 0 0 0 0 2 5 We see A5 is diagonizable as expected Part 2 Solutions for the Study Guide Lesson 22 Here are Maple s solutions for the problems in Lesson 22 The matrices are numbered in accord with the problem numbers You may note some trivial dilTerences between the solutions and the selected solutions in the Study Guide Keep in mind any nonzero multiple of an eigenvector is an eigenvector More generally any nonzero linear combination of eigenvectors corresponding to the same eigenvalue is an eigenvector A er studying the examples above you should be able to decipher Maple s output Explicitly the eigenvects function returns a list consisting of ordered triples of the form eigenvalue algebraic multiplicity linearly independent set of eigenvectors The number of vectors in the returned set of eigenvectors is the geometric multiplicity of the eigenvalue The matrix is diagonizable if and only if the algebraic and geometric multiplicities are equal for each eigenvalue gt M01matrix22 6215 M01396 2 39 1 5 41 1 11 7 12 11 gt M02matrix22 412 3 imp394 1 39 2 3 5515 4 115 2 1 121 gt eigenvects M01 gt M03matrix22 4 152 M03 4 391 39 5 2 gt eigenvects M03 1 2 1 2 3211g11 3 211 g11 Page 6 i l i i matrix22 51 5 3 MO4395 1 39 5 3 eigenvectsM04 4211112142111121 M05matrix22 622 3 M05 396 2 39 2 3 eigenvectsM05 2 1 12 7 1 2 11 M06matrix226223 6 2 M063972 3 2112a7a121 eigenvectsM06 M07matrix33522220202 5 2 2 AJOW 2 2 0 2 0 2 eigenvectsM07 311222101161411 M08matrix33533320302 5 3 3 A B 3 2 0 3 0 2 eigenvectsMOB 29 1 09391 8 1 1 3919 1 193919391 M09matrix33 7030 72325 7 0 3 M09 0 7 2 3 2 5 eigenvectsMOQ 3 13 3 61 31 81321 7110 M10matrix33 101o12125 1 AIME 0 1 NH0 1 2 5 eigenvectsMlO 69 0 13913929 1392 Page 7 Quadratic Forms Once we know the eigenvalues of the symmetric matrix of a quadratic form we can write down the diagonalized version of the quadratic form In general we are a er more We want to know an explicit change of variables that diagonalizes the quadratic form With a little coaching Maple can solve this sort of problem I39ll show you one way of doing it here but there may be better ways Ifyou use Maple you should experiment gt Q11 6xquot24xy 3yquot2 Q11 6x24xy 3y2 39gt M11hessianQ112 xy 6 2 M1172 3 39 gt D11diageigenvalsM11 D117 0 0 2 gt issimilar M11 D11 P11 true We would like to use the inverse of P11 to de ne our change of variables but we have to normalize the columns of the inverse first Let39s call the result 811 gt normalizecol inverseP11 1 normalizecolinverseP11 2 S11aug39mentmatrix20 2 1 v5 45 l 2 V5 45 5 5 gt sub112evalmS11 amp matrix21uv 2 l J uJ V sub11 1 2 g wg v gt subsxsub1111 ysub1121 Q11 Q11 expand 6 uem4 u vJG u vJ 3GWEN Q11 7u2 2v2 Page 8 We have success llly diagonalized the quadratic form Q11 Let39s try another one Q126x24xy3y2 gt M12 hessian Q122 xy 6 2 M12 gt Q12 6xquot24xy3yquot2 2 3 gt D12 diageigenvals M12 2 0 D12 39io 71 true lgt issimilar M12 D12 P12 gt We already knew that M12 and D12 are similar because M12 is symmetric and so diagonizable The only reason for asking Maple about it is that is a convenient way to assign the desired value to P12 just as above see also Example 1 above gt normalizecol inverseP12 1 normalizecolinverseP12 2 512aug39ment matrix20 1 2 4 2 1 5 El gt sub12evalmS12 amp matrix21 uv sub12 2 1 ug v gt subsxsub1211 ysub1221 Q12 Q12 expand 1 2 2 1 2 2 1 2 1 2 6 ug vj 4 g ug v ug v3E ug v Q122u27v2 That nishes the problems in the study guide but let s also consider a quadratic form in 3 variables gt Q3xquot212xy24yz 6xz6yquot26zquot2 Q3x212xy24yz 6xz6yz6z2 Page 9 gt MhessianQ2 xyz 3 6 3 M 6 6 12 3 12 6 39 gt D0diageigenvalsM 6 0 0 0 q85 0 D0 2 9 3 0 0 85 2 2 gt issimilarMD0P true 39 normalizecol inverseP 1 normalizecol inverseP 2 norma lizecolinverseP 3 Saugment matrix30 15 6 1 6 1 21 4850744g 4850745 1 217E1 2 171g1g S 21 21 17 J85074 17 Jim 745 iqE 15 85 1 5 4g 21 5485074 E 5 4850 7445 39gt subevalms amp matrix31uvw i39qEu 6 V 6 w 21 A85074 g 4850 745 1 217J J v 2 17J Ew with 21 17 J85074J 17 J850 7485 2E 154g4gv1 54g 85w u 21 5J85074J 5 Jim 745 39gt Tsubsxsub11 ysub21 zsub31 Q T35212542443 653642632 1850 74Jg 2850741g W 2 lawgng 1 5 ng 37 21 u 5 M 5 M 217J J v2 17JE1Ew 4i Eu 17 M 17 M V sub Page 10 0 5 4 4 21 6 V 6 w 0 u 2 1 4 2 4 1 It is not obvious that we have diagonalized Q so let39s expand T gt TexpandT 2 v2 4 85 w2 4 85 v2 T6u 1608 1 13260 85074q85 850 74485 85074q 85 2 W 13260 850 7411 85 Now it is obvious but it would be nice if we could collect the terms as well 39gt collectT uvW 85 1 6 u2 1608 13260 Jvz 850741 85 850741 85 1 85 2 13260 1608 w 850 74 8 85 850 7411 85 The coef cients do not look like the o ginal eigenvalues so let39s check them gt c2subs u0 v1 w0 T 85 1 c2 1608 13260 85074 q 85 85074 q 85 gt isc2D022 true gt c3subs u0 v0 w1 T 1 85 6313260 1608 850 74 q 85 850 74 q 85 gt isc3D033 true In spite of appearances the coef cients are correct Page 11

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