Matrix Eigenvalues and Eigenvectors September 13, 2017

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1 Mtri Eigenvlues nd Eigenvectors September, 7 Mtri Eigenvlues nd Eigenvectors Lrry Cretto Mechnicl Engineering 5A Seminr in Engineering Anlysis September, 7 Outline Review lst lecture Definition of eigenvlues nd eigenvectors: A = Use Det[A I] = to find eigenvlues Solve [A I] = to get eigenvectors to within multiplictive constnt Number of different eigenvlues nd eigenvectors Review Lst Lecture Guss elimintion for solving equtions nd determining rn number of linerly independent rows or columns Solution of A = b No solutions unless rn A = rn [A b] Unique if rn A = rn [A b] = number of unnowns infinite if rn < unnowns Homogenous equtions, A = : only solution is = unless Det A = sme s sying Rn A < n Uses of Eigenvlues In electricl nd mechnicl networs, provides fundmentl frequencies Shows coordinte trnsformtions pproprite for physicl problems Provides wy to epress networ problem s digonl mtri Trnsformtions bsed on eigenvectors used in some solutions of A = b Eigenvlues nd Eigenvectors Bsic definition A squre: A= is eigenvector, is eigenvlue Bsic ide is tht eigenvector is specil vector of mtri A; multipliction of by A produces multiplied by constnt A = => A = [A I] = Homogenous equtions; requires Det [A I] = for solution other thn = 5 Det[ A I] Det Det[A I] = n n n Det[A I] = produces n n th order eqution tht hs n roots for. My hve duplicte roots for eigenvlues. nn n n n ME 5A Seminr in Engineering Anlysis Pge

2 Mtri Eigenvlues nd Eigenvectors September, 7 Two-by-two Mtri Eigenvlues Qudrtic eqution with two roots for eigenvlues Eigenvlue solutions Two-by-two Mtri Eigenvlues Write Det A s Add the two solutions to get Multiply the two solutions to get Det A Det A Det A 7 8 Sum nd Product The results on the previous slide pply to ll mtri eigenvlues The sum of the eigenvlues is the sum of the digonl elements of the mtri, clled the trce of the mtri The product of the eigenvlues is the determinnt of the mtri n n Trce A i ii i i i n Det A i 9 Two-by-two Mtri Eigenvectors Two eigenvectors: = [ ] T nd = [ ] T = [ ] T Substitute ech eigenvlue solution,, into A I = to find ll components Nottion: y i is component i of vector y; z is one vector in set of vectors with components z i Two-by-two Eigenvectors II Eigenvector equtions re homogeneous, so eigenvectors re determined only within multiplictive constnt Pic = rbitrry Two-by-two Emple Find eigenvlues nd 5 eigenvectors of A A 5 Det[ A I] Det[ A I] 5 Solutions re = nd = ME 5A Seminr in Engineering Anlysis Pge

3 Mtri Eigenvlues nd Eigenvectors September, 7 Two-by-two Emple Continued Find components for = A Solve[A I] = for components 5 5 One eqution in two unnowns Pic = then = 5 from first eqution Eigenvector is [5 ] T 5 Two-by-two Emple Concluded Net find components for = Sme s pproch for finding A 5 5 Both equtions give = Pic = With =, cn be ny vlue nd still stisfy ech eqution = [ ] T 5 Chec Two-by-two Emple A A Eigenvector Fctors emple showed A = regrdless of choice of nd This is generl result We cn pic one eigenvector component; typicl choices re to me eigenvector simple or unit vector with unit length How Mny Eigenvlues? An n n mtri hs n distinct eigenvlues Algebric multiplicity of n eigenvlue, M, is the number of roots of Det[A I] = tht hve the sme root, Geometric multiplicity, m, of eigenvlue is number of linerly independent eigenvectors for this 7 Multiple Eigenvlue Emple A A I Det A I 8 ME 5A Seminr in Engineering Anlysis Pge

4 Mtri Eigenvlues nd Eigenvectors September, 7 ME 5A Seminr in Engineering Anlysis Pge 9 Multiple Eigenvlue Emple II I A Det Solutions re =,, = hs lgebric multiplicity of Find eigenvectors from A I = Loo t = Multiple Eigenvlue Emple III Apply Guss elimintion to these equtions Pic nd Multiple Eigenvlue Emple IV Two linerly independent eigenvectors for = - Pic = nd = => = Pic = nd = => = - Pic nd then Continue Emple for = 5 7 Apply Guss elimintion to these equtions I A Pic = => = - Emple Results Eigenvlues = -, = -, = hve eigenvectors shown below Eigenvector Liner Dependence Cn we hve + + = without = = =? Homogenous equtions hve = = = if mtri hs full rn

5 Mtri Eigenvlues nd Eigenvectors September, 7 Liner Dependence II Mtri hs full rn it its determinnt is not zero Det 5 Since determinnt is not zero, the only solution is = = =, so eigenvectors re linerly independent 5 AX In This Emple AX = XΛ X Λ is digonl mtri of eigenvlues Will show this s generl result net clss Qudrtic Forms Use generl nottion for trnspose of column vector Write generl qudrtic formul s vector formul with symmetric i A A A 7 Qudrtic Forms II Theorem in Chpter 8 of Kreyszig sys tht symmetric mtrices hve orthogonl eigenvectors For n orthogonl mtri, X - = X T Cn get orthonorml eigenvectors Theorem in Chpter 8 of Kreyszig shows tht if n n n mtri, A, hs bsis set of eigenvectors s columns in n n n mtri, X, D = X - AX is digonl mtri of eigenvlues 8 Qudrtic Forms III Qudrtic Forms IV If D = X - AX, then XD = XX - AX = AX nd XDX - = AXX - = A Qudrtic forms, Q = T A, will hve symmetric A mtri, which will hve n orthonorml eigenvlue set: X - = X T For Q = T A, with A = XDX - = XDX T if X is orthonorml, X - = X T so Q = T A = T XDX T Define y = X T = X - so = Xy 9 From Q = T XDX T nd y = X T : Q = T XDy We cn write T X = X T T = y T so Q = y T Dy y re principl coordintes ME 5A Seminr in Engineering Anlysis Pge 5

6 Mtri Eigenvlues nd Eigenvectors September, 7 Smple Problem Find principl coordintes for qudrtic eqution: Q = T A = 8, where Solution steps: Find eigenvlues, nd normlized eigenvector mtri X = [, ] for A Find principl coordintes, y = X T Solution is Q = y T Dy = Smple Problem II Find principl coordintes for qudrtic eqution: Q = A = λ 7 λ 5 = Find eigenvector for λ = 7 λ Both equtions give = λ Smple Problem III Find eigenvector for λ = 7 λ Both equtions give = - For both eigenvectors pic = to get eigenvectors = [ ] T ; = [- ] T Divide by for normlized = [/ / ] T ; = [-/ / ] T Smple Problem IV Get X from normlized eigenvectors nd find y = X T Q = 8 in terms of eigenvlues nd y s Q = y T Dy = Smple Problem V Result cn be modified into eqution for n ellipse = 5 ME 5A Seminr in Engineering Anlysis Pge

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