π 1 = tr(a), π n = ( 1) n det(a). In particular, when n = 2 one has
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1 Eigen Methods Math 246, Spring 2009, Professor David Levermore Eigenpairs Let A be a real n n matrix A number λ possibly complex is an eigenvalue of A if there exists a nonzero vector v possibly complex such that Av = λv Each such vector is an eigenvector associated with λ, and λ,v is an eigenpair of A Fact : If λ,v is an eigenpair of A then so is λ, αv for every complex α 0 In other words, if v is an eigenvector associated with an eigenvalue λ of A then so is αv for every complex α 0 In particular, eigenvectors are not unique Reason Because λ,v is an eigenpair of A you know that holds It follows that Aαv = αav = αλv = λαv Because the scalar α and vector v are nonzero, the vector αv is also nonzero Therefore λ, αv is also an eigenpair of A Recall that the characteristic polynomial of A is defined by 2 p A z = detzi A It has the form p A z = z n + π z n + π 2 z n π n z + π n, where the coefficients π, π 2,, π n are real In other words, it is a real monic polynomial of degree n One can show that in general In particular, when n = 2 one has π = tra, π n = n deta p A z = z 2 traz + deta Because detzi A = n deta zi, this definition of p A z coincides with the book s definition when n is even, and is its negative when n is odd Both conventions are common We have chosen the convention that makes p A z monic What matters most about p A z is its roots and their multiplicity, which are the same for both conventions Fact 2: A number λ is an eigenvalue of A if and only if p A λ = 0 In other words, the eigenvalues of A are the roots of p A z Reason If λ is an eigenvalue of A then by there exists a nonzero vector v such that λi Av = λv Av = 0 It follows that p A λ = detλi A = 0 Conversely, if p A λ = detλi A = 0 then there exists a nonzero vector v such that λi Av = 0 It follows that λv Av = λi Av = 0, whereby λ and v satisfy, which implies λ is an eigenvalue of A Fact 2 shows that the eigenvalues of a n n matrix A can be found if you can find all the roots of the characteristic polynomial of A You can then find all the eigenvectors associated with each eigenvalue by finding a general nonzero solution of
2 2 You can quickly find the eigenvectors for any 2 2 matrix A with help from the Cayley- Hamilton Theorem, which states that p A A = 0 The eigenvalues λ and λ 2 are the roots of p A z, so p A z = z λ z λ 2 Hence, by the Cayley-Hamilton Theorem 3 0 = p A A = A λ IA λ 2 I = A λ 2 IA λ I It follows that every nonzero column of A λ 2 I is an eigenvector associated with λ, and that every nonzero column of A λ I is an eigenvector associated with λ 2 Example Find the eigenpairs of A = 2 3 Solution The characteristic polynomial of A is p A z = z 2 6z + 5 = z z 5 By Fact 2 the eigenvalues of A are and 5 Because 2 2 A I =, A 5I = 2 2 Every column of A 5I has the form α for some α 0, while every column of A I has the form α for some α 0 It follows from 3 that the eigenpairs of A are,, Example Find the eigenpairs of A = 2 3 Solution The characteristic polynomial of A is 5, 2 2, 2 2 p A z = z 2 6z + 3 = z + 4 = z By Fact 2 the eigenvalues of A are 3 + i2 and 3 i2 Because i2 2 A 3 + i2i =, A 3 i2i = 2 i2 Every column of A 3 i2i has the form α for some α 0, i while every column of A 3 + i2i has the form α for some α 0 i It follows from 3 that the eigenpairs of A are 3 + i2,, 3 i2, i i i2 2 2 i2
3 Notice that in the above example the eigenvectors associated with 3 i2 are complex conjugates to those associated with 3 + i2 This illustrates is a particular instance of the following general fact Fact 3: If λ,v is an eigenpair of the real matrix A then so is λ,v Reason Because λ,v is an eigenpair of A you know by that Av = λv Because A is real, the complex conjugate of this equation is Av = λv, where v is nonzero because v is nonzero It follows that λ,v is an eigenpair of A Both examples given above illustrate particular instances of the following general facts Fact 4: Let λ be an eigenvalue of the real matrix A If λ is real then it has a real eigenvector If λ is not real then none of its eigenvectors are real Reason Let v be any eigenvector associated with λ, so that λ,v is an eigenpair of A Let λ = µ + iν and v = u + iw where µ and ν are real numbers and u and w are real vectors One then has Au + iaw = Av = λv = µ + iνu + iw = µu νw + iµw + νu, which is equivalent to Au µu = νw, and Aw µw = νu If ν = 0 then u and w will be real eigenvectors associated with λ whenever they are nonzero But at least one of u and w must be nonzero because v = u + iw is nonzero Conversely, if ν 0 and w = 0 then the second equation above implies u = 0 too, which contradicts the fact that at least one of u and w must be nonzero Hence, if ν 0 then w 0 Solutions of First-Order Systems We are now ready to use eigenvalues and eigenvectors to construct solutions of first-order differential systems with a constant coefficient matrix The system we study is 4 dx dt = Ax, where xt is a vector and A is a real n n matrix We begin with the following basic fact Fact 5: If λ,v is an eigenpair of A then a solution of 4 is 5 xt = e λt v Reason By direct calculation we see that whereby xt given by 5 solves 4 dx dt = d e λt v = e λt λv = e λt Av = A e λt v = Ax, dt If λ,v is a real eigenpair of A then recipe 5 will yield a real solution of 4 But if λ is an eigenvalue of A that is not real then recipe 5 will not yield a real solution However, if we also use the solution associated with the conjugate eigenpair λ, v then we can construct two real solutions 3
4 4 6 Fact 6: Let λ,v be an eigenpair of A with λ = µ + iν and v = u + iw where µ and ν are real numbers while u and w are real vectors Then two real solutions of 4 are x t = Re e λt v = e µt u cosνt w sinνt, x 2 t = Im e λt v = e µt w cosνt + u sinνt Reason Because λ,v is an eigenpair of A, by Fact 3 so is λ,v By recipe 5 two solutions of 4 are e λt v and e λt v, which are complex conjugates of each other Because equation 4 is linear, it follows that two real solutions of 4 are given by x t = Re e λt v = eλt v + e λt v, x 2 t = Im e λt v = eλt v e λt v 2 i2 Because λ = µ + iν and v = u + iw we see that e λt v = e µt cosνt + i sinνt u + iv = e µt[ u cosνt w sinνt + i w cosνt + u sinνt ], whereby x t and x 2 t are read off from the real and imaginary parts, yielding 6 Example Find two linearly independent real solutions of dx = Ax, where A = dt 2 3 Solution By a previous example we know that A has the real eigenpairs,, 5, By recipe 5 the equation has the real solutions x t = e t, x 2 t = e 5t These solutions are linearly independent because W[x,x 2 ]0 = det = 2 0 Example Find two linearly independent real solutions of dx dt = Ax, where A = 2 3 Solution By a previous example we know that A has the conjugate eigenpairs 3 + i2,, 3 i2, i i Because e 3+i2t i = e 3t cos2t + i sin2t cos2t + i sin2t = e i 3t, sin2t + i cos2t by recipe 6 the equation has the real solutions cos2t sin2t x t = e 3t, x sin2t 2 t = e 3t cos2t
5 These solutions are linearly independent because 0 W[x,x 2 ]0 = det = 0 0 Exponentials of Diagonalizable Matrices If recipe 5 yields n linearly independent solutions of the first-order system 4 then they can be used to construct the matrix exponential e ta The key to this construction is the following fact from linear algebra Fact 7: If a real n n matrix A has n eigenpairs, λ,v, λ 2,v 2,, λ n,v n, such that the eigenvectors v, v 2,, v n are linearly independent then 7 A = VDV, where V is the n n matrix whose columns are the vectors v, v 2,, v n ie 8 V = v v 2 v n, while D is the n n diagonal matrix λ λ 9 D = λ n Reason Underlying this result is the fact that AV = A v v 2 v n = Av Av 2 Av n = λ v λ 2 v 2 λ n v n λ = 0 λ v v 2 v 2 n 0 = VD 0 0 λ n Once we show that V is inverible then 7 follows upon multiplying the above relation on the left by V We claim that detv 0 because the vectors v, v 2,, v n are linearly independent Suppose otherwise Because detv = 0 there exists a nonzero vector c such that Vc = 0 This means that c 0 = Vc = c v v 2 v n 2 = c v + c 2 v c n v n c n Because vectors v, v 2,, v n are linearly independent, this implies c = c 2 = = c n = 0, which contradicts the fact c is nonzero Therefore detv 0 Hence, the matrix V is invertible and 7 follows upon multiplying relation 0 on the left by V We call a real n n matrix A diagonalizable when there exists an invertible matrix V and a diagonal matrix D such that A = VDV To diagonalize A means to find such a V and D Fact 7 states that A is diagonalizable when it has n linearly independent eigenvectors The converse of this statement is also true 5
6 6 Fact 8: If a real n n matrix A is diagonalizable then it has n linearly independent eigenvectors Reason Because A is diagonalizable it has the form A = VDV where the matrix V is invertible and the matrix D is diagonal Let the vectors v, v 2,, v n be the columns of V We claim these vectors are linearly independent Indeed, if 0 = c v + c 2 v c n v n then because V = v v 2 v n we see that 0 = c v + c 2 v c n v n = c v v 2 v n 2 = Vc c n Because V is invertible, this implies that c = 0 The vectors v, v 2,, v n are therefore linearly independent Because V = v v 2 v n and because A = VDV where D has the form 9, we see that Av Av 2 Av n = A v v 2 v n = AV = VDV V = VD λ 0 0 = 0 λ v v 2 v n λ n = λ v λ 2 v 2 λ n v n Because the vectors v, v 2,, v n are linearly independent, they are all nonzero It then follows from the above relation that λ,v, λ 2,v 2,, λ n,v n are eigenpairs of A, such that the eigenvectors v, v 2,, v n are linearly independent Example Show that A = is diagonalizable, and diagonalize it 2 3 Solution By a previous example we know that A has the real eigenpairs,, 5, Because we also know the eigenvectors are linearly independent, A is diagonalizable Then 8 and 9 yield 0 V =, D = 0 5 Because detv = 2, one has V = 2 It follows from 7 that A is diagonalized as A = VDV = 2 c 0 0 5
7 We are now ready to give a construction of the matrix exponential e ta Fact 9: If the real n n matrix A has n eigenpairs, λ,v, λ 2,v 2,, λ n,v n, such that the vectors v, v 2,, v n are linearly independent then e ta = Ve td V, where V and D are the n n matrices given by 8 and 9 Reason Set Φt = Ve td V It then follows that d dt Φt = d Ve td V = V d dt dt etd V = VDe td V = AVe td V = AΦt, whereby the matrix-valued function Φt satisfies d Φt = AΦt dt Moreover, because e 0D = I we see that Φt also satisfies the initial condition Φ0 = Ve 0D V = VIV = VV = I It follows that Φt = e ta, whereby follows Formula is the book s method for computing e ta when A is diagonalizable Because not every matrix is diagonalizable, it cannot always be applied When it can be applied, most of the work needed to apply it goes into computing V and V The matrix e td is simply given by e λ t e td 0 e = λ 2t e λnt Once you have V, V, and e td, formula requires two matrix multiplications Example Compute e ta for A = 2 3 Solution By a previous example we know that A has the real eigenpairs,, 5,, and that A is diagonalizable By 8 and 9 we also know that 0 V =, D =, V 0 5 = 2 By formulas and 2 we therefore have e ta = Ve td V = e t e 5t = e t e t 2 e 5t e 5t = e 5t + e t e 5t e t 2 e 5t e t e 5t + e t 7
8 8 Example Compute e ta for A = 2 3 Solution By a previous example we know that A has the conjugate eigenpairs 3 + i2,, 3 i2, i i By 8 and 9 we know that V =, D = i i Because detv = i2, we have By formula 2 we have e td = V = i2 i = i i i2 i i e 3+i2t 0 e 0 e 3 i2t = e 3t i2t 0 0 e i2t By formula we therefore have e ta = Ve td V = e3t e i2t 0 i 2 i i 0 e i2t i = e3t e i2t ie i2t 2 i i e i2t ie i2t = e3t 2 = e3t 2 2 cos2t 2 sin2t 2 sin2t 2 cos2t e i2t + e i2t ie i2t + ie i2t ie i2t ie i2t e i2t + e i2t = e 3t cos2t sin2t sin2t cos2t Remark Because A is real, e ta must be real As the above example illustrates, the matrices V and D may not be real, but will always combine in formula to yield the real result Remark While not every matrix is diagonalizable, most matrices are Here we give four criteria that insure a real n n matrix A is diagonalizable If A has n distinct eigenvalues then it is diagonalizable If A is symmetric A T = A then its eigenvalues are real λ j = λ j, and it will have n real eigenvectors v j that can be normalized so that v T j v k = δ jk With this normalization V = V T If A is skew-symmetric A T = A then its eigenvalues are imaginary λ j = λ j, and it will have n eigenvectors v j that can be normalized so that v j v k = δ jk With this normalization V = V If A is normal A T A = AA T then it will have n eigenvectors v j that can be normalized so that v jv k = δ jk With this normalization V = V Matrices that are either symmetric or skew-symmetric are also normal There are normal matrices that are neither symmetric nor skew-symmetric Because the normal criterion is harder to verify than the symmetric and skew-symmetric criteria, it should be checked last Both of the examples we have given above have distinct eigenvalues The first example is symmetric The second is normal, but is neither symmetric nor skew-symmetric
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