6 Eigenvalues Eigenvalues are a common part of our life: vibrations, light transmission, tuning guitar, design buildings and bridges, washing machine, Partial differential problems, water flow, The simplest linear operator is L(x) λx We wish to decompose a linear operator into such simple components (λ is an eigenvalue) Most matrices A are similar to a diagonal matrices having the diagonal entries eigenvalues Eigenvalues can be used to solve linear differential equations 6 Eigenvalues and Eigenvectors Def A R n n, a scalar λ R is an eigenvalue of A if there is a nonzero vector x (called an eigenvector) such that Ax λx Prop 6 λ is an eigenvalue of A Ax λx for some nonzero vector x Process to solve the eigenvalue problem: (A λi)x 0 has nonzero solution N(A λi) {0} det(a λi) 0 Get the characteristic polynomial det(a λi) Set up the characteristic equation det(a λi) 0 2 Solve det(a λi) 0 for λ to get eigenvalues λ,, λ k (with multiplicities) 3 For each eigenvalue λ i ( i k), solve (A λ i I)x 0 to get the eigenspace N(A λ i I) corresponding to the eigenvalue λ i Every nonzero vector x in N(A λ i I) is an eigenvector of A corresponding to the eigenvalue λ i Ex (Example 3, pp303 [ in 7th ] ed) Find the characteristic equation, eigenvalues, and 3 2 eigenvectors of A 3 2 Ex HW (k) (pp30 in 7th ed) Ex HW 2 (pp30 in 7th ed) Ex HW (j) (pp30 in 7th ed) 45
(skip) Application : Structures Buckling of a Beam (p304 in 7th ed) (skip) The characteristic equation det(a λi) 0 may have complex roots So a real matrix may have complex eigenvalues 0 0 Ex (skip) The eigenvalues of a permutation matrix A 0 0 are the roots of { 0 0 λ 3 + 0 That is, λ, 3 3 } i 2 2 2 2 Thm 62 If A and B are similar, then they have the same characteristic polynomial, the same characteristic equation, and the same eigenvalues Proof Suppose A SBS, then det(a λi) det[sbs λss ] det[sbs S(λI)S ] det[s(b λi)s ] det(s) det(b λi) det(s ) det(b λi) More properties (optional): The product of eigenvalues of A equals to det A 2 The sum of eigenvalues of A equals to tr(a) (the trace of A) That is, n n λ i a ii i i 3 Similar matrices have the same determinant and trace 6 Homework Sect 6, bil, 2, 3, 4, 5, 8, 9 62 Systems of Linear Differential Equations A system of linear differential equations has the form (y i are functions of t) y a y + a 2 y 2 + + a n y n y 2 a 2 y + a 22 y 2 + + a 2n y n y n a n y + a n2 y 2 + + a nn y n 46
Write A : [ a ij (a constant matrix) and Y : Y(t) ]n n function of t) Then Y n y y n (a vector-valued The linear differential system becomes Y AY Ex (n ) The solution of (t) ay(t) is y ce at (c R) Thm 63 Assume that A R n n has n distinct eigenvalues λ,, λ n Choose an eigenvector x i for the eigenvalue λ i Then [x,, x n ] is a basis of R n, and the linear differential system Y AY has the general solution: Y(t) c e λ t x + c 2 e λ 2t x 2 + + c n e λnt x n, c,, c n R Ex Verify that Y i e λ it x i is a solution of Y AY for i n Note: If such a basis {x n,, x n } exists, let X [x,, x n ] and D λ λ n, then AX A[x,, x n ] [Ax,, Ax n ] [λ x,, λ n x n ] λ [x,, x n ] λ n XD Therefore X AX D (or A XDX ) Such A is called diagonalizable Process to solve Y AY (if A has n distinct eigenvalues): () Find out the eigenvalues λ,, λ n of A (2) Find out the eigenspace N(A λ i I) and choose an eigenvector x i N(A λ i I) for each eigenvalue λ i of A (3) Get the general solution Y(t) c e λ t x + c 2 e λ 2t x 2 + + c n e λnt x n Ex HW a (pp323 in 7th ed) Def Initial value problem (IVP) Y AY, Y(0) Y 0 There is only one exact solution for the IVP: 47
() Find the general solution of Y AY (2) Use Y(0) Y 0 to find out the exact solution of the IVP Ex HW 2d (pp323 in 7th ed) (skip) Application : Mixture (p35 in 7th ed) 2 Higher order systems For a higher order differential equation y (n) + a n y (n ) + + a + a 0 y 0, denote then Y Y : y y (n 2) y (n ), y (n ) y (n ) y (n) a 0 y a a n y (n ) 0 0 0 y 0 0 0 : AY 0 0 0 y (n 2) a 0 a a 2 a n y (n ) Solve Y AY, we get y as the first entry of Y Ex Y Y Ex (brief) Example 3 (p39 in 7th ed) (skip) If we have a system consisting of higher order equations, we can similarly change the system to a linear differential system (cf p320 in 7th ed) 62 Homework Sect 62 bf, 2c, 3, Solve 4y 48
63 Diagonalization Def: A matrix A R n n is called diagonalizable if A is similar to a diagonal matrix D: A XDX In such case, D diag(λ,, λ n ) where λ,, λ n are eigenvalues of A, and the i-th column of X is an eigenvector of A corresponding to the eigenvalue λ i Most (but not all) square matrices are diagonalizable [ ] Ex The matrix is not diagonalizable 0 What matrices are diagonalizable? Thm 64 If A has n distinct eigenvectors λ,, λ n with corresponding eigenvectors x,, x n, then x,, x n are linearly independent The above A is diagonalizable by the following result Thm 65 A R n n is diagonalizable if and only if A has n linearly independent eigenvectors (ie R n has a basis consisting of eigenvectors of A) Proof A is diagonalizable A XDX for a diagonal matrix D diag(λ,, λ n ) AX XD, ie A[x,, x n ] [x,, x n ]diag(λ,, λ n ) [Ax,, Ax n ] [λ x,, λ n x n ] x i is an eigenvector of A corresponding to λ i Moreover, {x,, x n } is a basis of R n since X is nonsingular The proof suggests a process to diagonalize A () Find out the eigenvalues λ,, λ k of A (2) Get the eigenspace N(A λ i I) for each eigenvalue λ i of A If k dim[n(a λ i I)] n, then A is diagonalizable i (3) Select a basis for each eigenspace N(A λ i I) The union of these bases form a basis [x,, x n ] of R n Let X : [x,, x n ] then A XDX 49
[ ] [ ] 2 Ex has two eigenvectors (corresponding to eigenvalue 3) and 2 (corresponding to eigenvalue ) So [ ] [ ] [ ] [ ] 2 3 0 2 0 [ ] 2 Ex Find the eigenvectors and diagonalize the matrix 3 6 What can we do by diagonalizing a matrix? (App ) Powers and exponentials of diagonalizable matrices: Power: For every integer k, A XDX A k XD k X 2 Def: (Exponential) The power series of e x is e x + x + 2! x2 + We define the matrix exponential of A by For a diagonal matrix D e A : I + A + 2! A2 + 3! A3 + λ, n0 n! An [ ] n0 n! xn λ n e D I + D + 2! D2 + 3! D3 + e λ e λn If A XDX then A k XD k X and e A I + A + 2! A2 + 3! A3 + I + XDX + 2! XD2 X + 3! XD3 X + X(I + D + 2! D2 + 3! D3 + )X X e λ e λn X Xe D X 50
Ex Compute e A for A : [ ] [ ] [ ] [ ] 2 3 0 2 0 The solution for the initial value problem Y AY, Y(0) Y 0 is: Y e At Y 0 Proof We prove the result for a diagonalizable matrix A XDX Y AY XDX Y (X Y) D(X Y) z Let Z : X Y z n Then Z DZ (ie z i λ i z i, and D diag(λ,, λ n ) i,, n) z i e λit c i (ie Z e Dt c for c : Y XZ (Xe Dt X )(Xc ) : e At c Set t 0 in Y e At c We get Y(0) c Y 0 Therefore, Y e At Y 0 Ex The solution of ay, y(0) y 0 is y(t) e at y 0 Ex Example 8 (p340 in 7th ed) 63 Homework c c n ) Sect 63 de, 2de, 3de, 7, 27a (24a in 6th ed) 5