Math 211 Math 211 Lecture #20 November 7, 2000 2 Higher - PDF document

1 Math 211 Math 211 Lecture #20 November 7, 2000 2 Higher Dimensional Systems Higher Dimensional Systems x ′ = A x • A is an n × n real matrix. • If λ is an eigenvalue and v � = 0 is an associated eigenvector, then x ( t ) = e λt v is a solution. 3 Suppose that λ 1 , . . . , λ k are Proposition: distinct eigenvalues of A , and that v 1 , . . . , v k are associated nonzero eigenvectors. Then v 1 , . . . , v k are linearly independent. Return 1 John C. Polking

4 Suppose the n × n real matrix A Theorem: has n distinct eigenvalues λ 1 , . . . , λ n , and that v 1 , . . . , v n are associated nonzero eigenvectors. Then the exponential solutions x i ( t ) = e λ i t v i , 1 ≤ i ≤ n form a fundamental system of solutions for the system x ′ = A x . • Example 17 − 30 − 8   A = 16 − 29 − 8   − 12 24 7 Return 5 Complex Eigenvalues Complex Eigenvalues Suppose A is a real n × n matrix. • Suppose λ is a complex eigenvalue and w is an associated nonzero eigenvector. • Then λ is an eigenvalue and w is an associated nonzero eigenvector. • z ( t ) = e λt w and z ( t ) = e λt w are linearly independent complex valued solutions. • x ( t ) = Re( z ( t )) and y ( t ) = Im( z ( t )) are linearly independent real valued solutions. Return 6 Example Example 21 10 4   A = − 70 − 31 − 10   30 10 − 1 • The theorem applies if some of the eigenvalues are complex and we replace complex conjugate pairs of solutions by their real and imaginary parts. Theorem 2 John C. Polking

7 Repeated Eigenvalues – Example 1 Repeated Eigenvalues – Example 1 − 5 − 10 6   A = 8 19 − 12   12 30 − 19 • p ( λ ) = ( λ + 3)( λ + 1) 2 • Eigenspace for the eigenvalue λ 1 = − 3 has dimension 1 ⇒ one exponential solution  − 1 / 3  x 1 ( t ) = e − 3 t 2 / 3   1 Example 1a Example 2 Example 2a Analysis Return 8 Repeated Eigenvalues – Example 1 Repeated Eigenvalues – Example 1 • Eigenspace for the eigenvalue λ 2 = − 1 has dimension 2 ⇒ two linearly independent exponential solutions − 5 / 2 3 / 2     x 2 ( t ) = e − t x 3 ( t ) = e − t 1 & 0     0 1 Example 1 Example 2 Example 2a Analysis Return 9 Repeated Eigenvalues – Example 2 Repeated Eigenvalues – Example 2 1 2 − 1   A = − 4 − 7 4   − 4 − 4 1 • p ( λ ) = ( λ + 3)( λ + 1) 2 • Eigenspace for the eigenvalue λ 1 = − 3 has dimension 1 ⇒ one exponential solution  − 1 / 2  x 1 ( t ) = e − 3 t 3 / 2   1 Example 1 Example 1a Example 2a Analysis Return 3 John C. Polking

10 Repeated Eigenvalues – Example 2 Repeated Eigenvalues – Example 2 • Eigenspace for the eigenvalue λ 2 = − 1 has dimension 1 ⇒ only one exponential solution − 1 / 2   x 2 ( t ) = e − t 1   1 • Need a third solution. Example 1 Example 1a Example 2 Analysis Return 11 Multiplicities Multiplicities A an n × n matrix • Distinct eigenvalues λ 1 , . . . , λ k . • The characteristic polynomial is p ( λ ) = ( λ − λ 1 ) q 1 ( λ − λ 2 ) q 2 · . . . · ( λ − λ k ) q k . • The algebraic multiplicity of λ j is q j . • The geometric multiplicity of λ j is d j , the dimension of the eigenspace of λ j . Return 12 Multiplicities (cont.) Multiplicities (cont.) We always have: • q 1 + q 2 + · · · + q k = n . • 1 ≤ d j ≤ q j . • There are d j linearly independent exponential solutions corresponding to λ j . • If d j = q j for all j we have n linearly independent solutions. Multiplicities Return 4 John C. Polking

13 Examples Examples • In both p ( λ ) = ( λ + 1) 2 ( λ + 3) . • In both λ 2 = − 1 has algebraic multiplicity 2. • In Ex. 1 λ 2 = − 1 has geom. multiplicity 2. • In Ex. 2 λ 2 = − 1 has geom. multiplicity 1. • Problems arise when d j < q j . Return Example 1 Example 1a Example 2 Example 2a Multiplicities.a 14 New Approach New Approach • D = 1 x ′ = ax ⋄ Solution x ( t ) = Ce at . • D > 1 x ′ = A x ⋄ Tried x ( t ) = e λt v . ⋄ Why not x ( t ) = e tA v ? • But what is e tA ? 15 Exponential of a Matrix Exponential of a Matrix The exponential of the n × n Definition: matrix A is the n × n matrix e A = I + A + 1 2! A 2 + 1 3! A 3 + · · · ∞ 1 � n ! A n . = 0 5 John C. Polking

16 Examples Examples � r 1 0 � • A = 0 r 2 � e r 1 0 � e A = . e r 2 0 • e λI = e λ I. • e 0 I = I. Return 17 Properties 1 Properties 1 • A commutes with e A . • If A and B commute, then e A + B = e A · e B . • The inverse of e A is e − A . d dte tA = Ae tA . • Return 18 Properties 2 Properties 2 • The solution to the intial value problem x ′ = A x with x (0) = v is x ( t ) = e tA v . Return 6 John C. Polking

19 The Key Idea The Key Idea Let λ be a number (an eigenvalue), and A an n × n matrix. • A = λI +( A − λI ) ; λI & A − λI commute. e tA = e t [ λI +( A − λI )] = e tλI · e t ( A − λI ) = e λt · e t ( A − λI ) = e λt · [ I + t ( A − λI ) + t 2 2!( A − λI ) 2 + · · · ] Examples Prop 1 Return 20 Eigenvector Eigenvector Let λ be an eigenvalue and v an associated eigenvector. ⇒ ( A − λI ) v = 0 . e tA v = e λt · e t ( A − λI ) v = e λt [ I + t ( A − λI ) + t 2 2!( A − λI ) 2 + · · · ] v = e λt [ v + t ( A − λI ) v + t 2 2!( A − λI ) 2 v + · · · ] = e λt v Key Idea Return 21 Matrices with One Eigenvalue Matrices with One Eigenvalue A an n × n matrix with characteristic polynomial p ( λ ) = ( λ − λ 1 ) n . • Cayley-Hamilton Theorem: If p ( λ ) is the characteristic polynomial of the matrix A then p ( A ) = 0 I . • In our case ( A − λ 1 I ) n = 0 I. Key Idea Return 7 John C. Polking

22 Matrices with One Eigenvalue Matrices with One Eigenvalue (cont.) (cont.) e tA = e λ 1 t · [ I + t ( A − λ 1 I ) + t 2 2!( A − λ 1 I ) 2 + · · · t n − 1 ( n − 1)!( A − λ 1 I ) n − 1 ] + Key Idea Return 23 Example 1 Example 1 � − 3 1 � A = − 1 − 1 • p ( λ ) = ( λ + 2) 2 . � − 1 1 � ( A + 2 I ) 2 = 0 I A + 2 I = − 1 1 e tA = e − 2 t [ I + t ( A + 2 I )] � 1 − t t � = e − 2 t . − t 1 + t Key Idea Solution 1 E Return 24 Example 2 Example 2 0 − 9 27   A = − 2 3 − 18   − 1 3 − 12 ( A + 3 I ) 2 = 0 I. • p ( λ ) = ( λ + 3) 3 . e tA = e − 3 t [ I + t ( A + 3 I )] 1 + 3 t − 9 t 27 t   = e − 3 t  . − 2 t 1 + 6 t − 18 t  − t 3 t 1 − 9 t 1 E Key Idea Solution Return 8 John C. Polking

25 Example 3 (a) Example 3 (a) 1 2 − 1   A = − 4 − 7 4   − 4 − 4 1 • Earlier example. • p ( λ ) = ( λ + 3)( λ + 1) 2 • Distinct eigenvalues λ 1 = − 3 & λ 2 = − 1 • Different from previous two examples. Example Return 26 Example 3 (b) Example 3 (b) • Eigenspace for the eigenvalue λ 1 = − 3 has dimension 1 ⇒ one exponential solution x 1 ( t ) = e λ 1 t v 1 − 1 / 2   = e − 3 t 3 / 2   1 Return 27 Example 3 (c) Example 3 (c) • Eigenspace for the eigenvalue λ 2 = − 1 has dimension 1 ⇒ only one exponential solution x 2 ( t ) = e λ 2 t v 2 − 1 / 2   = e − t 1   1 • However, null(( A − λ 2 I ) 2 ) has dimension 2. Return 9 John C. Polking

28 Example 3 (d) Example 3 (d) • If v ∈ null(( A − λ 2 I ) 2 ) then e tA v = e λ 2 t [ I + t ( A − λ 2 I ) + t 2 2!( A − λ 2 I ) 2 + · · · ] v = e λ 2 t [ v + t ( A − λ 2 I ) v + t 2 2!( A − λ 2 I ) 2 v + · · · ] = e λ 2 t [ v + t ( A − λ 2 I ) v ] . Eigenvector Key Idea Return 29 Example 3 (e) Example 3 (e) • null( A + I ) 2 has basis 0 1     1 and v 3 = 0     1 0 • Third solution: x 3 ( t ) = e tA v 3 = e − t [ v 3 + t ( A + I ) v 3 ] Key Idea 3(d) 30 Example 3 (f) Example 3 (f) x 3 ( t ) = e − t [ v 3 + t ( A + I ) v 3 ] = e − t [ v 3 − 4 t v 2 ] 1 + 2 t    . = e − t − 4 t  − 4 t 3(b) 3(c) 10 John C. Polking

31 Summary Summary • In Examples 1 & 2 the matrix has one eigenvalue. ⋄ The series for e t ( A − λI ) truncated to a finite sum. • In Example 3 the matrix had two eigenvalues. ⋄ The series for e t ( A − λ 2 I ) does not truncate. ⋄ The series for e t ( A − λ 2 I ) v does truncate if ( A − λ 2 I ) 2 v = 0 . Key Idea 1 2 3(a) 3(d) Return 32 Generalized Eigenvectors Generalized Eigenvectors If λ is an eigenvalue of A and Definition: ( A − λI ) p v = 0 for some integer p ≥ 1 , then v is called a generalized eigenvector associated with λ. • The series for e t ( A − λI ) v truncates to a finite sum if v is a generalized eigenvector associated with λ. • We can compute e tA v . Key Idea Summary 33 Generalized Eigenvectors Generalized Eigenvectors Theorem: If λ is an eigenvalue of A with algebraic multiplicity q , then there is an integer p ≤ q such that null(( A − λI ) p ) has dimension q . • For each generalized eigenvector v we can compute e tA v . • We can find q linearly independent solutions this way. Key Idea 11 John C. Polking