Eigenvalues and Eigenvectors

Transcription

1 Eigenvalues and Eigenvectors Definition 0 Let A R n n be an n n real matrix A number λ R is a real eigenvalue of A if there exists a nonzero vector v R n such that A v = λ v The vector v is called an eigenvector of A belonging to λ Suppose that A R n n and λ R Then E A λ = { v R n A v = λ v} is a subspace of R n λ is an eigenvalue of A if and only if E A λ { 0} If λ is an eigenvector of A, then E A λ is called an eigenspace of A Definition 02 Suppose that A R n n and t is an indeterminate The characteristic polynomial of A is χ A t = DettI n A Expanding the determinant, we see that χ A t = t n + lower order terms in t is a polynomial in t of degree n The following theorem gives a method of computing the eigenvalues of a matrix Theorem 03 Suppose that A R n n Then λ R is a real eigenvalue of A if and only if λ is a real root of χ A t = 0 Proof λ R is a real eigenvalue of A if and only if λ R and E A λ = KernelλI n NλI n A { 0}, which holds if and only if λ R and χ A λ = DetλI n 0 Corollary 04 A matrix A R n n has at most n distinct eigenvalues This follows from the fact that a polynomial of degree n has at most n distinct roots Since DetA ti n = n χ A t, eigenvalues may also be computed as the roots of DetA ti n = 0 Example 05 Let Compute the real eigenvalues, and a basis of each of the eigenspaces of A We compute DetA ti 2 = 3 t 2 t = t2 3t + 2 = t t 2, which has the real roots t = and t = 2 Thus the real eigenvalues of A are λ = and λ = 2 We now compute a basis of the eigenspace E of A We have that E = NA I A I 2 =

2 is the RRE form of A I 2 The standard form solution of the associated homogeneous system is x = t with t R We write to see that x = t x 2 { is a basis of E A We compute a basis of the eigenspace E A 2 of A We have that E A 2 = NA 2I A 2I 2 = 2 is the RRE form of A 2I 2 The standard form solution of the associated homogeneous system is x = 2t with t R We write to see that is a basis of E A 2 } x 2 = t x 2 { 2 Definition 06 Suppose that V is a real vector space and L : V V is a linear mapping A number λ R is a real eigenvalue of L if there exists a nonzero vector v V such that L v = λ v The vector v is called an eigenvector of L belonging to λ } The eigenspace E L λ of an eigenvalue λ of L is E L λ is a subspace of V E L λ = { v V L v = λ v} Example 07 Let C R be the infinitely differentiable functions on R Let L : C R C R be differentiation: Lf = df dx for f C R Then every real number is an eigenvalue of L For λ R, {e λx } is a basis of E L λ If A R n n, then the eigenvalues and eigenspaces of the matrix A and of the linear mapping L A : R n R n are the same Theorem 08 Suppose that L : V V is a linear mapping and λ,, λ r are distinct eigenvalues of L Suppose that {v i, vi 2,, vi d i } are bases of the eigenspaces Eλ i for i r, with d i = dimeλ i Then are linearly independent {v,, v d, v 2,, v 2 d 2,, v r,, v r d r } 2

3 We prove this in a special case; we suppose that λ is an eigenvalue of A with eigenvector v, λ 2 is an eigenvalue of A with eigenvector v 2, and λ λ 2 We will show that {v, v 2 } are linearly independent Suppose that c, c 2 R and c v + c 2 v 2 = 0 Then 2 0 = L 0 = Lc v + c 2 v 2 = c Lv + c 2 Lv 2 = c λ v + c 2 λ 2 v 2 subtracting equation 2 from λ times equation, we obtain 0 = λ c v + c 2 v 2 c λ v + c 2 λ 2 v 2 = c 2 λ λ 2 v 2 Since λ λ 2 0 and v 2 0, we have that c 2 = 0 Now going back to, we see that c v = 0 Since v 0, we have that c = 0 As c = c 2 = 0 is the only solution to, we have that {v, v 2 } are linearly independent Definition 09 Suppose that L : V V is a linear mapping of finite dimensional vector spaces L is diagonalizable if there exists a basis {v,, v n } of V consisting of eigenvectors of L The word diagonalizable in Definition 09 is explained by the following theorem Theorem 00 Suppose that L : V V is a linear mapping, and there exists a basis β = {v,, v n } of V consisting of eigenvectors of L Then M β β L is a diagonal matrix Proof Let c,, c n R be the eigenvalues of the v i, so that Lv i = c i v i for i n We then have c 0 0 M β β L = Lv 0 c 2 0 β, Lv 2 β,, Lv n β = c n is a diagonal matrix Definition 0 Matrices A, B R n n are similar if there exists an invertible matrix C R n n such that B = C AC A matrix A is diagonalizable over the reals if A is similar to a diagonal matrix D Theorem 02 The matrix A R n n is diagonalizable over the reals if and only if the linear mapping L A : R n R n is diagonalizable We will prove the most interesting direction, that L A diagonalizable implies A is diagonalizable Let β = {v,, v n } be a basis of R n consisting of eigenvectors of A Let c,, c n be the corresponding eigenvalues, so that Lv i = c i v i for i n Then D = M β β L = Lv β, Lv 2 β,, Lv n β = 3 c c 2 0 c n

4 is a diagonal matrix Let β be the standard basis of R n, and let C = M β β = v, v 2,, v n We have C AC = M β β M β β L A M β β = M β β L D Thus A is similar to a diagonal matrix Theorem 02 gives an algorithm to determine if a matrix is diagonalizable, and if it is, how to diagonalize it Example 03 Let Determine if A is diagonalizable over the reals 2 If A is diagonalizable, find an invertible matrix C and a diagonal matrix D such that C AC = D We calculated in Example 05 that the eigenvalues of A are λ = and λ 2 = 2, and that a basis of E A is {, T } and a basis of E A 2 is { 2, } A is diagonalizable over the reals since and 2 are the real eigenvalues of A and dim E A + dim E A 2 = 2 = sizea 2 Set and C = D = Then C AC = D by the algorithm of Theorem 02 Example 04 Let 0 Determine if A is diagonalizable over the reals 2 If A is diagonalizable, find an invertible matrix C and a diagonal matrix D such that C AC = D The characteristic polynomial of A is χ A t = DettI 2 t 0 t = t2 Thus the only eigenvalue of A is λ = 0 We compute a basis of the eigenspace E0 = NA 0I 2 = NA 0 4

5 is the RRE form of A The standard form solution of the associated homogeneous system is x = t x 2 = 0 with t R Writing x, x 2 T = t, 0 T, we see that {, 0 T } is a basis of E A 0 A is not diagonalizable over the reals since 0 is the only eigenvalue of A and dim E A 0 = < 2 = sizea The complex numbers C are defined by adjoining the imaginary number i = to R; that is i 2 = and C = {a + bi a, b R} We define C n to be the n column vectors with complex coefficients, C n to be the n row vectors with complex coefficients, and C m n to be the m n matrices with complex coefficients Almost everything that we have done in this class is valid if we replace R with C; for instance we have real vector spaces over the reals R and complex vector spaces over the complex numbers C The only exception is that an inner product must be defined differently on a complex vector space Complex numbers are important for us because they have the important property that they are algebraically closed Theorem 05 Fundamental Theorem of Algebra Suppose that ft = t n + a t n + +a n is a polynomial with complex coefficients, and n Then ft = 0 has a complex root α Definition 06 Let A C n n be an n n complex matrix A number λ C is a complex eigenvalue of A if there exists a nonzero vector v C n such that A v = λ v The vector v is called an eigenvector of A belonging to λ Suppose that A C n n and λ C Then the eigenspace of λ, E A λ = { v C n A v = λ v} is a subspace of C n Since the reals are contained in the complex numbers, any real matrix is also a complex matrix, any real eigenvalue is also a complex eigenvalue and any real eigenvector is a complex eigenvector A complex matrix A C n n is diagonalizable over C if A is similar over C to a complex diagonal matrix D; that is, there exists an invertible matrix C C n n such that D = C AC Example 07 Consider the matrix 0 0 A is not diagonalizable over the reals 2 A is diagonalizable over the complex numbers 5

6 We compute χ A t = DettI 2 t t = t2 + = t it + i A has no real eigenvalues, so A is not diagonalizable over the reals However, A has two complex eigenvalues, λ = i and λ 2 = i We find a complex basis of Ei = NA ii 2 i A ii 2 = i i i i is the RRE form of A ii 2 The last operation was the elementary row operation of adding i times the first row to the second row, as i, i = i, i 2 = i, The standard form solution of the associated homogeneous system is x = it with t R Writing x, x 2 T = t i, T, we see that { } i is a basis of Ei Now we find a complex basis of E i = NA + ii 2 i i i A + ii 2 = i i is the RRE form of A ii 2 The last operation was the elementary row operation of adding i times the first row to the second row, as i, i = i, 2 i 2 = i, The standard form solution of the associated homogeneous system is x = it with t R Writing x, x 2 T = ti, T, we see that { } i is a basis of E i i and i are the eigenvalues of A and dim E A i + dim E A i = 2 = sizea so that A is diagonalizable over the complex numbers Set C = i i and D = Then C AC = D by the algorithm of Theorem 02 i 0 0 i Going back to the matrix 0 6,

7 which we showed was not diagonalizable over R, we see that the only complex eigenvalue of A is 0 since χ A t = t 2, and {, 0 T } is a basis of the complex eigenspace E A 0 0 is the only complex eigenvalue of A and dim E A 0 = < 2 = sizea so that A is not diagonalizable over the complex numbers 7