Eigenvalues and Eigenvectors

Transcription

1 Eigenvalues and Eigenvectors week -2 Fall 26 Eigenvalues and eigenvectors The most simple linear transformation from R n to R n may be the transformation of the form: T (x,,, x n ) (λ x, λ 2,, λ n x n ), x λ x T λ 2 x n λ n x n Example The linear transformation T : R 2 R 2, defined by [ [ [ 2 x T (x), is to dilate the first coordinate two times and the second coordinate three times Example 2 Let T : R 2 R 2 be a linear transformation defined by [ [ 4 2 x T (x) What can we say about T geometrically? Consider the basis B {u, u 2 } of R 2, where [ [ 2 u, u 2 Then [ 6 T (u ) [ 2 T (u 2 ) 6 For any vector v c u + c 2 u 2, we have [v B Thus [ c [ 2 [ 2 c 2, and u, 2u 2 T (v) c T (u ) + c 2 T (u 2 ) c u 2c 2 u 2, [ [T (v) B [ c 2c 2 2 [ c c 2 If the one uses the basis B to describe vector v with coordinate vector v, then the coordinate vector of T (v) under the basis B simply described as [ [ [v T (v) B 2 B This means that the matrix of T relative to the basis B is as simple as a diagonal matrix

2 The above discussion demonstrates that the nonzero vectors v satisfying the condition T (v) λv () for scalars λ is important to describe a linear transformation T Definition Given a linear transformation T : R n R n, T (x) Ax A nonzero vector v in R n is called an eigenvector of T (the matrix A) if there exists a scalar λ such that T (v) Av λv (2) The scalar λ is called an eigenvalue of T (the matrix A) and the nonzero vector v is called an eigenvector of T (of the matrix A) corresponding to the eigenvalue λ [ [ [ 6 6 Example Let A Then u is an eigenvector of A However, but v is not an eigenvector of A Proposition 2 For any n n matrix A, the value is an eigenvalue of A det A Proof Note that the set of eigenvectors of A corresponding to the zero eigenvalue is the set Nul A {}; and A is invertible if and only if Nul A {} The theorem follows from the two facts Theorem If v, v 2,, v p be eigenvectors of a matrix A corresponding to distinct eigenvalues λ, λ 2,, λ p, respectively, then v, v 2,, v p are linearly independent Proof Let k be the smallest positive integer such that v, v 2,, v k are linearly independent If k p, nothing is to be proved If k < p, then v k+ is a linear combination of v,, v k ; that is, there exist constants c, c 2,, c k such that v k+ c v + c 2 v c k v k Applying the matrix A to both sides, we have Av k+ λ k+ v k+ λ k+ (c v + c 2 v c k v k ) c λ k+ v + c 2 λ k+ v c k λ k+ v k ; Thus Av k+ A(c v + c 2 v c k v k ) c Av + c 2 Av c k Av k c λ v + c 2 λ 2 v c k λ k v k c (λ k+ λ )v + c 2 (λ k+ λ 2 )v c k (λ k+ λ k )v k Since v, v 2,, v k are linearly independent, we have Note that the eigenvalues are distinct Hence c (λ k+ λ ) c 2 (λ k+ λ 2 ) c k (λ k+ λ k ) c c 2 c k, which implies that v k+ is the zero vector This is contradictory to that v k+ 2

3 2 How to find eigenvectors? To find eigenvectors, it is meant to find vectors x and scalar λ such that that is, Ax λx, (2) (λi A)x (22) Since x is required to be nonzero, the system (22) is required to have nonzero solutions; we thus have Expanding the det(λi A), we see that det(λi A) (2) p(λ) det(λi A) is a polynomial of degree n in λ, called the characteristic polynomial of A To find eigenvalues of A, it is meant to find all roots of the polynomial p(λ) The polynomial equation (2) about λ is called the characteristic equation of A For an eigenvalue λ of A, the system (λi A)x is called the eigensystem for the eigenvalue λ; its solution set Nul (λi A) is called the eigenspace corresponding to the eigenvalue λ Theorem 2 The eigenvalues of a triangular matrix are the entries on its main diagonal Example 2 The matrix has the characteristic polynomial p(λ) λ 2 λ 5 λ 2 (λ 2)2 (λ 5) Then there are two eigenvalues λ 2 and λ 2 5 For λ 2, the eigensystem λ 2 x λ 5 λ 2 x has two linearly independent eigenvectors v For λ 2 5, the eigensystem λ 2 2 λ 2 5 λ 2 2, v 2 x x has one linearly independent eigenvector v

4 Example 22 The matrix has the characteristic polynomial p(λ) λ 2 λ 5 λ 2 (λ 2)2 (λ 5) We obtain two eigenvalues λ 2 and λ 2 5 For λ 2 (though it is of multiplicity 2), the eigensystem x has only one linearly independent eigenvector v For λ 2 5, eigen-system has one linearly independent eigenvector x v Example 2 Find the eigenvalues and the eigenvectors for the matrix A The characteristic equation of A is det(λi A) (λ ) λ λ λ λ λ + 2 (λ + 2) λ λ + 2 (λ + 2) λ (R 2 R ) λ + 2 (λ + 2) Then A has two eigenvalues λ 2 and λ 2 7 For λ 2 (its multiplicity is 2), the eigen-system x (λ )(λ + 2)(λ 4) 8(λ + 2) (λ + 2) 2 (λ 7) 4

5 has two linearly independent eigenvectors v, v 2 For λ 7, the eigen-system x has one linearly independent eigenvector v Theorem 22 Let λ, µ and ν be distinct eigenvalues of a matrix A Let u, u 2,, u p be linearly independent eigenvectors for the eigenvalue λ; v, v 2,, v q be linearly independent eigenvectors for the eigenvalue µ; and w, w 2,, w r be linearly independent eigenvectors for the eigenvalue ν Then the vectors together are linearly independent u, u 2,, u p, v, v 2,, v q, w, w 2,, w r Proof Suppose there are scalars a,, a p, b,, b q, c,, c r such that (a u + + a p u p ) + (b v + + b q v q ) + (c w + + c r w r ) (24) It suffices to show that all the scalars a,, a p, b,, b q, c,, c r are Set Note that u a u + + a p u p, v b v + + b q u q, w c w + + c r w r Au a Au + + a p Au p a λu + a p λu p λu Similarly, Av µv and Aw νw If u, then the linear independence of u,, u p implies that a a p Similarly, v implies b b q, and w implies c c r Now we claim that u v w If not, there are following three types Type : u, v w Since v w, it follows from (24) that u, a contradiction Type 2: u, v, w Then u is the eigenvector of A for the eigenvalue λ and v the eigenvector of A for the eigenvalue µ; they are eigenvectors for distinct eigenvalues So u and v are linearly independent But (24) shows that u + v, which means that u and v are linearly dependent, a contradiction Type : u, v, w This means that u, v, w are eigenvectors of A for distinct eigenvalues λ, µ, ρ respectively So they are linearly independent However, (24) shows that u + v + w, which means that u, v, w are linearly dependent, a contradiction again Note The above theorem is also true for more than three distinct eigenvalues Diagonalization Definition An n n matrix A is said to be similar to an n n matrix B if there is an invertible matrix P such that P AP B 5

6 Theorem 2 Similar matrices have the same characteristic polynomial and hence have the same eigenvalues Note Similar matrices may have different eigenvectors For instance, the matrices A 2 5 and B have the same eigenvalues λ 2 and λ 2 5; but A and B have different eigenvectors A square matrix A is called diagonal if all non-diagonal entries are zero, that is, a a 2 A a n It is easy to see that for any k, A k a k a k 2 a k n Definition A square matrix A is said to be diagonalizable if A is similar to a diagonal matrix, that is, there exists an invertible matrix P and a diagonal matrix D such that P AP D Theorem 4 (Diagonalization Theorem) An n n matrix A is diagonalizable iff A has n linearly independent eigenvectors Proof We demonstrate the proof for the case n If A is diagonalizable, there exist an invertible matrix P and a diagonal matrix D such that P AP D, where P [u, v, w u v w u 2 v 2 w 2, D λ µ u v w ν Note that P AP D is equivalent to AP P D Since AP A[u, v, w [Au, Av, Aw and P D u v w u 2 v 2 w 2 λ µ λu µv νw λu 2 µv 2 νw 2 [λu, µv, νw, u v w ν λu µv νw we have [Au, Av, Aw [λu, µv, νw Thus Au λu, Av µv, Aw νw Since P is invertible, the vectors u, v, w are linearly independent It follows that u, v, w are three linearly independent eigenvectors of A Conversely, if A has three linear independent eigenvectors u, v, w corresponding to the eigenvalues λ, µ, ν respectively Then Au λu, Av µv, Aw νw Let P [u, v, w u v w u 2 v 2 w 2 u v w, D λ µ ν 6

7 Then AP A[u, v, w [Au, Av, Aw [λu, µv, νw λu µv νw λu 2 µv 2 νw 2 λu µv νw u v w u 2 v 2 w 2 λ µ P D u v w ν Since u, v, w are linearly independent, thus the matrix P is invertible Therefore This means that A is diagonalizable P AP D Example Diagonalize the matrix A and compute A 8 The characteristic polynomial of A is λ λi A λ R 2 + R λ R (λ )R (λ 2) (λ 2) 2 λ 2 λ 2 λ (λ 2)2 (λ ) There are two eigenvalues λ 2 and λ 2 For λ 2, the eigensystem x has two linearly independent eigenvectors v, v 2 For λ 2, the eigensystem 2 x has one linearly independent eigenvector Set P v, D 2 2 7

8 Then or equivalently, Thus P AP 2 P DP A D, A 8 (P DP )(P DP ) (P DP ) }{{} Example 2 Compute the matrix A 8, where A 2 P D 8 P 2 The characteristic polynomial of A is λ λi A λ 2 λ (λ ) λ λ (λ )(λ 2 4λ) + 2λ λ(λ 2)(λ ) + 2 λ We have eigenvalues λ, λ 2 2, and λ For λ, x, v 2 Set For λ 2 2, For λ, x x P [v, v 2, v, v 2 2, v /2 2 /2 8

9 Then Thus P AP 2 A 8 (P DP )(P DP ) (P DP ) }{{} 8 2 / D P 2 8 /2 /2 /2 / P 4 Complex eigenvalues Theorem 4 For a 2 2 matrix, if one of the eigenvalues of A is not a real number, then the other eigenvalue must be conjugate to this complex eigenvalue Let λ a bı with b be a complex eigenvalue of A and let x u + ıv be a complex eigenvector of A for λ, that is, A(u + ıv) (a bı)(u + ıv) Let P [u, v Then Proof [ P a b AP b a Ax Au + ıav, Ax λx (a bı)(u + ıv) (au + bv) + ı( bu + av) It follows that Thus Au au + bv, [ a b AP A[u, v [u, v b a Av bu + av [ a b P b a [ 5 2 Example 4 Let A Find a matrix P such that P AP is diagonal or antisymmetric [ λ 5 2 Solution Since λ λ 2 8λ + 7, the eigenvalues of A are complex numbers λ 8 ± For λ 4 ı, we have the eigensystem [ ı 2 ı 4 ± ı [ [ x 9

10 Solving the system, we have the eigenvector [ [ [ x ı Let P [ [ Then P [ [ 5 2 [ [ + ı [ 4 4