Eigenvalues and Eigenvectors

Transcription

1 MATH 307 Eigenvalues and Eigenvectors Dr. Neal, WKU Let T : R n R n be a linear transformation with n n matrix representation A. A non-zero vector x R n is called an eigenvector of T if there exists a scalar c such that T( x ) = c x. The scalar c is then called an eigenvalue. In matrix form, we say that a nonzero X is an eigenvector if A X = c X. The eigenvectors are simply those vectors in R n that are expanded or contracted in the same or opposite direction by the transformation T. Example 1. Let T : R 2 R 2 be given by T ((x, y) ) = ( 5 x, 3 y). Then A = For non-zero vectors of the form x = (x, 0), we have T( x ) = T((x, 0)) = ( 5 x, 0) = 5 x. Thus, any non-zero vector of the form (x, 0) is an eigenvector of T with associated eigenvalue c = 5. Since T( x ) = 5 x, such a vector x = (x, 0) is simply expanded by 5 in the opposite direction. These eigenvectors (x, 0) form a one-dimensional eigenspace with basis (1, 0). Similarly, non-zero vectors of the form (0, y) are eigenvectors of T with associated eigenvalue c = 3. These vectors are expanded by a factor of 3 by the transformation T. These eigenvectors form another one-dimensional eigenspace with basis (0, 1). In this case, the two eigenspace basis vectors (1, 0) and (0, 1) form a basis for all of R 2. Finding the Eigenvalues To find all possible eigenvalues, we must find the scalars c such that A X = c X for some non-zero X. This equation can be re-written in the following ways: Thus the equation A X c X = 0 A X c I n X = 0 ( A c I n ) X = 0. ( A c I n ) X = 0 must have a non-trivial solution X. In other ( ) must equal 0. (If the determinant were not equal to 0, then there words, det A c I n would be only the trivial solution of X = 0.) To find the eigenvalues c, we must solve the equation det ( A c I n ) = 0. Example 1 (continued): Because T has matrix representation A = 5 0 c 0 5 c 0 = =. So det A c I c 0 3 c n ( ) = ( 5 c)(3 c). 5 0, then A c I 0 3 n Solving ( 5 c)(3 c) = 0, we obtain c = 5 and c = 3 as the only eigenvalues.

2 Dr. Neal, WKU Finding the Eigenvectors Once we have an eigenvalue c, we must find all (non-zero) solutions to the homogeneous system ( A c I n ) X = 0. These solutions are called the eigenspace for the eigenvalue c. The eigenspace is a subspace of R n ; thus, we also can find a basis for each eigenspace. Example 1 (continued): When c = 5, then A c I n = A c I n eigenspace is (1, 0) The general solution to 0 8 ( ) X = 0 is x 1 = t and x 2 = 0 (i.e., vectors of the form (t, 0) So a basis for this For c = 3, then A c I n = A c I n eigenspace is the vector (0, 1) Now, the general solution to the system ( ) X = 0 is x 1 = 0 and x 2 = t (i.e., vectors of the form (0, t ) So a basis for this ( ) = Example 2. Let T : R 2 R 2 be a counterclockwise rotation by 90. Then T (1, 0) 0 1 c 1 (0, 1) and T ((0, 1) ) = ( 1, 0). So A =. Then A c I 1 0 n = and det A c I 1 c n ( ) = c Solving c 2 +1 = 0, we see that there are no real solutions. Thus, there are no eigenvalues or eigenvectors. No vector in R 2 is magnified along the same line by this transformation. Every vector is rotated Example 3. Let T : R 3 R 3 be defined by the matrix A = Find the eigenvalues and bases for the associated eigenspaces. Determine if the basis vectors form a basis for all of R 3. 1 c 0 1 Solution. We look at the matrix A c I n = 0 1 c 0 and we must compute c det ( A c I n ). To do so, we will use the weave method for finding 3 3 determinants: 1 c c 0 1 c c c 0

3 Dr. Neal, WKU [(1 c )( 1 c )(1 c ) ] [1( 1 c )1 + (1 c ) (1 c )] = (1 c )( 1 c )(1 c ) ( 1 c ) = ( 1 c )[(1 c) 2 1]. So we must solve ( 1 c )[(1 c) 2 1] = 0 c = 1, or (1 c) 2 = 1 1 c = ± = c c = 0 or c = 2. So we have three eigenvalues c = 1, c = 0, and c = If c = 1, then A c I n = rref x 1 = 0, x 2 = t, x 3 = 0. The one-dimensional eigenspace for c = 1 has basis (0, 1, 0). Any non-zero vector of the form t (0, 1, 0) = (0, t, 0) is an eigenvector for eigenvalue c = If c = 0, then A c I n = rref x 1 = t, x 2 = 0, x 3 = t. The one-dimensional eigenspace for c = 0 has basis ( 1, 0, 1) If c = 2, then A c I n = rref x 1 = t, x 2 = 0, x 3 = t. The one-dimensional eigenspace for c = 2 has basis (1, 0, 1). Consider the three eigenspace basis vectors S = {(0, 1, 0), ( 1, 0, 1), (1, 0, 1)}. Do they form a basis for all of R 3? Let P = Then det (P) = 2 0; thus, S does form a basis for R Diagonalizing the Matrix A Suppose the n n matrix A has n eigenvectors that form a basis for all of R n. Let P be the associated matrix representation of these eigenvectors (det (P) 0; P 1 exists). Then the matrix D = P 1 A P is diagonal with the corresponding eigenvalues down the main diagonal. In this case, we also obtain the following results:

4 det (P 1 A P) = det (A ) and Tr( P 1 A P ) = Tr( A ). Dr. Neal, WKU (These results actually hold for any invertible matrix P and are proven below.) Example 3 (continued): Here A = and P = We see that P 1 AP = 0 0 0, which shows the three eigenvalues c = 1, c = 0, and c = 2 down the main diagonal. Finally we can check that det (A ) = 0 = det (P 1 A P) and Tr( A ) = 1 = Tr( P 1 A P ). Definition. Let A, B be n n matrices. We say that A is similar to B if there exists an invertible n n matrix P such that A = P 1 B P. Theorem 4.1. Similarity is an equivalence relation on the set of n n matrices. That is, similarity is refelxive, symmetric, and transitive. Reflexive: A is similar to A for all n n matrices A. Proof. Let P = I n. Then P 1 = I n also, and A = I n 1 A In = P 1 AP ; so A is similar to A. Symmetric: If A is similar to B, then B is similar to A. Proof. Assume A is similar to B. Then there exists an invertible n n matrix P such that A = P 1 B P. But Q = P 1 is also invertible and Q 1 = P ; thus, Hence, B is similar to A. B = PA P 1 = Q 1 AQ. Transitive: If A is similar to B and B is similar to C, then A is similar to C. Proof. Assume A is similar to B and B is similar to C. Then there exist invertible n n matrices P and Q such that A = P 1 B P and B = Q 1 CQ = C. Then R = PQ is also an invertible n n matrix and thus, A is similar to C. A = P 1 B P = P 1 Q 1 CQP = (QP) 1 C(QP) = R 1 C R ;

5 Theorem 4.2 Similar matrices have the same determinant and the same trace. Dr. Neal, WKU Proof. If A is n n and P is an invertible n n matrix with A = P 1 B P, then det(a) = det(p 1 B P) = det(p 1 ) det(bp) = det(b P) det(p 1 ) = det(b PP 1 ) = det(b I n ) = det(b). Thus, similar matrices have the same determinant. For the trace operator, recall that Tr(A B) = Tr(B A); thus, Tr(A) = Tr(P 1 B P) = Tr(B PP 1 ) = Tr(B I n ) = Tr(B). Computing A k Let A be an n n having n distinct eigenvectors that form a basis for R n. Let P be the associated matrix representation of these eigenvectors (det (P) 0; P 1 exists). Then the matrix D = P 1 A P is diagonal with the corresponding eigenvalues down the main diagonal. Using D = P 1 A P, we can solve for A to obtain A = P DP 1. It is now possible to compute A k by A k = (P DP 1 ) k = P DP 1 P DP 1... P DP 1 = P D k P 1. However D is diagonal of the form c c D = c n where the c i are the eigenvalues corresponding to the eigenvectors of each column of P. So k c k c k k 0 c c D k = and thus A k = P P k k c n c n The next examples show the usefulness of this form of A k :

6 Dr. Neal, WKU Example 4. Currently 30% of the adult population is on a diet to lose weight. But each week, 10% of those on a diet drop off, and 5% of those not on a diet start one. Let T be the transition matrix. (i) If the process maintains fixed transition probabilities, then what is the limit state? (ii) Prove that the limiting state is independent of the initial state. Solution. We first shall find the eignevalues and eigenvectors, and use them to find T k. Then we shall compute lim k T k. There are two states (On Off), with initial state A = ( ). The transition matrix T is Initial State On Off On Off State After Transition = T λ 0.1 Then T λ I 2 = and λ det(t λ I 2 ) = (0.9 λ)(0.95 λ) = λ λ = (λ 1)(λ 0.85). Solving det(t λ I 2 ) = 0, we obtain the two eigenvalues λ=1 and λ= When λ=1, we have T λ I 2 = and rref (T λ I ) = 1 1. Thus, (T λ I )X = 0 has the solution form x 1 = x 2 = t, with a basis being the vector (1, 1) When λ= 0.85, we have T λ I 2 = and rref (T λ I ) = 1 2. Thus, 0 0 (T λ I 2 )X = 0 has the solution x 1 = 2t, x 2 = t, with a basis being the vector ( 2, 1). We now write the eigenvectors (1, 1) and ( 2, 1) in column form to create the matrix P = 1 2. Then P 1 T P = 1 0 and T = P 1 0 P 1. Therefore, we have T k = P 1k k P 1 and lim T k = P 1 0 P 1 1/3 2 /3 =. k 0 0 1/3 2 /3 Then for any initial state matrix A = ( p 1 p 2 ) with p 1 + p 2 = 1, we have lim AT k 1/3 2/3 = ( p 1 p 2 ) = 1 k 1/3 2/3 3 p p p p 2 = Thus, no matter what the initial state is, these transition probabilities force an eventual steady-state of 1/3 On a Diet and 2/3 Off a Diet.

7 Dr. Neal, WKU Example 5. Genotypes AA, Aa, aa are always bred with the genotype AA. The initial generation is 10% AA, 30% Aa and 60% aa. The resulting successive generations continue to be bred with AA. Let T be the transition matrix. (i) Compute lim T k. (ii) What is the limiting state of the genotypes regardless of the k state of the initial generation? Solution. There are three states (AA Aa aa), with initial state I = ( ). The transition matrix T is given by Breed AA with AA Aa aa Next generation AA Aa = T aa λ 0 0 Then T λ I 3 = λ 0 and det(t λ I 3) = (1 λ)(0.5 λ)( λ). 0 1 λ Solving det(t λ I 3 ) = 0, we obtain the three eigenvalues λ=1, λ= 0.5, and λ = 0. When λ=1, we have T λ I 3 = and rref (T λ I 3 ) = Thus, (T λ I 3 )X = 0 has the solution x 1 = x 2 = x 3 = t, with a basis being the vector (1, 1, 1) When λ= 0.5, we have T λ I 3 = and rref (T λ I 3) = Thus, (T λ I 3 )X = 0 has the solution x 1 = 0, x 2 = 0.5x 3, x 3 = t, with a basis being the vector (0, 0.5, 1). When λ= 0, we have T λ I 3 = and rref (T λ I 3 ) = Thus, (T λ I 3 )X = 0 has the solution x 1 = 0, x 2 = 0, x 3 = t, with a basis being the vector (0, 0, 1). We now write the eigenvectors in column form to create the matrix P =

8 Then P 1 T P = and T = P P 1. Therefore, we have 1 k 0 0 T k = P k 0 P 1 and lim T k = P k P 1 = Dr. Neal, WKU ( ) with p 1 + p 2 + p 3 =1, we have Then for any initial state matrix I = p 1 p 2 p 3 lim AT k = ( p 1 p 2 p 3 ) k = ( p 1 + p 2 + p 3 0 0) = ( 1 0 0). Thus, no matter what the initial state is, after many iterations of breeding successive generations solely with AA, the entire population becomes AA. We conclude with a final result: Theorem 4.3. (a) Let T : R n R n be a linear transformation. If T is one-to-one and onto and c is an eigenvalue of T, then c 0 and 1 / c is an eigenvalue of T 1. (b) Let A be invertible. If c is an eigenvalue of A, then 1 / c is an eigenvalue of A 1. Proof. (a) Because c is an eigenvalue of T, there exists a non-zero vector x such that T( x ) = c x. If c = 0, then T( x ) = 0 for a non-zero vector x, which cannot happen because T is one-to-one. Thus, c 0. Because T is one-to-one and onto, T 1 exists. And because T( x ) = c x, we have x = T 1 (c x ) = c T 1 ( x ). Thus, T 1 ( x ) = 1 x and 1 / c is an c eigenvalue of T 1. (b) Suppose c is an eigenvalue of A. Then A X = c X for some non-zero eigenvector X. Because A is invertible, we have X = A 1 (c X) = c A 1 X. Because X is a non-zero vector, then c cannot equal 0 or else c A 1 X = X would be 0. So c 0. Now because c A 1 X = X, we have A 1 X = 1 c X. eigenvector X. So 1 / c is an eigenvalue of A 1 with the same

9 Exercises Dr. Neal, WKU (i) A = (ii) A = (iii) A = (iv) A = For each of the above matrices A, (a) Find the eigenvalues of A. (b) For each eigenvalue, find the eigenspace and a basis for the eigenspace. (c) Determine if there is a basis of eigenvectors. If so, show how to convert A into a diagonal matrix that is similar to A. Then verify that the similar matrix has the same determinant and the same trace as A. (d) If A is invertible, then find the eigenvalues of A Successive generations of genotypes AA, Aa, aa are always bred with the same type of genotype (AA always with AA, Aa always with Aa, and aa always with aa). (a) Give the matrix of transition probabilities T. (b) Find the eigenvalues of T. For each eigenvalue, find the eigenspace and a basis for the eigenspace. (c) Compute T k and lim T k. k (d) Given an initial state of I = ( p 1 p 2 p 3 ), what is the limiting distribution? Prove or disprove that the limiting state depends on the initial state.

10 Dr. Neal, WKU Answers (i) c = 1, basis ( 1, 1, 0), ( 1, 0,1); c = 2, basis (1, 1, 1). Let P = det (P) = 3 0, so there is a basis of eigenvectors. A is similar to P 1 AP = Also, det (A) = 2 = det (P 1 A P) and Tr(A) = 0 = Tr(P 1 A P). The eigenvalues of A 1 are 1 and 1/ (ii) c = 0, basis ( 1, 1, 0); c = 1, basis (0, 0, 1); c = 2, basis (1, 1, 0). Let P = det (P = 2 0, so there is a basis of eigenvectors. A is similar to P 1 AP = and det (A) = 0 = det(p 1 A P)) and Tr(A) = 3 = Tr(P 1 A P). (iii) c = 2, basis ( 1, 0,1); c = 4, basis (1, 2, 1); c = 0, basis (1, 2, 1) Let P = det (P) = 8 0, so there is a basis of eigenvectors. A is similar to P 1 AP = and det (A) = 0 = det(p 1 A P) and Tr(A) = 6 = Tr(P 1 A P) (iv) c = 4, basis (1, 0, 0), (0, 0, 1); c = 2, basis (1, 2, 1). Let P = ; det (P) = 2 0, so there is a basis of eigenvectors. A is similar to P 1 AP = and det (A) = 32 = det (P 1 A P) and Tr(A) = 10 = Tr(P 1 A P). The eigenvalues of A 1 are 1/4 and 1/2.