Eigenvalues and eigenvectors of a matrix


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1 Eigenvalues and eigenvectors of a matrix Definition: If A is an n n matrix and there exists a real number λ and a nonzero column vector V such that AV = λv then λ is called an eigenvalue of A and V is called an eigenvector corresponding to the eigenvalue λ. Example: Consider the matrix A = ( )( ) ( ) 2 1. Note that 2 3 = 4 ( ) 1 2 So, the number 4 is an eigenvalue of A and the column vector eigenvector corresponding to 4. ( ) 1 is an 2 Dana Mackey (DIT) Numerical Methods II 1 / 23
2 Finding eigenvalues and eigenvectors for a given matrix A 1. The eigenvalues of A are calculated by solving the characteristic equation of A: det(a λi ) = 0 2. Once the eigenvalues of A have been found, the eigenvectors corresponding to each eigenvalue λ can be determined by solving the matrix equation AV = λv Example: Find the eigenvalues of A = Dana Mackey (DIT) Numerical Methods II 2 / 23
3 Example: The eigenvalues of A = quadratic equation ( ) 2 1 are 1 and 4, which is seen by solving the λ λ = λ2 5λ + 4 = 0 To find the eigenvectors corresponding to 1 we write V = ( )( V1 V 2 ) = 1 ( V1 V 2 ) so ( V1 V 2 2V 1 + V 2 = V 1 2V 1 + 3V 2 = V 2 ) and solve This gives a ( single ) independent equation V 1 + V 2 = 0 so any vector of the 1 form V = α is an eigenvector corresponding to the eigenvalue 1. 1 Dana Mackey (DIT) Numerical Methods II 3 / 23
4 Matrix diagonalisation The eigenvalues and eigenvectors of a matrix have the following important property: If a square n n matrix A has n linearly independent eigenvectors then it is diagonalisable, that is, it can be factorised as follows A = PDP 1 where D is the diagonal matrix containing the eigenvalues of A along the diagonal, also written as D = diag[λ 1,λ 2,...,λ n ] and P is a matrix formed with the corresponding eigenvectors as its columns. For example, matrices whose eigenvalues are distinct numbers are diagonalisable. Symmetric matrices are also diagonalisable. Dana Mackey (DIT) Numerical Methods II 4 / 23
5 This property has many important applications. For example, it can be used to simplify algebraic operations involving the matrix A, such as calculating powers: A m = ( PDP 1) (PDP 1) (PDP 1) = PD m P 1 Note that if D = diag[λ 1,λ 2,...,λ n ] then D m = diag[λ m 1,λm 2,...,λm n ], which is very easy to calculate. ( ) 2 1 Example: The matrix A = has eigenvalues 1 and 4 with 2 3 ( ) ( ) 1 1 corresponding eigenvectors and. Then it can be diagonalised 1 2 as follows ( )( )( ) /3 1/3 A = /3 1/3 Dana Mackey (DIT) Numerical Methods II 5 / 23
6 The power and inverse power method Overview: The Gerschgorin circle theorem is used for locating the eigenvalues of a matrix. The power method is used for approximating the dominant eigenvalue (that is, the largest eigenvalue) of a matrix and its associated eigenvector. The inverse power method is used for approximating the smallest eigenvalue of a matrix or for approximating the eigenvalue nearest to a given value, together with the corresponding eigenvector. Dana Mackey (DIT) Numerical Methods II 6 / 23
7 The Gerschgorin Circle Theorem Let A be an n n matrix and define R i = n j=1 j i a ij for each i = 1,2,3,...n. Also consider the circles C i = {z Cl : z a ii R i } 1 If λ is an eigenvalue of A then λ lies in one of the circles C i. 2 If k of the circles C i form a connected region R in the complex plane, disjoint from the remaining n k circles then the region contains exactly k eigenvalues. Dana Mackey (DIT) Numerical Methods II 7 / 23
8 Example Consider the matrix A = The radii of the Gerschgorin circles are R 1 = = 1; R 2 = = 2; R 3 = = 3 and the circles are C 1 = {z Cl : z 1 1}; C 2 = {z Cl : z 5 2}; C 3 = {z Cl : z 9 3} Since C 1 is disjoint from the other circles, it must contain one of the eigenvalues and this eigenvalue is real. As C 2 and C 3 overlap, their union must contain the other two eigenvalues. Dana Mackey (DIT) Numerical Methods II 8 / 23
9 The power method The power method is an iterative technique for approximating the dominant eigenvalue of a matrix together with an associated eigenvector. Let A be an n n matrix with eigenvalues satisfying λ 1 > λ 2 λ 3... λ n The eigenvalue with the largest absolute value, λ 1 is called the dominant eigenvalue. Any eigenvector corresponding to λ 1 is called a dominant eigenvector. Let x 0 be an nvector. Then it can be written as x 0 = α 1 v 1 + α n v n where v 1,...,v n are the eigenvectors of A. Dana Mackey (DIT) Numerical Methods II 9 / 23
10 Now construct the sequence x (m) = Ax (m 1), for m 1. x (1) = Ax (0) = α 1 λ 1 v 1 + α n λ n v n x (2) = Ax (1) = α 1 λ 2 1v 1 + α n λ 2 nv n. x (m) = Ax (m 1) = α 1 λ m 1 v 1 + α n λ m n v n and hence which gives x (m) λ m 1 ( ) m ( λ2 λn = α 1 v 1 + α 2 + α n λ 1 λ 1 x (m) lim m λ m = α 1 v 1 1 ) m v n Hence, the sequence x(m) λ m 1 dominant eigenvalue. converges to an eigenvector associated with the Dana Mackey (DIT) Numerical Methods II 10 / 23
11 The power method implementation Choose the initial guess, x (0) such that max i x (0) i = 1. For m 1, let y (m) = Ax (m 1), x (m) = y(m) y (m) p m where p m is the index of the component of y (m) which has maximum absolute value. Note that and since x (m 1) p m 1 y (m) = Ax (m 1) λ 1 x (m 1) = 1 (by construction), it follows that y (m) p m 1 λ 1 Dana Mackey (DIT) Numerical Methods II 11 / 23
12 Note: Since the power method is an iterative scheme, a stopping condition could be given as λ (m) λ (m 1) < required error where λ (m) is the dominant eigenvalue approximation during the m th iteration, or max x (m) i i x (m 1) i < required error Note: If the ratio λ 2 /λ 1 is small, the convergence rate is fast. If λ 1 = λ 2 then, in general, the power method does not converge. Example Consider the matrix A = Show that its eigenvectors are λ 1 = 12, λ 2 = 3 and λ 3 = 3 and calculate the associated eigenvectors. Use the power method to approximate the dominant eigenvalue and a corresponding eigenvector. Dana Mackey (DIT) Numerical Methods II 12 / 23
13 Matrix polynomials and inverses Let A be an n n matrix with eigenvalues λ 1,λ 2,...λ n and associated eigenvectors v 1,v 2,...v n. 1 Let p(x) = a 0 + a 1 x + + a m x m be a polynomial of degree m. Then the eigenvalues of the matrix B = p(a) = a 0 + a 1 A + + a m A m are p(λ 1 ), p(λ 2 ),...p(λ n ) with associated eigenvectors v 1, v 2,...v n. 2 If det(a) 0 then the inverse matrix A 1 has eigenvalues 1 λ 1, 1 λ 2,... 1 λ n with associated eigenvectors v 1, v 2,...v n. Dana Mackey (DIT) Numerical Methods II 13 / 23
14 The inverse power method Let A be an n n matrix with eigenvalues λ 1, λ 2,...λ n and associated eigenvectors v 1, v 2,...v n. Let q be a number for which det(a qi ) 0 (so q is not an eigenvalue of A). Let B = (A qi ) 1 so the eigenvalues of B are given by µ 1 = 1 λ 1 q, µ 2 = 1 λ 2 q,... µ m = 1 λ m q. If we know that λ k is the eigenvalue that is closest to the number q, then µ k is the dominant eigenvalue for B and so it can be determined using the power method. Dana Mackey (DIT) Numerical Methods II 14 / 23
15 Example Consider the matrix A = Use Gerschgorin circles to obtain an estimate for one of the eigenvalues of A and then get a more accurate approximation using the inverse power method. (Note: The eigenvalues are, approximately, , and ) Dana Mackey (DIT) Numerical Methods II 15 / 23
16 The Gershgorin circles are z + 4 2; z 3 1 and z 15 2 so they are all disjoint! There is one eigenvalue in each of the three circles so they lie close to 4, 3 and 15. Let λ be the eigenvalue closest to q = 3 and let B = (A 3I ) 1. Then µ = 1 λ 3 is the dominant eigenvalue for B and can be determined using the power method, together with an associated eigenvector, v. Note that, once the desired approximation, µ (n), v (n) has been obtained, we must calculate λ (n) = µ (n) The associated eigenvector for λ is the same as the one calculated for µ. Dana Mackey (DIT) Numerical Methods II 16 / 23
17 Smallest eigenvalue To find the eigenvalue of A that is smallest in magnitude is equivalent to find the dominant eigenvalue of the matrix B = A 1. (This is the inverse power method with q = 0.) Example: Consider the matrix A = Use the inverse power method to find an approximation for the smallest eigenvalue of A. (Note: The eigenvalues are 3, 4 and 5.) Dana Mackey (DIT) Numerical Methods II 17 / 23
18 The QR method for finding eigenvalues We say that two n n matrices, A and B, are orthogonally similar if there exists an orthogonal matrix Q such that AQ = QB or A = QBQ T If A and B are orthogonally similar matrices then they have the same eigenvalues. Proof: If λ is an eigenvalue of A with eigenvector v, that is, then Av = λv BQ T v = λq T v which means λ is an eigenvalue of B with eigenvector Q T v. Dana Mackey (DIT) Numerical Methods II 18 / 23
19 To find the eigenvalues of a matrix A using the QR algorithm, we generate a sequence of matrices A (m) which are orthogonally similar to A (and so have the same eigenvalues), and which converge to a matrix whose eigenvalues are easily found. If the matrix A is symmetric and tridiagonal then the sequence of QR iterations converges to a diagonal matrix, so its eigenvalues can easily be read from the main diagonal. Dana Mackey (DIT) Numerical Methods II 19 / 23
20 The basic QR algorithm We find the QR factorisation for the matrix A and then take the reverse order product RQ to construct the first matrix in the sequence. A = QR = R = Q T A A (1) = RQ = A (1) = Q T AQ Easy to see that A and A (1) are orthogonally similar and so have the same eigenvalues. We then find the QR factorisation of A (1) : A (1) = Q (1) R (1) = A (2) = R (1) Q (1) This procedure is then continued to construct A (2), A (3), etc. and (if the original matrix is symmetric and tridiagonal) this sequence converges to a diagonal matrix. The diagonal values of each of the iteration matrices A (m) can be considered as approximations of the eigenvalues of A. Dana Mackey (DIT) Numerical Methods II 20 / 23
21 Example Consider the matrix A = whose eigenvalues are , , Some of the iterations of the QR algorithm are shown below: A (2) = A (10) = Note that the diagonal elements approximate the eigenvalues of A to 3 decimal places while the offdiagonal elements are converging to zero. Dana Mackey (DIT) Numerical Methods II 21 / 23
22 Remarks 1 Since the rate of convergence of the QR algorithm is quite slow (especially when the eigenvalues are closely spaced in magnitude) there are a number of variations of this algorithm (not discussed here) which can accelerate convergence. 2 The QR algorithm can be used for finding eigenvectors as well but we will not cover this technique now Exercise: Calculate the first iteration of the QR algorithm for the matrix on the previous slide. Dana Mackey (DIT) Numerical Methods II 22 / 23
23 More examples 1. Let A = Calculate the eigenvalues of A (using the definition) and perform two iterations of the QR algorithm to approximate them. 2. Let A = whose eigenvalues are , , Perform one iteration of the QR algorithm. Dana Mackey (DIT) Numerical Methods II 23 / 23
24 Applications of eigenvalues and eigenvectors 1. Systems of differential equations Recall that a square matrix A is diagonalisable if it can be factorised as A = PDP 1, where D is the diagonal matrix containing the eigenvalues of A along the diagonal, and P (called the modal matrix) has the corresponding eigenvectors as columns. Consider the system of 1st order linear differential equations x (t) = a 11 x(t) + a 12 y(t), y (t) = a 21 x(t) + a 22 y(t) written in matrix form as X (t) = AX(t), where X = (x,y) T and A = (a ij ) i.j=1,2. Dana Mackey (DIT) Numerical Methods II 24 / 23
25 If A is diagonalisable then A = PDP 1, where D = diag(λ 1,λ 2 ) so the system becomes P 1 X (t) = DP 1 X(t) If we let X(t) = P 1 X(t) then X (t) = D X(t) which can be written as x (t) = λ 1 x(t), ỹ (t) = λ 2 ỹ(t) which is easy to solve: x(t) = C 1 e λ 1t, ỹ(t) = C 2 e λ 2t, where C 1 and C 2 are arbitrary constants. The final solution is then recovered from X(t) = P X(t). Dana Mackey (DIT) Numerical Methods II 25 / 23
26 Example: Consider the following linear predatorprey model, where x(t) represents a population of rabbits at time t and y(t) are foxes x (t) = x(t) 2y(t), y (t) = 3x(t) 4y(t) Solve the system using eigenvalues and determine how the two populations are going to evolve. Assume x(0) = 4, y(0) = 1. Dana Mackey (DIT) Numerical Methods II 26 / 23
27 Applications of eigenvalues and eigenvectors 2. Discrete agestructured population model The Leslie model describes the growth of the female portion of a human or animal population. The females are divided into n age classes of equal duration (L/n, where L is the population age limit). The initial number of females in each age group is given by ( x (0) = x (0) 1,x (0) 2 ) (0) T,...x n where x (0) 1 is the number of females aged 0 to L/n years, x (0) 2 is the number of females aged L/n to 2L/n etc. This is called the initial age distribution vector. Dana Mackey (DIT) Numerical Methods II 27 / 23
28 The birth and death parameters which describe the future evolution of the population are given by a i = average no. of daughters born to each female in age class i b i = fraction of females in age class i which survive to next age class Note that a i 0 for i = 1,2,...n and 0 < b i 1 for i = 1,2,...n 1. Let x (k) be the age distribution vector at time t k (where k = 1,2,... and t k+1 t k = L/n). The Leslie model states that the distribution at time t k+1 is given by x (k+1) = Lx (k) where L, the Leslie matrix, is defined as Dana Mackey (DIT) Numerical Methods II 28 / 23
29 This matrix equation is equivalent to a 1 a 2 a 3... a n b L = 0 b b n 1 0 x (k+1) 1 = a 1 x (k) 1 + a 2 x (k) 2 + a n x n (k) x (k+1) i+1 = b i x (k) i, i = 1,2,...n 1 The Leslie matrix has the following properties: 1. It has a unique positive eigenvalue λ 1, for which the corresponding eigenvector has only positive components; 2. This unique positive eigenvalue is the dominant eigenvalue; 3. The dominant eigenvalue λ 1 represents the population growth rate, while the corresponding eigenvector gives the limiting age distribution. Dana Mackey (DIT) Numerical Methods II 29 / 23
30 Example Suppose the age limit of a certain animal population is 15 years and divide the female population into 3 age groups. The Leslie matrix is given by L = If there are initially 1000 females in each age class, find the age distribution after 15 years. Knowing that the dominant eigenvalue is λ 1 = 3 2, find the long term age distribution. Dana Mackey (DIT) Numerical Methods II 30 / 23
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