MATHEMATICS FOR ENGINEERS BASIC MATRIX THEORY TUTORIAL 3 - EIGENVECTORS AND EIGENVALUES

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1 MATHEMATICS FOR ENGINEERS BASIC MATRIX THEORY TUTORIAL 3 - EIGENVECTORS AND EIGENVALUES This is the third tutorial on matrix theory. It is entirely devoted to the subject of Eigenvectors and Eigenvalues which are used to solve many types of problems in engineering such as the frequency of vibrating systems with several degrees of freedom. INTRODUCTION Suppose we have a square matrix A and that there is a vector x such that A x = λx λ is a scalar called the eigenvalue of A and x is the eigenvector. The eigenvector must not be a zero vector but λ can be zero. The problem is how to find the eigenvector and eigenvalue for a given matrix. This work only applies to a square matrix. THE POWER METHOD for FINDING THE EIGENVECTOR 6 Consider the matrix A 4 1 Step 1 Choose x o 1 Step 2 Evaluate Ax o and call it x 1 Ax o = x 1 = = 4 5 Reduce by dividing by 5 and call it x 1 Evaluate Ax 1 and call it x 2 Reduce by dividing by 5.8 and call it x 2 Evaluate Ax 2 and call it x 3 Reduce by dividing by and call it x 3 Evaluate Ax 3 and call it x 4 Reduce by dividing by and call it x 4 Evaluate Ax 4 and call it x 5 Reduce by dividing by and call it x Ax 2 = x Ax 3 = x Ax 4 = x x 1 x 1 x 2 x 3 x 4 x 5 D.J.Dunn 1

2 If we kept going we would see that the values converge on 2 and 1 and this is the eigenvector We have A x Ax So λx = λ so λ = 6 6 This may also be found from the relationship called the Rayleigh Quotient 6 [ 2 1] T x Ax 30 = = = 6 T x x 5 [ 2 1] 1 There must be another eigenvalue as a 4 x 4 matrix always yields two. The power method does not work for all cases. SECOND METHOD USING THE IDENTITY MATRIX You might recall the identity matrix or unit matrix is a special case of the diagonal matrix in which all the leading elements are 1 and that any array multiplied by the unit matrix is unchanged. The unit matrix is usually designated I. It follows that A X = λix A X λix = 0 (A λi)x = 0 Although X = 0 is a solution we cannot have a zero eigenvector. We need a non-zero solution. If one exists then det(a λi) = 0 This is called the CHARACTERISTIC EQUATION. Lets try it on the previous example. det(a λi)x = (3 - λ)(4 - λ) (6)(1) = λ 2-7λ + 6 = (λ - 6)(λ - 1) so λ = 6 or 1 The eigenvectors are the vector X that make (A λi) X = 0. Take λ = 6 (A λi)x = 0 The characteristic equation matrix is X is a single column vector with coefficients a and b (A λi)x gives us simultaneous equations. We must solve the simultaneous equation resulting. The equations are -3a + 6 b = 0 and a -2b = 0. This is not difficult to solve and show a = 2 and b = 1 but in more difficult cases you might use the augmented matrix. An eigenvector is α where α is any scalar and this is the same one found previously. In Cartesian coordinates the vector could be α(2x, y) Now repeat the process and find an eigenvector corresponding to λ = 1 The characteristic equation is D.J.Dunn 2

3 We must solve the simultaneous equations The equations are 2a + 6b = 0 and a + 3 b = 0. Again this is not difficult to solve and a = -3, b = 1. 3 An eigenvector is β Where β is any scalar number. In cartesian form this might be β(-3x, y) and this is the same one found previously. WORKED EXAMPLE No.1 Find the eigenvalues and eigenvectors of matrix SOLUTION The characteristic equation is A - λi = The determinant is (4-λ)(3-λ) (1)(2) = λ 2-7λ = λ 2-7λ + 10 Equate to zero and solve λ λ 2-7λ + 10 = 0 (λ - 5)(λ - 2) = 0 λ = 5 or 2 and these are the eigenvalues To find the eigenvectors we must solve the simultaneous equations for each eigenvalue. Take λ = 5 The characteristic equation is We must solve The equations to be solved are a + b = 0 and 2a -2b = 0 from which it is apparent that a = b so one eigenvector is : α is any scalar value. In Cartesian form this might be α(x, y) Take λ = 2 The characteristic equation is We must solve The equations to be solved are 2a + b = 0 and 2a + b = 0 from which it is apparent that b = -2a so one eigenvector is : β is any scalar value. In Cartesian form this might be β(x, 2y) D.J.Dunn 3

4 WORKED EXAMPLE No.2 Find the eigenvalues and eigenvectors for the matrix SOLUTION First find the characteristic equation. Next find det(m - λi) First find the minors The cofactors go so A 11 = λ 2-2λ +1 A 12 = 4-4λ A 13 = 4-4λ From the matrix a 11 = - λ a 12 = 2 a 13 = 1 The determinant is a 11 A 11 + a 12 A 12 + a 13 A 13 = (- λ)(λ 2-2λ + 1) + (2)( 4-4λ) + (1)(4-4λ) The determinant is -λ 3 + 2λ 2-11λ + 12 Equate to zero and change sign for convenience 0 = λ 3-2λ 2-11λ + 12 = (λ + 3) (λ -4) (λ - 1) The eigenvalues are -3, 4 and 1 For eigenvalue λ = -3 (M-λI)X = 0 This has reduced the equations to 3a + 2b +c = 0 and a + b = 0 hence b = -a and 3a 2a + c = 0 so c = -a The eigenvector is γ(a, -a, -a) = γa(1, -1,-1)or in Cartesian form α(x, -y, -z) α is any scalar value. D.J.Dunn 4

5 For eigenvalue λ = 4 (M-λI)X = 0 We have reduced the equations to -4a + 2b +c = 0 4a 3b = 0 8a = 0 Once solution is a = b = c =0 but the eigenvector can not be the zero vector. Another solution is: b = 4a/3, c = 4a/3 The eigenvector is then γ(a, -4a/3, 4a/3) = γa(1, -4/3, 4/3) or in Cartesian form β(x, -4y/3, 4z/3) where β is any scalar value. For eigenvalue λ = 1 (M-λI)X = 0 This reduces to -1a + 2 b + c = 0 and 4a = 0 in which case c = -2b and the eigenvector is: γ(0, b, -2b) or (γ/b)(0, 1, -2) and in Cartesian form µ(0, y, -2z) where µ is any scalar value. WORKED EXAMPLE No.3 Find the eigenvalues and eigenvectors for the matrix SOLUTION First find the characteristic equation. Next find det(m - λi) First find the minors D.J.Dunn 5

6 The cofactors go so A 11 = λ 2-3λ - 4 A 12 = 2λ + 2 A 13 = 4 + 4λ From the matrix a 11 = 3 - λ a 12 = 2 a 13 = 4 The determinant is a 11 A 11 + a 12 A 12 + a 13 A 13 = (3 - λ)(λ 2-3λ - 4) + (2)( 2λ + 2) + (4)(4 + 4λ) The determinant is λ 3 + 6λ 2 +15λ + 8 Equate to zero 0 = -λ 3 + 6λ λ +8 = (λ +1) (λ + 1) (λ - 8) The eigenvalues are -1, -1 and 8 For eigenvalue λ = -1 (M-λI)X = 0 This should yield simultaneous equations to be solved but all three are multiples of the same equation so we must find values of a, b and c that satisfy 2a + b + 2c = 0. There are infinite solutions to this. For eigenvalue λ = 8 (M-λI)X = 0 This yields simultaneous equations to be solved. Manipulate as follows. Now we have -5a +2 b +4 c = 0 and -2b + c = 0 From this c = 2b The eigenvector is -5a +2b + 8b = 0 b = a/2 hence c = a In Cartesian form the vector is β(2x, y, 2z) D.J.Dunn 6

7 SELF ASSESSMENT EXERCISE Find the eigenvalues and eigenvectors for the following matrixes. Answers -2 and 5 α(-x, 3y/2) and β(x, 2y) Answers 1, -1 and 5, (0, 0, z), (-x, 2y, 5z/2) and (x, y, 5z/4) Answers -6, 1 and 5, (2x, -y, z), (x, 3y, -3z) and (x, 3y, z) D.J.Dunn 7