# matrices

week_09_matrices_part_1 Tuesday, 22 September 2020 week_09_m atrices_pa… 7:53 pm week_09_matrices_part_2 Tuesday, 22 September 2020 week_09_m atrices_pa… 8:35 pm week_09_matrices_part_3 Wednesday, 23 September 2020 week_09_m atrices_pa… 4:25 pm week_09_matrices_part_4 Wednesday, 23 September 2020 week_09_m atrices_pa… 5:39 pm week_07_matrices_part_4 Thursday, 10 September 2020 week_07_m atrices_pa… 3:22 pm week_08_matrices_part_1 Tuesday, 15 September 2020 week_08_m atrices_pa… 2:22 pm week_08_matrices_part_2 Tuesday, 15 September 2020 week_08_m atrices_pa… 3:42 pm week_08_matrices_part_3 Tuesday, 15 September 2020 week_08_m atrices_pa… 5:10 pm week_06_matrices_part_3 Thursday, 27 August 2020 week_06_m atrices_pa… 9:42 pm week_06_matrices_part_4 Friday, 28 August 2020 week_06_m atrices_pa… 9:16 am week_07_matrices_part_1 Wednesday, 9 September 2020 week_07_m atrices_pa… 9:21 pm week_07_matrices_part_2 Wednesday, 9 September 2020 week_07_m atrices_pa… 10:06 pm week_07_matrices_part_3 Thursday, 10 September 2020 week_07_m atrices_pa… 2:09 pm week_09_complex_part_4 Wednesday, 6 May 2020 week_09_co mplex_par… 2:57 pm week_06_matrices_part_1 Thursday, 27 August 2020 week_06_m atrices_pa… 8:12 pm week_06_matrices_part_2 Thursday, 27 August 2020 week_06_m atrices_pa… 9:07 pm School of Science mathematics and statics Student# Name family name given names Class Exercise 5 Matrices 0 1 2 1. If A = 1 2 3 determine A−1 using elementary row operations. 2 3 0 Verify that the inverse you have found is correct by calculating A−1 A. (Show all working and label all row operations.) Week 11 Class Exercise 5 Matrices Week 11 2. (a) Find all values of λ for which −λ 2 −12 10 − λ = 0. (b) Hence, or otherwise, find all values of λ for which −λ 0 2 0 5−λ 0 −12 = 0. 0 10 − λ 1 2 61 8 −11 8 22 8 , X = 1 and Y = −4 . 3. Let A = 2 1 −11 8 61 (a) Calculate AX. Hence, find the special λ such that AX = λX . (b) Does a λ value exist such that AY = λY ? Justify your answer. (4 + (1 21 + 1 12 ) + (1 12 + 1 12 ) = 10 marks) week_10_matrices_part_2 Thursday, 1 October 2020 week_10_m atrices_pa… 2:53 pm ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. mathematics and statics Topic A2.7 Eigenvalues and Eigenvectors Screen 1 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. Upon successful completion of this topic, you will be able to • recognise that eigenvectors are vectors which do not change direction under a linear mapping, i.e. if X is an eigenvector, AX = λX for some scalar λ • recognise the significance of the eigenvalue λ as a measure of how much X is “stretched” by the mapping • verify that a given vector is an eigenvector of a given matrix and determine the corresponding eigenvalue • determine the eigenvalues and corresponding eigenvectors of a given 2 × 2 or 3 × 3 matrix A • determine a matrix P such that P −1 AP is a diagonal matrix D • describe the relationship between the columns of P and the diagonal elements of D • determine whether a given square matrix A can be diagonalised Screen 2 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. Special Directions As we have seen in the previous topic, a matrix which represents a rotation in 3D has a special direction, namely the direction vector of the axis of rotation. All points X along this axis are multiples of the direction vector and have the property that AX = X, i.e. they do not change position under the transformation. In fact, all matrices have special directions which satisfy the matrix equation AX = λX (1) where λ is a scalar constant associated with each particular X. The special direction vectors X are called eigenvectors and the associated special λ values are called eigenvalues. Since any multiple of X clearly satisfies (1) we can think of the eigenvectors as describing invariant lines under the given mapping, i.e lines which do not move under the mapping. It is clear that for rotations in 3D there is an eigenvector associated the axis of rotation, and for reflections in 2D an eigenvector is associated with the mirror line. Consider the following example: Screen 3 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. Worked Example 1 Determine the eigenvalues and eigenvectors for the matrix ” A= 1 4 4 −5 # and interpret the mapping defined by A physically. Firstly, det A = −21 so that some reflection and stretching is involved. (Remember a pure reflection has a determinant of −1.) The equation defining the eigenvalues and eigenvectors is AX = λX so that we have ” 1 4 4 −5 #” x y # ” =λ x y # . Screen 4 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. Writing this a system of equations we have x + 4y = λx 4x − 5y = λy and collecting like terms on one side we have (1 − λ)x + 4y = 0 4x + (−5 − λ)y = 0 Note that this system of equations is equivalent to the matrix system ” 1−λ 4 4 −5 − λ #” x y # ” = 0 0 # or in other words (A − λI)X = O. Here we strike our first hurdle. We have 2 equations and 3 unknowns! There seems to be no way to find a unique solution for x, y and λ, however the task is not quite as difficult as it seems. In fact we don’t really want a unique solution at all here! Screen 5 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. For almost every choice of λ, the matrix A − λI will have an inverse. This means that we may solve for X using matrix algebra: X = (A − λI)−1 O = O so that there is a unique solution X = O. Now, the solution X = O is obvious by inspection and is called the trivial solution. If we are looking for special directions we definitely do not want the trivial solution so these λ values are no good to us. Non-trivial solutions will only occur for the values of λ such that A − λI has no inverse. From the determinants topic we know that the matrix A − λI will have no inverse if det(A − λI) = 0. This is called the characteristic equation of A. Now, for the current example, 1−λ 4 4 −5 − λ = (1 − λ)(−5 − λ) − 16 = λ2 + 4λ − 21 = (λ − 3)(λ + 7) det(A − λI) = Screen 6 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. Thus det(A − λI) = 0 if λ = 3 or λ = −7. For each of these special λ values, the eigenvalues, we may now solve the system of equations to determine the associated interesting, non-trivial solution: We return to the system (A − λI)X = O and solve this system for each λ using an augmented matrix. For λ = 3 we have ” 4 0 -2 4 −8 0 # and, following the first row operation: ” -2 4 0 0 0 0 # R2 + 2R1 → R2 If we now let y = t we have x = 2t, so that the solution is ” X=t for any t. 2 1 # Screen 7 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. For λ = −7 we have ” 8 4 0 4 2 0 # and, following the first row operation: ” 8 4 0 0 0 0 # 2R2 − R1 → R2 If we now let y = t we have x = −(1/2)t, so that the solution is ” X=t for any t∗ . − 21 1 # ” =t ∗ −1 2 # Screen 8 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. Geometrical Interpretation y y = −2x P0 Q0 y = (1/2)x Q x P Screen 9 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. For λ = 3 the corresponding solution is x = 2t, y = t. If we eliminate t we have y = (1/2)x. For λ = −7 the corresponding solution is x = −t, y = 2t. If we eliminate t we have y = −2x. Points on these lines will always remain on these lines under the mapping defined by A. For example consider point P on y = −2x. Since λ = −7, points on this line will satisfy AX = −7X, so that point P will move to P 0 on the other side of the origin and the distance from P 0 to the origin will be 7 times that of P . Similarly, for point Q on y = (1/2)x, we have λ = 3. All points on this line will satisfy AX = 3X, and point Q will move to Q0 on the same side of the origin but at a distance from the origin three times greater. From the sketch below it is clear that y = (1/2)x behaves something like a mirror while y = −2x is a line perpendicular to the “mirror”. Screen 10 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. Summary Given a matrix A we determine the eigenvalues and eigenvectors by • solving the characteristic equation det(A − λI) = 0. This is a polynomial equation in terms of λ. For 2 × 2 matrices the characteristic polynomial det(A − λI) is a quadratic and for 3 × 3 matrices it is a cubic. • for each λ, determining the corresponding eigenvector. Screen 11 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. Worked Example 2 Determine the eigenvalues and eigenvectors for the matrix 2 0 2 A = 0 7 0 . 2 0 5 We begin by solving the characteristic equation det(A − λI) = 0: det(A − λ) = 2−λ 0 2 0 7−λ 0 2 0 5−λ 2−λ 2 2 5−λ = (7 − λ) expanding by second row = (7 − λ) [(2 − λ)(5 − λ) − 4] h = (7 − λ) λ2 − 7λ + 6 i = (7 − λ)(λ − 1)(λ − 6) Screen 12 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. Thus, det(A − λI) = 0 if λ = 1, 6 or 7. For λ = 1 we solve (A − λI)X = O: 1 0 2 0 0 6 0 0 . 2 0 4 0 After the first set of row operations we have 1 0 0 0 2 0 R2 /6 → R2 1 0 0 R3 − 2R1 → R3 0 0 0 We let z = t. The second equation tells us that y = 0, while the first says x = −2t. Hence the solution is −2 X = t 0 . 1 Screen 13 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. For λ = 6 we solve (A − λI)X = O: -4 0 2 0 0 1 0 0 . 2 0 −1 0 After the first set of row operations we have -4 0 0 0 2 0 1 0 0 2R3 + R1 → R3 0 0 0 We let z = t. The second equation tells us that y = 0, while the first says x = (1/2)t. Hence the solution is 1 1 2 X = t 0 = t∗ 0 . 1 2 Screen 14 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. For λ = 7 we solve (A − λI)X = O: -5 0 2 0 0 0 0 0 . 2 0 2 0 Swap rows 2 and 3: -5 0 2 0 2 0 2 0 R2 ↔ R3 0 0 0 0 Zero in first column: -5 0 0 0 0 0 2 0 14 0 5R2 + 2R1 → R2 0 0 We let y = t. The first and second equations tell us that x = 0 and z = 0. Hence the solution is 0 X = t 1 . 0 Screen 15 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. The three eigenvalues are 1, 6 and 7. The corresponding eigenvectors are −2 0 , 1 1 0 2 and 0 1 . 0 These three eigenvectors are described as linearly independent since none of them can be obtained as a linear combination of the others. There is a theorem which states that if A is a symmetric matrix with distinct real eigenvalues then the eigenvectors are not only linearly independent, but mutually orthogonal. Screen 16 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. Worked Example 3 Determine with reasons which, if any, of the following vectors 0 0 , 0 1 −1 1 1 2 1 and are eigenvectors of the matrix 1 1 0 A= 1 2 1 . 0 1 1 0 Firstly, 0 is never an eigenvector. The zero vector is the trivial solution. 0 Secondly, 0 1 1 1 1 0 1 A −1 = 1 2 1 −1 = 0 = 0 −1 . 1 0 1 1 1 0 1 Screen 17 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. 1 Thus −1 is an eigenvector associated with the eigenvalue λ = 0. 1 Lastly, 1 1 1 0 1 3 1 6 2 1 2 1 2 A = = = 3 2 . 3 1 1 0 1 1 1 1 Thus 2 is an eigenvector associated with the eigenvalue λ = 3. 1 Screen 18 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. Diagonalisation Suppose we have a matrix A which has eigenvalues λ1 , λ2 and λ3 , and corresponding eigenvectors v1 , v2 and v3 . If we construct a matrix P with the eigenvectors of A as columns, i.e. P = h v1 v2 v3 i then the product P −1 AP is a diagonal matrix, i.e. λ1 0 0 0 P −1 AP = 0 λ2 0 0 λ3 (note the order of the eigenvalues along the diagonal). Note that only those matrices with a full set of eigenvectors may be diagonalised. Often the set of eigenvectors of a matrix A provides a natural (and better) set of coordinate axes for the mapping represented by A. The process of diagonalisation is closely related to the process of rotating the axes to line up with the eigenvectors of A (e.g. lining up the coordinate axes with the axis of rotation, or the mirror line). Screen 19 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. Worked Example 4 ” Find a matrix P which diagonalises the matrix A = From worked example 1, A has eigenvectors ” 2 1 # ” and −1 2 # . We construct P using the eigenvectors as columns: ” P = 2 −1 1 2 # . Now, the inverse of P is easy to write down using the formula: P −1 1 = 5 ” 2 1 −1 2 # . # 1 4 . 4 −5 Screen 20 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. Check that P diagonalises A by calculating P −1 AP : ” P −1 1 AP = 5 ” 1 = 5 ” 1 = 5 ” = 2 1 −1 2 #” 6 3 7 −14 #” 15 0 0 −35 3 0 0 −7 # . 1 4 4 −5 # #” 2 −1 1 2 # 2 −1 1 2 # Screen 21 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. Cayley-Hamilton Theorem The Cayley-Hamilton theorem states simply that a matrix satisfies its own characteristic equation. ” # 1 4 For example, the matrix A = from worked example 1 has characteristic (polynomial) 4 −5 equation λ2 + 4λ − 21 = 0 and hence satisfies the matrix equation A2 + 4A − 21I = O. This equation provides a cheap way of finding the inverse of A, for if we multiply through by A−1 , we obtain A−1 A2 + 4A−1 A − 21A−1 I = A−1 O which implies that A + 4I − 21A−1 = O. Rearranging gives A−1 = 1 (A + 4I). 21 Activities You should read pages 363 to 368 of the text and attempt problems 19 – 21 before moving on to the next topic. week_10_matrices_part_1 Thursday, 1 October 2020 week_10_m atrices_pa… 2:32 pm ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. mathematics and statics Topic A2.7 Eigenvalues and Eigenvectors Screen 1 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. Upon successful completion of this topic, you will be able to • recognise that eigenvectors are vectors which do not change direction under a linear mapping, i.e. if X is an eigenvector, AX = λX for some scalar λ • recognise the significance of the eigenvalue λ as a measure of how much X is “stretched” by the mapping • verify that a given vector is an eigenvector of a given matrix and determine the corresponding eigenvalue • determine the eigenvalues and corresponding eigenvectors of a given 2 × 2 or 3 × 3 matrix A • determine a matrix P such that P −1 AP is a diagonal matrix D • describe the relationship between the columns of P and the diagonal elements of D • determine whether a given square matrix A can be diagonalised Screen 2 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. Special Directions As we have seen in the previous topic, a matrix which represents a rotation in 3D has a special direction, namely the direction vector of the axis of rotation. All points X along this axis are multiples of the direction vector and have the property that AX = X, i.e. they do not change position under the transformation. In fact, all matrices have special directions which satisfy the matrix equation AX = λX (1) where λ is a scalar constant associated with each particular X. The special direction vectors X are called eigenvectors and the associated special λ values are called eigenvalues. Since any multiple of X clearly satisfies (1) we can think of the eigenvectors as describing invariant lines under the given mapping, i.e lines which do not move under the mapping. It is clear that for rotations in 3D there is an eigenvector associated the axis of rotation, and for reflections in 2D an eigenvector is associated with the mirror line. Consider the following example: Screen 3 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. Worked Example 1 Determine the eigenvalues and eigenvectors for the matrix ” A= 1 4 4 −5 # and interpret the mapping defined by A physically. Firstly, det A = −21 so that some reflection and stretching is involved. (Remember a pure reflection has a determinant of −1.) The equation defining the eigenvalues and eigenvectors is AX = λX so that we have ” 1 4 4 −5 #” x y # ” =λ x y # . Screen 4 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. Writing this a system of equations we have x + 4y = λx 4x − 5y = λy and collecting like terms on one side we have (1 − λ)x + 4y = 0 4x + (−5 − λ)y = 0 Note that this system of equations is equivalent to the matrix system ” 1−λ 4 4 −5 − λ #” x y # ” = 0 0 # or in other words (A − λI)X = O. Here we strike our first hurdle. We have 2 equations and 3 unknowns! There seems to be no way to find a unique solution for x, y and λ, however the task is not quite as difficult as it seems. In fact we don’t really want a unique solution at all here! Screen 5 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. For almost every choice of λ, the matrix A − λI will have an inverse. This means that we may solve for X using matrix algebra: X = (A − λI)−1 O = O so that there is a unique solution X = O. Now, the solution X = O is obvious by inspection and is called the trivial solution. If we are looking for special directions we definitely do not want the trivial solution so these λ values are no good to us. Non-trivial solutions will only occur for the values of λ such that A − λI has no inverse. From the determinants topic we know that the matrix A − λI will have no inverse if det(A − λI) = 0. This is called the characteristic equation of A. Now, for the current example, 1−λ 4 4 −5 − λ = (1 − λ)(−5 − λ) − 16 = λ2 + 4λ − 21 = (λ − 3)(λ + 7) det(A − λI) = Screen 6 of 21 ©2019 School of Science (Mathematical Sciences), RMIT. All rights reserved. Thus det(A − λI) = 0 if λ = 3 or λ = −7. For each of these special λ values, the eigenvalues, we may now solve the system of equations to determine the associated interesting, non-trivial solution: We return to the system (A − λI)X = O and solve this system for each λ using an augmented matrix. For λ = 3 we have ” 4 0 -2 4 −8 0 # and, following the first row operati…

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