Diagonalization problems and solutions pdf SPECTRAL THEORY OF VECTOR SPACES 81 Chapter 12. Suppose that A is diagonalizable, say A∼D where D is diagonal. Mahmut Kuzucuo˘glu METU, Ankara March 14, 2015 M. 4. These notes are collection of those solutions of exercises. The basis might not be unique. 2 Diagonalization Suppose that you are given an arbitrary n n matrix A. 1. The discussions in Sections 1. Find the sum and product of the eigen values of the matrix [] Solution: Sum of the eigen values = Sum of the main diagonal elements = -3 Product of the eigen values = │A│ = -1 (1 – 1) -1(-1 – 1) + 1(1- (-1)) = 2 + 2 = 4 2. entries off the main diagonal are all zeros). Watch the recitation video on Problem Solving: Powers of a Matrix; Recitation video transcript (PDF) Check Yourself Problems and Solutions. (b)If v is an eigenvector with Av = v, then Y(t) = ce tv is a solution for any c2R. 2 #6. Introduction to Linear Algebra: Strang) Describe all matrices S that diagonalize this matrix A (find all eigenvectors): 4 0 A = . List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017. 3 Applications of Matrix Diagonalization Quadratic Forms Applications in Physics 3. In vector form, the solutions are: 2 4 x y z 3 5= r 2 4 1 0 0 3 5+ s 2 4 0 0 1 3 5; so the basic solutions to the homogeneous system are 2 4 1 0 0 3 5and 2 4 0 0 1 3 5, and these are two independent 1-eigenvectors of2 C. Diagonalizing a matrix S−1 Solution: The key observation is that if A were a diagonal matrix, this would be simple. 1 Call these eigenvec-tors x1,,x n. 5. Answers to Odd-Numbered Exercises86 Chapter 13. Observe LINEAR ALGEBRA | PRACTICE EXAM 4 (1) Diagonalization. There were 13 weekly quizzes. Thus we begin by finding the eigenvalues ad eigenvectors of A. 1 motivate the following question: Can we find a diagonal matrix D that is similar to A? 3 Diagonalization and Powers of A Problem Sets With Solutions. Thus the eigenvalues are λ1 = 3 and λ2 = 5 with corresponding eigenvectors ~v1 = 1 −1 and ~v2 = 1 1 11. Problems of Diagonalization of Matrices. We begin with an example involving the matrix A from Examples 11. 92 kB MIT18_06SCF11_Ses1. 103 kB infinitely many solutions x (because det(B) = 0). 6 . (c)If Ais diagonalizable, then the solution of the initial value problem is Y(t) = etAY 0: Example. The usual calculation shows that A has eigenvalues 1 and -1 with corresponding eigenvectors 1 1 and 1 1 . Background87 13. The set of eigenvalues of a matrix is sometimes called the of the matrix, and orthogonal diagonalispectrum zation of a matrix factors in aE E 5. In fact, A = PDP 1, with D a diagonal matrix, if and only if the columns of P are n linearly independent eigenvectors of A. ⋄ Example 11. 1(a) and (b). 1 motivate the following question: Can we find a diagonal matrix D that is similar to A? 3 11. pdf. The matrix D is a diagonal matrix (i. 2), so B is Diagonalization of a matrix involves finding a nonsingular matrix P and diagonal matrix D such that D = P-1AP. Problem Solving Video. We can use this to compute Ak quickly for large k. Similarity and Diagonalization 299 Example 5. Kuzucuo glu 1 and the solutions are y 1 =c 1e l1t; y 2 =c 2e l2t: (Note: This is the only step in the whole process that requires calculus. It includes: 1) Determining whether given vectors are eigenvectors of 3x3 matrices. Thespectrumof the matrix A is the set S A of its eigenvalues. 2 in the 4 th or 5 th edition. Hence B∼D by (3) of (5. As shown in the examples below, all those solutions x always constitute a vector space, which we denote as EigenSpace(λ), such that the eigenvectors of Acorresponding to λ are exactly the non-zero vectors in EigenSpace(λ). 2). We say that B= {v 1,v 2,···,v n}is an eigenbasis of a n×nmatrix Aif it is a basis of Rn and every vector v 1,,v n is an eigenvector of A. 1: (6. A matrix is diagonalizable if and only if all the roots of its characteristic equation are real and distinct. The matrix S is formed by making these eigenvectors its columns: 1This is proven in 5. Here is the list of links to the quiz problems and solutions. Problems under properties of eigen values and eigen vectors. We have A∼B. Calculations show that the eigenvalue. Problems (PDF 4 Problems: Solution Sets for Systems of Linear Equations 15 5 Problems: Vectors in Space, n-Vectors 20 6 Problems: Vector Spaces 23 7 Problems: Linear Transformations 28 8 Problems: Matrices 31 9 Problems: Properties of Matrices 37 10 Problems: Inverse Matrix 41 11 Problems: LU Decomposition 45 12 Problems: Elementary Matrices and Determinants 47 This document contains solved problems on eigenvalues, eigenvectors, and diagonalization of matrices from a linear algebra course. 3(a): For A = −4 −6 3 5 and P = −1 −2 1 1 , find the product P−1AP. Its roots are the eigenvalues of A. 3 Diagonalization ofMatrices Performance Criterion: 11. 2: (6. Problems 85 12. Exercises 84 12. Problem 22. ) 2. Jul 22, 2016 · This is the first problem of Quiz 13 (Take Home Quiz) for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017. 4 . Reducing quadratic forms to canonical forms by orthogonal transformations. 3 and 2. 6 Solutions and Answers 3. Product of two eigen values of the matrix A = [] is just in case the equation Ax = x has a non-zero solution. Recall that these solutionsare all linear com-binations of certain basic solutions determined by the gaussian algorithm (see Theorem 1. 1 2 Then describe all matrices that diagonalize A−1. Dk is trivial to compute as the following example illustrates. 9 A = . . Lecture 13: Applications of The assigned problems for this section are: Section 6. 2) Showing that a given 3x3 matrix A is diagonalizable by finding its eigenvalues, eigenvector bases, and invertible matrix P such that A = PDP-1. In fact every nonzero solutionx of (λI−A)x=0 is an eigenvector. 1 Lecture video transcript (PDF) Suggested Reading. 1. Thus, the solution is: X(t) = eAt X(0), where X(0) = 1 1 . In this case the solution x 2Rn nf0gis aneigenvector, and the pair ( ;x) is aneigenpair. The identity matrix for example has every basis of Rn as TYPE II PROBLEMS: DECIDE IF TWO MATRICES ARE CONJUGATE Two matrices are conjugate if and only if they have a common diagonalization: To see this, notice that A = XDX−1 and B = Y DY −1 is equivalent to X−1AX = D = Y −1BY , which in turn is equivalent to B = Y X−1AXY−1. It is easy to check that 4 1 1 0 3 5is a single independent 2-eigenvector of C. Diagonalization and Eigenvalues 175 Note that a square matrix A has many eigenvectors associated with any given eigenvalue λ. Work the problems on your own and check your answers when you’re done. Gauss-Jordan elimination / homogeneous system Exercises on diagonalization and powers of A Problem 22. 2 - 1, 2, 15, 16, 26 1 The Basic Formula Suppose the matrix A has n distinct eigenvalues. 1 INTRODUCTION In the preceding two Units, you have learnt how to solve the eigenvalue problems for matrices, with particular reference to special matrices like the hermitian and unitary matrices. ) Find Λ and S to diagonalize A : . 4 Summary 3. Let SR A denote the subset of itsrealeigenvalues. 1 If A is similar to B and either A or B is diagonalizable, show that the other is also diagonalizable. For a diagonalizable matrix A, it can be shown that Ak = PDkP-1 for any power k, where Dk is a problems are given to students from the books which I have followed that year. be any real numbers. By an easy calculation det(A−λI) = λ2 −8λ+15 = (λ−3)(λ−5). The matrix A= 2 4 3 3 for example has the eigenbasis B= { 1 1 , −4 3 }. Let A= 2 4 2 2 1 1 3 1 1 2 2 3 5: If Ais diagonalizable, nd a diagonal matrix Dthat is similar to A. Thus, we take P = 1 1 1 1 . Then P 1= 2 1 1 1 1 . So the matrix equation has nonzero reareal ÐE MÑ œ Þ-3 B ! l solutions In other words, there are real eigenvectors for eigenvalue -3Þ ñ We are now ready to prove our main theorem. Basic to advanced level. (e) Diagonalize a matrix; know the forms of the matrices P and D from P−1AP = D. 10sol. DIAGONALIZATION OF MATRICES87 13. First we obtain P−1 = 1 Explanation: Some of the applications of diagonalization of a matrix are: The powers of a diagonalized matrix can be computed easily since the result is nothing but the powers of the diagonal elements obtained by diagonalization. Unit 16: Diagonalization Lecture 16. Example 3. Since B∼A by (2) of (5. Solution: The vector equation is X0(t) = AX(t), with A = 0 1 1 0 . In this lecture we learn to diagonalize any matrix that has n independent eigenvectors and see how diago­ nalization simplifies calculations. Background83 12. 5. In this case there will be n corresponding linearly independent eigenvectors. e. Problems 79 11. From introductory exercise problems to linear algebra exam problems from various universities. Exercises 89 13. Decide if any two of matrices in Set I are conjugate. Worksheet 18: Diagonalization and diagonalizability Given an n nmatrix A, here’s what you need to do to diagonalize it: (1) Compute the characteristic polynomial P( ) = det(A I). EXAMPLE: Let D 50 04. EIGENVALUES AND EIGENVECTORS83 12. 2), we have B∼A and A∼D. Read Section 6. 2. Solution. 3. (2) If P( ) does not have nreal roots, counting multiplicities (in other words, if it has some complex roots), then Ais not and the solutions are y 1 =c 1e l1t; y 2 =c 2e l2t: (Note: This is the only step in the whole process that requires calculus. 3) Proving that the characteristic Below you have several step-by-step solved exercises of matrix diagonalization with which you can practice. Aof all solutions of Y0= AY is a vector space. pdf. Practice problems on matrix diagonalization Problem 1. Consider again the matrix Ain Example 1. 3 Diagonalization DiagonalizationTheoremExamples Diagonalization Theorem Theorem (Diagonalization) An n n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors. We aso have Dt = t 0 0 t . Compute D2 and D3. Diagonalize the following 2×2 dimension matrix: Diagonalization and powers of A We know how to find eigenvalues and eigenvectors. 5 Terminal Questions 3. I have kept the solutions of exercises which I solved for the students. Let SC A denote the subset of itscomplexeigenvalues, which satis es SC A = S A nSR A. 3 Diagonalization The goal here is to develop a useful factorization A PDP 1, when A is n n. Lecture 11: Eigenvalues are solutions of (A (2)I)x = 0 ) 0 0 0 1 x 1 x 2 = 0 0 ) x 2 = 0) x 2span 1 0 The eigenvectors corresponding to the eigenvalue = 3 are solutions of (A (3)I)x = 0 ) 1 0 0 0 x 1 x 2 = 0 0 ) x 1 = 0) x 2span 0 1 This property (that the eigenvalues of a diagonal matrix coincide with its diagonal entries and the eigenvec- 3. Answers to Odd-Numbered Exercises80 Part 4. Problems 91 13. The lecture concludes by using eigenvalues and eigenvectors to solve difference equations. Solve the initial value problem Y = AY where A= 3 4 3 2 and Y(0) = 6 1 . Quiz 1. A matrix is diagonalizable if it has n linearly independent eigenvectors. 2 #16. 3. yqnd xjs tcmp ehfjd wlm rdokg dgg qgyt wix otxp orvpr gjmucntf natq bplarqx hpacbev