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Section 5.1 Eigenvectors and Eigenvalues Eigenvectors and Eigenvalues •Useful throughout pure and applied mathematics. •Used to study difference equations and continuous dynamical systems. •Provide critical information in engineering design •Arise naturally in such fields as physics and chemistry. •Used in statistics to analyze multicollinearity Example: 3 3 •Let A 1 5 •Consider 3 v1 1 Av for some v 1 v2 1 1 v3 1 Definition: •The Eigenvector of Anxn is a nonzero vector x such that Ax=λx for some scalar λ. •λ is called an Eigenvalue of A Statistics (multicollinearity) yi 0 1 x1 2 x2 ....... i xi i •Where y is the dependent response vector and the x’s are the independent explanatory vectors •The β’s are least squares regression coefficients •εi are errors •We desire linear independence between x vectors •Can use Eigen analysis to determine From Definition: •Ax = λx = λIx •Ax – λIx = 0 •(A – λI)x = 0 •Observations: 1. 2. 3. λ is an eigenvalue of A iff (A – λI)x= 0 has non-trivial solutions A – λI is not invertible IMT all false The set {xεRn: (A – λI)x= 0} is the nullspace of (A – λI)x= 0, A a subspace of Rn The set of all solutions is called the eigenspace of A corresponding to λ Example Show that 2 is an eigenvalue of 3 3 A 1 5 and find the corresponding eigenvectors. Comments: Warning:The method just used (row reduction) to find eigenvectors cannot be used to find eigenvalues. Note: The set of all solutions to (A-λI)x =0 is called the eigenspace of A corresponding to λ. Example Let 1 2 2 A 3 2 1 0 1 1 An eigenvalue of A is λ=3. Find a basis for the corresponding eigenspace. Theorem The eigenvalues of a triangular matrix are the entries on its main diagonal. Proof of 3x3 case a11 Let A 0 0 a12 a22 0 a13 a23 a33 a11 So (A-λI) = A 0 0 a12 a22 0 a11 0 0 a13 0 0 a23 0 0 a33 0 0 a12 a22 0 a13 a23 a33 Proof of 3x3 case By definition λ is an eigenvalue iff (A-λI)x=0 has non-trivial solutions so a free variable must exist. a11 0 0 a12 a22 0 a13 a23 a33 This occurs when a11=λ or a22=λ or a33=λ Example Consider the lower triangular matrix below. 4 0 0 A 0 0 0 1 0 3 λ= 4 or 0 or -3 Addtion to IMT Anxn is invertible iff s. The number 0 is not an eigenvalue t. det A≠0 (not sure why author waits until not to add this) Theorem If eigenvectors have distinct eigenvalues then the eigenvectors are linearly independent This can be proven by the IMT Section 5.2 The Characteristic Equation Finding Eigenvalues 1. We know (A-λI)x=0 must have nontrivial solutions and x is non-zero. That is free variables exist. 2. So (A-λI) is not invertible by the IMT 3. Therefore det(A-λI)=0 by IMT Characteristic Equation det(A-λI)=0 Solve to find eigenvalues Note: det(A-λI) is the characteristic polynomial. Previous Example: •Let 3 3 A 1 5 find eigenvalues Example Find Eigenvalues 1 2 1 A 0 5 0 1 8 1 Example Find characteristic polynomial and eigenvalues 3 2 3 A 0 6 10 0 0 2 Example a. b. c. Find the characteristic polynomial Find all eigenvalues Find multiplicity of each eigenvalue 2 5 A 9 1 0 3 1 2 0 0 0 0 3 0 5 1 Recap a. b. c. λ is an eigenvalue of A if (A-λI)x=0 has non-trivial solutions (free variables exist). Eigenvectors (eigenspace )are found by row reducing (A-λI)x=0. Eigenvalues are found by solving det(A-λI)=0. Section 5.3 Diagonalization Diagonalization •The goal here is to develop a useful factorization A=PDP-1, when A is nxn. •We can use this to compute Ak quickly for large k. •The matrix D is a diagonal matrix (i.e. entries off the main diagonal are all zeros). Example Find a formula for Ak given A=PDP-1 & 6 1 A 2 3 1 1 P 1 2 5 0 D 0 4 2 1 P 1 1 1 Diagonalizable Matrix “A” is diagonalizable if A=PDP-1 where P is invertible and D is a diagonal matrix. Note: AP=PD When is a matrix diagonalizable? Let’s examine eigenvalues and eigenvectors of A 6 1 A 2 3 The Diagonalization Theorem If Anxn & has n linearly independent eigenvectors. Then 1. A=PDP-1 2. Columns of P are eigenvectors 3. Diagonals of D are eigenvalues. Example Diagonalize 2 0 0 A 1 2 1 1 0 1 We need to find P & D Theorem If Anxn has n distinct eigenvalues then A is diagonalizable. Example Diagonalize 2 4 6 A 0 2 2 0 0 4 Example Diagonalize 2 0 0 A 2 6 0 3 2 1