MTH 464: Computational Linear Algebra

MTH 464: Computational Linear Algebra Lecture Outlines Exam 4 Material Prof. M. Beauregard Department of Mathematics & Statistics Stephen F. Austin State University April 15, 2018 Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 1 / 40 Contents I 1 Lecture 1 2 Lecture 2 3 Lecture 3 4 Lecture 4 5 Lecture 5 6 Review Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 2 / 40

Lecture 1 Overview Goal for today: Learn to diagonalize a matrix [section 5.3] Outline: 1 Diagonalization: What and Why? 2 Diagonalization: How? 3 Diagonalization: When? Assignment (5.3): Read: section 5.5 (skip 5.4) Work: section 5.3 (p. 288) #1, 5, 7, 11, 14, 17, 21, 22, 23, 25 Extra practice: #3, 9, 13, 15, 19, 27 Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 3 / 40 Lecture 1 Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 4 / 40

Lecture 1 Diagonalization: What and Why? What is diagonalization? Represent eigenvalues/vectors with a matrix factorization: A = PDP 1 D = P 1 AP where D records eigenvalues and P records eigenvectors of A Why diagonalize a matrix? Computing powers of matrices Decouple a discrete dynamical system Decouple a system of differential equations Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 5 / 40 Lecture 1 Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 6 / 40

Diagonalization: How? Lecture 1 Method (page 285-286): To diagonalize an n n matrix A 1 Find the eigenvalues λ 1,..., λ n of A 2 Find corresponding eigenvectors v 1,..., v n [ v1... v 3 Construct P from the eigenvectors: P = n ] 4 Construct a diagonal matrix D from the eigenvalues: λ 1... 0 D =..... 0... λ n 5 If P is invertible (the eigenvectors are linearly independent) then A = PDP 1 D = P 1 AP 6 To check the calculation, verify that AP = PD and P is invertible. Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 7 / 40 Lecture 1 Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 8 / 40

Diagonalization: When? Theorem 5 (page 284) Lecture 1 An n n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors. In more detail: A = PDP 1 with D diagonal if and only if the columns of P are n linearly independent eigenvectors of A; then the diagonal entries of D are the corresponding eigenvalues of A. Theorem 6 (page 286) If A is n n and has n distinct eigenvalues, then A is diagonalizable. Warning: The inverse of this statement is not true (in general): If a matrix has a repeated eigenvalue, it may still be diagonalizable (to find out whether it is, you must check the eigenvectors). Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 9 / 40 Lecture 1 Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 10 / 40

Lecture 1 Examples Determine if the following matrices are diagonalizable. If so, determine D and P. 5 8 1 A = 0 0 7 0 0 2 [ ] 5 1 A = 5 1 [ ] 3 2 A = 0 3 Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 11 / 40 Lecture 1 Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 12 / 40

Lecture 1 Diagonalization: More Precise Characterization Theorem 7 (page 287 slightly modified) Let A be an n n matrix A whose distinct eigenvalues are λ 1,..., λ p with p n. For each k let: m(λ k ) denote the algebraic multiplicity of λ k d(λ k ) denote the dimension of the eigenspace for λ k. (This is called the geometric multiplicity of eigenvalue λ k ) Then: a. For 1 k p, d(λ k ) m(λ k ) b. A is diagonalizable iff for 1 k p, d(λ k ) = m(λ k ) (and the characteristic polynomial factors completely into linear factors) p c. A is diagonalizable iff d(λ k ) = n k=1 d. If A is diagonalizable and for each k, B k is a basis for the eigenspace for λ k, then the total collection of vectors in the sets B 1,..., B p forms an eigenvector basis for R n. Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 13 / 40 Lecture 1 Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 14 / 40

Lecture 2 Overview Goal for today: Outline: Learn to find and interpret complex eigenvalues/eigenvectors [section 5.5] 1 Diagonalization (review) 2 Complex Eigenvalues and Eigenvectors 3 Interpretation of Complex Eigenvalues and Eigenvectors Assignment (5.5): Read: section 5.6 Work: section 5.5 (p. 302) #3, 5, 7, 10, 13, 16, 21 Extra practice: #1, 9, 11, 15, 17 Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 15 / 40 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 16 / 40

Lecture 2 Diagonalization (review) To diagonalize an n n matrix A: Find eigenvalues λ 1,..., λ n and eigenvectors v 1,..., v n If eigenvectors are linearly independent then [ ] λ 1... 0 A = PDP 1 v1... v where P = n, D =..... 0... λ n Theorem 7 Let A be a real n n matrix whose distinct eigenvalues are λ 1,..., λ p. For each eigenvalue λ k : dim(eigenspace for λ k ) multiplicity of λ k and A is diagonalizable if and only if equality holds for each k = 1,..., p. [See text for other parts of this theorem.] Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 17 / 40 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 18 / 40

Lecture 2 Observations Suppose a matrix A is diagonalizable. A is similar to a diagonal matrix D of its eigenvalues. Remember that A can be viewed as a linear transformation. The diagonal matrix D is also a linear transformation matrix. If λ i are real then it describes dilation in the v i eigenvector direction. The eigenvector matrix P is a basis for which the linear transformation A can be described in. Question: What happens if the eigenvalues are complex? Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 19 / 40 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 20 / 40

Lecture 2 Example Consider a mass-spring (m = k = 1) with no damping. Let the initial displacement be y(0) = 1 and velocity be zero. This is modeled by, y (t) + y(t) = 0. The solution is y(t) = cos(t). Hence the displacement oscillates. Notice the differential equation can be modeled by a linear system, [ ] [ ] [ ] d y(t) 0 1 y(t) = dt v(t) 1 0 v(t) Compactly, v = A v. Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 21 / 40 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 22 / 40

Example - continued Lecture 2 What type of transformation is A? [ 0 1 A e 1 = 1 0 [ 0 1 A e 2 = 1 0 ] e 1 = e 2 ] e 2 = e 1 So the matrix A describes a clockwise rotation by π/2. What are the eigenvalues? λ = ±i Using A = PDP 1 the solution in matrix form is [ ] e it 0 v = P 0 e it P 1 v(0) So complex eigenvalues are related to rotation. Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 23 / 40 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 24 / 40

Lecture 2 Complex Eigenvalues and Eigenvectors Methods: Same as real case Fact: If matrix is real, then eigenvalues/eigenvectors are conjugate pairs: Av = λv A v = λ v where the bar denotes the complex conjugate Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 25 / 40 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 26 / 40

Lecture 2 Interpretation: Simple Case The matrix [ ] a b C = b a (where a and b are real numbers) has eigenvalues λ = a ± bi. Set r = λ = a 2 + b 2 and φ = arg(λ) so that a = r cos(φ) and b = r sin(φ) Then [ ] cos(φ) sin(φ) C = r sin(φ) cos(φ) Interpretation: C represents rotation by φ and scaling by r Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 27 / 40 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 28 / 40

Lecture 2 Simple Example Example 8 Let A = [ 1 2 2 1 A = ]. This can diagonalized as [ ] [ ] [ ] i i 1 + 2i 0 i/2 1/2 1 1 0 1 2i i/2 1/2 This does not help provide a geometric interpretation of the matrix mapping between R 2 and R 2. The eigenvalue does. It indicates the matrix rotates vectors contour clockwise by arctan(2) 63.43 and stretches by 5. Question: What if the matrix A has complex eigenvalues but is not in this special form? How can we interpret the matrix transformation? Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 29 / 40 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 30 / 40

Lecture 2 Interpretation: General Case Recall diagonalization of an n n matrix A: [ A = PDP 1 v1... v where P = n ] λ 1... 0, D =..... 0... λ n Interpretation: P represents change of basis, D represents scaling Theorem 9 (page 299) Let A be a real 2 2 matrix with a complex eigenvalue λ = a bi (b 0) and an associated eigenvector v in C 2. Then A may be factored as [ ] [ ] A = PCP 1 Re v Im v a b where P =, C = b a Interpretation: P represents change of basis, C rotation and scaling. Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 31 / 40 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 32 / 40

Lecture 2 Example 10 Let A = [ 5 5 1 1 ]. This has eigenvalues and vectors of [ 5 λ = 3 ± i, v + = 2 i [ ] 5 0 2 1 A = PCP 1 = [ 3 1 1 3 ] [ 5, v = ] 2 + i [ 1/5 0 2/5 1 This provides a geometric interpretation. Let x R 2. Then A x: [ ] [ ] 1 5 0 Maps x into the basis and 2 1 2 Rotates the vector by arctan(1/3) 18.43 3 Stretches it by 10 4 Maps transformed vector back into the standard (euclidean) basis. ] ] Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 33 / 40 Lecture 2 Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 34 / 40

Lecture 3 Overview Goal for today: Learn to solve discrete dynamical systems [section 5.6] Outline: 1 Complex Eigenvalues and Eigenvectors (review) 2 Discrete Dynamical Systems 3 Solutions from Eigenvalues and Eigenvectors 4 Behavior of Solutions Assignment (5.6): Read: section 5.7 Work: section 5.6 (p. 311) #1, 3, 7, 11, 14, 15 Extra practice: #5, 9, 13 Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 35 / 40 Lecture 3 Complex Eigenvalues and Eigenvectors (review) Simple case: This matrix has eigenvalues λ = a ± bi: [ ] [ ] a b cos(φ) sin(φ) C = = r b a sin(φ) cos(φ) where r = λ = a 2 + b 2 and φ = arg(λ) [a = r cos(φ), b = r sin(φ)] Interpretation: C represents rotation by φ and scaling by r General case (Theorem 9): If A is a real 2 2 matrix with a complex eigenvalue λ = a bi (b 0) and an associated eigenvector v in C 2, then A may be factored as [ ] [ ] A = PCP 1 Re v Im v a b where P =, C = b a Interpretation: P represents change of basis, C rotation and scaling Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 36 / 40

Lecture 3 Discrete Dynamical Systems Formulation: Given an n n matrix A and a vector x 0 in R n, solve x k+1 = Ax k, k = 0, 1, 2,... (also known as a difference equation) This is the simplest such problem: Homogeneous (no forcing term) Constant coefficients (A does not depend on k) Linear (so solutions may be superimposed) Solution: x k = A k x 0, k = 0, 1, 2,... (but this provides no insight into the long-term behavior) Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 37 / 40 Lecture 3 Solving a Discrete Dynamical System Method: To solve x k+1 = Ax k (A is n n) 1 Find the eigenvalues λ 1,..., λ n of A 2 Find corresponding eigenvectors v 1,..., v n 3 The general solution has the form x k = c 1 λ k 1 v 1 + + c n λ k nv n 4 Solve for c 1,..., c n from the initial vector x 0 : x 0 = c 1 v 1 + + c n v n Note: This works when A has n linearly independent eigenvectors (i.e., when A is diagonalizable). If that is not the case, then the solution involves generalized eigenvectors. Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 38 / 40

Lecture 3 Discrete Dynamical Systems: Behavior of Solutions As k, λ k = λ k { 0, if λ < 1, if λ > 1 For the 2 2 problem with general solution x k = c 1 λ k 1 v 1 + c 2 λ k 2 v 2 λ 1 < 1, λ 2 < 1: origin is an attractor: x k 0 as k λ 1 > 1, λ 2 > 1: origin is a repeller: x k as k λ 1 < 1, λ 2 > 1 (or vice-versa): origin is a saddle point Remark: Complex eigenvalues result in rotation (stable/unstable spirals or centers) Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 39 / 40 Overview Lecture 4 Goal for today: Learn to solve continuous dynamical systems (ODEs) [section 5.7] Outline: 1 Discrete Dynamical Systems: Phase Plane 2 ODE Systems 3 Solving ODE Systems 4 Continuous Dynamical Systems: Phase Plane Assignment (5.7): Read: section 5.7 (again) Work: section 5.7 (p. 319) #1, 2, 3, 5 Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 40 / 40

Lecture 4 Discrete Dynamical Systems: Phase Plane For the discrete dynamical system x k+1 = Ax k when A is 2 2: General solution in terms of eigenvalues and eigenvectors: x k = c 1 λ k 1 v 1 + c 2 λ k 2 v 2 where c 1 and c 2 are determined by the initial condition x 0 Long-term behavior (as k ) depends on eigenvalues To visualize, plot the trajectory x 0, x 1, x 2,... in the phase plane Classification of the fixed point at the origin: λ 1 < 1, λ 2 < 1: origin is an attractor: x k 0 as k λ 1 > 1, λ 2 > 1: origin is a repeller: x k as k λ 1 < 1, λ 2 > 1 (or vice-versa): origin is a saddle point Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 41 / 40 A remark Lecture 4 Not just a remark buta Huge Remark: The long-term behavior of discrete dynamical system is completely determined by the absolute values of the eigenvalues and NOT the sign of the real part of the eigenvalues. Don t be confused by your differential equations background. Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 42 / 40

Lecture 4 Continuous Dynamical Systems (ODEs) Variables: x 1 (t),..., x n (t) Equations (ordinary differential equations): x 1 = a 11x 1 + + a 1n x n. x n = a n1 x 1 + + a nn x n This is the simplest form: homogeneous, constant coefficients, linear Matrix form: x (t) = Ax(t), x(0) = x 0 where: x 1 (t) x 1 x(t) =, x (t) (t) =, A =.. x n (t). x n(t) a 11 a n1... a 1n.... a nn Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 43 / 40 Solving ODE Systems Lecture 4 Method: To solve x (t) = Ax(t), x(0) = x 0 1 Find the eigenvalues λ 1,..., λ n of A (where A is n n) 2 Find corresponding eigenvectors v 1,..., v n 3 The general solution has the form x(t) = c 1 e λ 1t v 1 + + c n e λ nt v n 4 Solve for c 1,..., c n from the initial vector x 0 : x 0 = c 1 v 1 + + c n v n Note: This works when A has n linearly independent eigenvectors (i.e., when A is diagonalizable). If that is not the case, then the solution involves generalized eigenvectors. Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 44 / 40

Lecture 4 Continuous Dynamical Systems (ODEs): Phase Plane For the ODE system x (t) = Ax(t) when A is 2 2: General solution in terms of eigenvalues and eigenvectors: x(t) = c 1 e λ1t v 1 + c 2 e λ2t v 2 where c 1 and c 2 are determined by the initial condition x(0) = x 0 Long-term behavior (as k ) depends on eigenvalues To visualize, plot the trajectory x(t) as a curve in the phase plane Classification of the fixed point at the origin (when λ 1, λ 2 are real): λ 1 < 0, λ 2 < 0: origin is an attractor: x(t) 0 as t λ 1 > 0, λ 2 > 0: origin is a repeller: x(t) as t λ 1 < 0, λ 2 > 0 (or vice-versa): origin is a saddle point Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 45 / 40 Example Lecture 4 [ ] 2 3 Consider x (t) = x(t). Classify the fixed point at the origin. 1 2 What is the direction of the greatest repulsion/attraction? [ ] [ ] x(t) = c 1 e t 3 + c 1 2 e t 1 1 [ ] 3 λ 1 = 1 repels orbits away from the origin in the direction of v 1 = 1 [ ] 1 λ 1 = 1 attracts orbits to the origin in the direction of v 2 = 1 [ ] 1 Given x(0) =. How do we determine c 2 1 and c 2? c 1 = 3/4 and c 2 = 5/4. Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 46 / 40

Lecture 5 Overview Goal for today: Outline: Learn more about continuous dynamical systems (ODEs) [section 5.7] 1 Phase Plane 2 Complex Eigenvalues 3 Review of Dynamical Systems: Discrete vs. Continuous Assignment: Read: review chapters 3 and 5 Work: section 5.7 (p. 319) #7, 9, 11, 13 Review: Supp. Exercise #1 for Chapters 3 and 5 for next class Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 47 / 40 Lecture 5 Continuous Dynamical Systems (ODEs): Phase Plane For the ODE system x (t) = Ax(t) when A is 2 2: General solution in terms of eigenvalues and eigenvectors: x(t) = c 1 e λ 1t v 1 + c 2 e λ 2t v 2 where c 1 and c 2 are determined by the initial condition x(0) = x 0 Long-term behavior (as k ) depends on eigenvalues To visualize, plot the trajectory x(t) as a curve in the phase plane Classification of the fixed point at the origin (when λ 1, λ 2 are real): λ 1 < 0, λ 2 < 0: origin is an attractor: x(t) 0 as t λ 1 > 0, λ 2 > 0: origin is a repeller: x(t) as t λ 1 < 0, λ 2 > 0 (or vice-versa): origin is a saddle point Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 48 / 40

Lecture 5 ODE Systems with Complex Eigenvalues For complex eigenvalues λ = a ± ib (where a and b are real): e λt = e (a±ib)t = e at e ±ibt = e at [cos(bt) ± i sin(bt)] Real part a = Re(λ) determines amplitude: a > 0: amplitude grows (trajectory spirals out from origin) a < 0: amplitude decays (trajectory spirals in toward origin) a = 0: amplitude constant (trajectory is an ellipse) Imaginary part b = Im(λ) determines frequency of oscillation Solutions corresponding to eigenvalue λ = a + bi and eigenvector v: Complex form: Real form (see box, page 316): x(t) = c 1 x 1 (t) + c 2 x 1 (t) = c 1 e λt v + c 2 e λt v y 1 (t) = Re x 1 (t) = e at [(Re v) cos(bt) (Im v) sin(bt)] y 2 (t) = Im x 1 (t) = e at [(Re v) sin(bt) + (Im v) cos(bt)] Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 49 / 40 Example Lecture 5 Consider x (t) = [ ] 3 2 x(t). Determine the solution x(t). 1 1 x(t) = c 1 e ( 2+i)t [ = e 2t ( c 1 e it [ 2 1 + i 2 1 + i ] [ ] + c 2 e ( 2 i)t 2 1 i ] [ ]) + c 2 e it 2 1 i In real form, x(t) = e 2t ( a 1 [ cos(t) + sin(t) cos(t) ] + a 2 [ sin(t) cos(t) sin(t) ]) What happens as t? Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 50 / 40

Lecture 5 Dynamical Systems: Discrete vs. Continuous For a 2 2 matrix A with eigen(values,vectors) (λ 1, v 1 ) and (λ 2, v 2 ): Discrete Continuous Equation: x k+1 = Ax k x (t) = Ax(t) Solution: x k = c 1 λ k 1 v 1 + c 2 λ k 2 v 2 x(t) = c 1 e λ1t v 1 + c 2 e λ2t v 2 Attractor: λ 1 < 1, λ 2 < 1 Re λ 1 < 0, Re λ 2 < 0 Repeller: λ 1 > 1, λ 2 > 1 Re λ 1 > 0, Re λ 2 > 0 Saddle point: λ 1 < 1, λ 2 > 1 Re λ 1 < 0, Re λ 2 > 0 Key: size of λ sign of Re λ Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 51 / 40 Review Goal for today: Review Outline: 1 Diagonalization (review) 2 Diagonalization and Dynamical Systems 3 True/False Questions Assignment: Read: 6.1 Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 52 / 40

Review Diagonalization (review) To diagonalize an n n matrix A: Find eigenvalues λ 1,..., λ n and eigenvectors v 1,..., v n If eigenvectors are linearly independent then [ ] λ 1... 0 A = PDP 1 v1... v where P = n, D =..... 0... λ n An n n matrix A is diagonalizable: Thm 5: if and only if A has n linearly independent eigenvectors Thm 6: if A has n distinct eigenvalues (converse not true!) Thm 7: if and only if for each eigenvalue λ: dim(eigenspace for λ) = multiplicity of λ Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 53 / 40 Review Dynamical Systems (Review) For a 2 2 matrix A with eigen(values,vectors) (λ 1, v 1 ) and (λ 2, v 2 ): Discrete Continuous Equation: x k+1 = Ax k x (t) = Ax(t) Solution: x k = c 1 λ k 1 v 1 + c 2 λ k 2 v 2 x(t) = c 1 e λ1t v 1 + c 2 e λ2t v 2 Attractor: λ 1 < 1, λ 2 < 1 Re λ 1 < 0, Re λ 2 < 0 Repeller: λ 1 > 1, λ 2 > 1 Re λ 1 > 0, Re λ 2 > 0 Saddle point: λ 1 < 1, λ 2 > 1 Re λ 1 < 0, Re λ 2 > 0 Key: size of λ sign of Re λ Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 54 / 40

Review Chapter 5 Supplemental Exercise 1 (a) If A is invertible and 1 is an eigenvalue for A, then 1 is also an eigenvalue for A 1. (b) If A is row-equivalent to the identity matrix I, then A is diagonalizable. (c) If A contains a row or column of zeros, then 0 is an eigenvalue of A. (d) Each eigenvalue of A is also an eigenvalue of A 2. (e) Each eigenvector of A is also an eigenvector of A 2. (f) Each eigenvector of an invertible matrix A is also an eigenvector of A 1. Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 55 / 40 Review Chapter 5 Supplemental Exercise (g) Eigenvalues must be nonzero scalars. (h) Eigenvectors must be nonzero vectors. (i) Two eigenvectors corresponding to the same eigenvalue are always linearly dependent. (j) Similar matrices always have exactly the same eigenvalues. (k) Similar matrices always have exactly the same eigenvectors. (l) The sum of two eigenvectors of a matrix A is also an eigenvector of A. Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 56 / 40

Review Chapter 5 Supplemental Exercise (m) The eigenvalues of an upper-triangular matrix A are exactly the nonzero entries on the diagonal of A. (n) The matrices A and A T have the same eigenvalues, counting multiplicities. (o) If a 5 5 matrix A has fewer than 5 distinct eigenvalues, then A is not diagonalizable. (p) There exists a 2 2 matrix that has no eigenvectors in R 2. (q) If A is diagonalizable, then the columns of A are linearly independent. (r) A nonzero vector cannot be an eigenvector corresponding to two different eigenvalues of A. Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 57 / 40 Review Chapter 5 Supplemental Exercise (s) A square matrix A is invertible if and only if there is a coordinate system in which the transformation x Ax is represented by a diagonal matrix. (t) If each vector e j in the standard basis for R n is an eigenvector of A, then A is a diagonal matrix. (u) If A is similar to a diagonalizable matrix B, then A is also diagonalizable. (v) If A and B are invertible n n matrices, then AB is similar to BA. (w) An n n matrix with n linearly independent eigenvectors is invertible. (x) If A is an n n diagonalizable matrix, then each vector in R n can be written as a linear combination of eigenvectors of A. Linear Algebra (MTH 464) Lecture Outlines April 15, 2018 58 / 40