Math 22 – Linear Algebra and its applications - Lecture 25 - Instructor: Bjoern Muetzel GENERAL INFORMATION ▪ Office hours: Tu 1-3 pm, Th, Sun 2-4 pm in KH 229 Tutorial: Tu, Th, Sun 7-9 pm in KH 105 ▪ Homework 8: due Wednesday at 4 pm outside KH 008. There is only Section B,C and D. 5 Eigenvalues and Eigenvectors 5.1 EIGENVECTORS AND EIGENVALUES Summary: Given a linear transformation 푇: ℝ푛 → ℝ푛, then there is always a good basis on which the transformation has a very simple form. To find this basis we have to find the eigenvalues of T. GEOMETRIC INTERPRETATION 5 −3 1 1 Example: Let 퐴 = and let 푢 = 푥 = and 푣 = . −6 2 0 2 −1 1.) Find A푣 and Au. Draw a picture of 푣 and A푣 and 푢 and A푢. 2.) Find A(3푢 +2푣) and 퐴2 (3푢 +2푣). Hint: Use part 1.) EIGENVECTORS AND EIGENVALUES ▪ Definition: An eigenvector of an 푛 × 푛 matrix A is a nonzero vector x such that 퐴푥 = 휆푥 for some scalar λ in ℝ. In this case λ is called an eigenvalue and the solution x≠ ퟎ is called an eigenvector corresponding to λ. ▪ Definition: Let A be an 푛 × 푛 matrix. The set of solutions 푛 Eig(A, λ) = {x in ℝ , such that (퐴 − 휆퐼푛)x = 0} is called the eigenspace Eig(A, λ) of A corresponding to λ. It is the null space of the matrix 퐴 − 휆퐼푛: Eig(A, λ) = Nul(퐴 − 휆퐼푛) Slide 5.1- 7 EIGENVECTORS AND EIGENVALUES 16 Example: Show that 휆 =7 is an eigenvalue of matrix A = 52 and find the corresponding eigenspace Eig(A,7). EIGENVECTORS AND EIGENVALUES 4− 1 6 ▪ Example: Let A = 2 1 6 . An eigenvalue of A is 휆 =2. 2− 1 8 Find a basis for the corresponding eigenspace Eig(A,2). EIGENVECTORS AND EIGENVALUES 1 −3 ▪ The eigenspace Eig(A,2) = Span{ 2 , 0 } is a subspace of ℝ3. 0 1 Slide 5.1- 11 THEOREMS ABOUT EIGENVALUES ▪ Theorem 1: The eigenvalues of a triangular matrix are the entries on its main diagonal. ▪ Warning: We can not find the eigenvalues of a matrix A by row reducing to echelon form U. As A and U have usually different eigenvalues. ▪ Proof of Theorem 1: THEOREMS ABOUT EIGENVALUES THEOREMS ABOUT EIGENVALUES ▪ Theorem 2: If v1, …, vr are eigenvectors that correspond to distinct eigenvalues λ1, …, λr of an 푛 × 푛 matrix A, then the set {v1, …, vr} is linearly independent. ▪ Proof: Suppose S = {v1, …, vr} is linearly dependent. ▪ Then there is a subset of S={v1, …, vr} , say {v1, …, vp} that is a basis for Span(S) and a vector, say vp+1 that is a linear combination of these vectors. ▪ Then there exist scalars c1, …, cp such that 푐1푣1 + ⋯ + 푐푝푣푝 = 푣푝+1 (1) ▪ Multiplying both sides of (1) by A and using the fact that 퐴푣푘 = 휆푘푣푘 for each k, we obtain by the linearity of A or (2) ▪ Multiplying both sides of (1) by 휆푝+1 and substituting the result in the right hand side of (2), we obtain or = 0. (3) ▪ Since {v1, …, vp} is linearly independent, the weights in (3) are all zero. But none of the factors 휆푖 − 휆푝+1 are zero, because the eigenvalues are distinct. Hence 푐푖 = 0 for 푖 = 1, … , 푝. ▪ But then (1) states that 푣푝+1 = 0, which is impossible. Slide 5.1- 16 EIGENVECTORS AND DIFFERENCE EQUATIONS Application to a recursive sequence in ℝ푛 Let A be an 푛 × 푛 matrix and consider the recursive sequence {xk} in 푛 푛 ℝ given by 푥0 = 푢 in ℝ and 푥푘+1 = 퐴푥푘 for 푘 = 0,1,2,3,… , Definition: We call a solution of this equation an explicit description of {xk} whose formula for each xk does not depend directly on A or on the preceding terms in the sequence other than the initial term 풙ퟎ = 풖 . Note: It follows that 푘 푘 푥푘+1 = 퐴 푥0 = 퐴 푢, However, this is not explicit enough to be a solution. Proof of the Note: EIGENVECTORS AND DIFFERENCE EQUATIONS ▪ Example: Let A be an 푛 × 푛 matrix such that 1 A푏 = 2푏 and A푏 = 푏 where 푏 , 푏 ≠ 0. 1 1 2 3 2 1 2 2 2 1.) Calculate 퐴 푏1 and 퐴 푏2. 푘 푘 2.) Calculate 퐴 푏1 and 퐴 푏2 and describe geometrically what 푘 푘 happens to 퐴 푏1 and 퐴 푏2. 푘 3.) Find a formula for 퐴 (4푏1+5푏2). .