1. Day 2 Eigenvectors neither stretched nor compressed, its eigenvalue is 1. All vectors with the same vertical direction—i.e., parallel to this vector—are also eigenvectors, with the same eigenvalue. Together with the zero-vector, they form the eigenspace for this eigenvalue. In this shear mapping of the Mona Lisa, the picture was deformed in such a way that its central vertical axis (red vector) has not changed direction, but the diagonal vector (blue) has changed direction. Hence the red vector is an eigenvector of the transformation and the blue vector is not. Since the red vector wasshear mapping Mona Lisa

2. Similar Matrices Two square matrices and A and B that are related bysquare matrices B = X -1 AX where X is a square nonsingular matrix are said to be similar. A transformation of the form X - 1 AX is called a similarity transformation, or conjugation by X.nonsingular matrixsimilarity transformation http://mathworld.wolfram.com/SimilarMatrices.html www.math.uiuc.edu/~franklan/Math416_SimilarMatrices.pdf

3. S -1 AS =Λ Suppose an n x n matrix A has n independent eigenvectors. Put those vectors in the column of a matrix called S (called the eigenvector matrix) If you multiply AS = A = = Hence: AS = SΛ or S -1 AS =Λ x 1 x 2 …x n λ 1 x 1, λ 2 x 2… λ n x n x 1 x 2 …x n λ 1 0 … 0 0 λ 2 …0 … 0,0…0 λ n Eigenvalue matrix, Λ Capital λ

4. S -1 AS =Λ Recall from our discussion of similar matrices that A and Λ are similar matrices. They represent the same transformation with regards to a different basis. We will use this to find out about powers (and later exponentials) of a matrix.

5. Diagonal Matrix D An n x nis similar to diagonal matrix if & only if A has n linearly independent eigenvectors. The elements on the diagonals are the eigenvalues of A. Theorem 7.4

6. Diagonalize the matrix if possible Diagonalization problem 1

7. Diagonalize problem 1 solution Note: A= SΛS -1

8. Diagonalize the matrix if possible

10. When can we diagonalize a matrix? If a matrix has n independent eigenvectors then we can diagonalize the matrix. (Th. 7.4) We can do this in all times when there are n different eigenvalues. (Th. 7.5) If there are repeated eigenvalues we may or may not find n independent eigenvectors

11. What are the Eigenvalues and eigenvectors of A 2, A 3 If Ax = λx multiply both sides of the equation by A A 2 x = Aλx = λAx = λ 2 x This tells us that the eigenvectors of A 2 x = λ 2 x. If we square A then we square the eigenvalues of a A but the eigenvectors are the same as the eigenvectors of A.

12. We can get the same information from the formula S -1 AS =Λ S -1 AS = Λ AS = S Λ A = S Λ S -1 Therefore A 2 = S Λ S -1 S Λ S -1 A 2 = S Λ 2 S -1 Note this is telling us the same information that we found before.

13. Raise matrix to a power Applications For use in transition matrices from one state to another. Waiting time / queuing theory. To study long term behavior. Markov matrices.

14. Raise matrix to a power In a previous slide we were able to diagonalize the matrix. Find A 4

15. A k = S Λ k S -1 To Find A 4 Multiply S D 4 S -1 We usually do not use this equation in this form. However, this demonstrates that diagonalization tells us about powers of a matrix. Note: we found the eigenvalues and eigenvectors on slide 7.

16. What Eigenvectors and Eigenvalues tell us about the powers of a matrix When do the powers of a matrix go to zero A k → 0 What would be true about A? Recall: A = S Λ S -1 A k = S Λ k S -1 A matrix that goes to zero when raised to the k power as k → ∞ is called stable.

17. A system that approaches zero as t → ∞ is said to be stable. A difference equation is stable if: The absolute value of each eigenvalue is less than one. Recall: The eigenvalues may be complex. Note: This approach only works if there are n independent eigenvectors.

18. A system that approaches a (non zero) constant as t→∞ is called a steady state What are the eigenvalues for a matrix that describes a transformation that will become a steady state regardless of the initial conditions? What assumptions are built into this?

19. A system that approaches a (non zero) constant as t→∞ is called a steady state What are the eigenvalues for a matrix that describes a transformation that will become a steady state regardless of the initial conditions? One or more eigenvalues that are 1 and the others have an absolute value of less than 1. What assumptions are built into this? n independent eigenvectors.

20. How does this usually get applied? u k+1 = A u k With starting condition u 0 u 1 = Au 0, u 2 = A 2 u 0, u 3 = A 3 u 0, u k = A k u 0 To solve Write u 0 as a combination of eigenvectors. u 0 = c 1 x 1 + c 2 x 2 + … + c n x n Multiply both sides of the equation by A (Note: the x’s are eigenvectors) Au 0 = c 1 λ 1 x 1 + c 2 λ 2 x 2 + … + c n λ n x n A k u 0 = c 1 λ 1 k x 1 + c 2 λ 2 k x 2 + … + c n λ n k x n

21. Key Formula for Difference Equations Note: this formula is based on having n independent eigenvectors. If that is not true then this approach can not be used.

22. The Fibonacci Sequence The sequence 0, 1, 1, 2, 3, 5, 8, … is called the Fibonacci Sequence. Each term is found by adding the two previous terms Equation: y(k+2) = y(k+1) + y(k) How fast is this sequence growing? How could we approximate the 100 th term of the sequence?

23. We need to change this into matrix language Equation: y(n+2) = y(n+1) + y(n) Trick create another equation y(n+1) = y(n+1) This system can be expressed at the matrix system

24. Find the determinant of A - λI 1- λ 1 = λ 2 – λ – 1 by the quadratic formula 1- λ λ 1 = ½ (1 + √5) λ 2 = ½ (1 - √5) λ 1 ≈ 1.618 λ 2 ≈ -0.618 What do the eigenvalues tell us about the growth of the Fibonacci Sequence?

25. Fibonacci solution λ 2 – λ – 1 = 0 Find the eigenvectors: Recall: λ 2 – λ – 1 = 0 for these two values

26. Fibonacci solution To find eigenvectors Recall: λ 2 – λ – 1 = 0 for these two values

27. Fibonacci solution Use our key equation Plug in λ 1 ≈ 1.618 λ 2 ≈ -0.618 And the corresponding eigenvectors for x 1 and x 2. Then use the initial conditions to find the c 1 and c 2.

28. Fibonacci solution A 100 u 0 = c 1 ½ (1 + √5) 100 x 1 + c 2 (1 -√5) 100 x 2 x 1 and x 2 are the eigenvectors by inspections the eigenvectors are Recall: λ 2 – λ – 1 = 0 for these two values.

29. Fibonacci solution A k u 0 = c 1 (λ 1 ) k λ 1 + c 2 (λ 2 ) k λ 2 1 1 λ 1 = ½ (1 + √5) λ 2 = ½ (1 - √5) Now use the initial conditions to find c 1 and c 2 u 0 = 1 u 0 = c 1 x 1 + c 2 x 2 0 [ ] λ 1 c 1 + λ 2 c 2 = 1 1 c 1 + 1 c 2 = 0 c 1 = √5/5 c 2 = -√5/5

30. Homework wkst 7.2 p. 461 6-11 all Customer: "How much is a large order of Fibonachos?" Cashier: "It's the price of a small order plus the price of a medium order."

31. More info www.math.hawaii.edu/~pavel/fibonacci.pdf‎