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Lesson 16 The Spectral Theorem and Applications

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Lesson 16 The Spectral Theorem and Applications

  1. 1. Lesson 16 (S&H, Section 14.6) The Spectral Theorem and Applications Math 20 October 26, 2007 Announcements Welcome parents! Problem Set 6 on the website. Due October 31. OH: Mondays 1–2, Tuesdays 3–4, Wednesdays 1–3 (SC 323) Prob. Sess.: Sundays 6–7 (SC B-10), Tuesdays 1–2 (SC 116)
  2. 2. Outline Hatsumon Concept Review Eigenbusiness Diagonalization The Spectral Theorem The split case The symmetric case Iterations Applications Back to Fibonacci Markov chains
  3. 3. A famous math problem “A certain man had one pair of rabbits together in a certain enclosed place, and one wishes to know how many are created from the pair in one year when it is the nature of them in a single month to bear another pair, and in the second month those born to bear also. Because the abovewritten pair Leonardo of Pisa in the first month bore, you (1170s or 1180s–1250) will double it; there will be a/k/a Fibonacci two pairs in one month.”
  4. 4. Diagram of rabbits f (0) = 1
  5. 5. Diagram of rabbits f (0) = 1 f (1) = 1
  6. 6. Diagram of rabbits f (0) = 1 f (1) = 1 f (2) = 2
  7. 7. Diagram of rabbits f (0) = 1 f (1) = 1 f (2) = 2 f (3) = 3
  8. 8. Diagram of rabbits f (0) = 1 f (1) = 1 f (2) = 2 f (3) = 3 f (4) = 5
  9. 9. Diagram of rabbits f (0) = 1 f (1) = 1 f (2) = 2 f (3) = 3 f (4) = 5 f (5) = 8
  10. 10. An equation for the rabbits Let f (k) be the number of pairs of rabbits in month k. Each new month we have The same rabbits as last month Every pair of rabbits at least one month old producing a new pair of rabbits
  11. 11. An equation for the rabbits Let f (k) be the number of pairs of rabbits in month k. Each new month we have The same rabbits as last month Every pair of rabbits at least one month old producing a new pair of rabbits So f (k) = f (k − 1) + f (k − 2)
  12. 12. Some fibonacci numbers k f (k) 0 1 1 1 2 2 3 3 4 5 5 8 Question 6 13 Can we find an explicit formula for f (k)? 7 21 8 34 9 55 10 89 11 144 12 233
  13. 13. Outline Hatsumon Concept Review Eigenbusiness Diagonalization The Spectral Theorem The split case The symmetric case Iterations Applications Back to Fibonacci Markov chains
  14. 14. Concept Review Definition Let A be an n × n matrix. The number λ is called an eigenvalue of A if there exists a nonzero vector x ∈ Rn such that Ax = λx. (1) Every nonzero vector satisfying (??) is called an eigenvector of A associated with the eigenvalue λ.
  15. 15. Diagonalization Procedure Find the eigenvalues and eigenvectors. Arrange the eigenvectors in a matrix P and the corresponding eigenvalues in a diagonal matrix D. If you have “enough” eigenvectors (that is, one for each column of A), the original matrix is diagonalizable and equal to PDP−1 . Pitfalls: Repeated eigenvalues Nonreal eigenvalues
  16. 16. Outline Hatsumon Concept Review Eigenbusiness Diagonalization The Spectral Theorem The split case The symmetric case Iterations Applications Back to Fibonacci Markov chains
  17. 17. Question Under what conditions on A would you be able to guarantee that A is diagonalizable?
  18. 18. Theorem (Baby Spectral Theorem) Suppose An×n has n distinct real eigenvalues. Then A is diagonalizable.
  19. 19. Theorem (Spectral Theorem for Symmetric Matrices) Suppose An×n is symmetric, that is, A = A. Then A is diagonalizable. In fact, the eigenvectors can be chosen to be pairwise orthogonal with length one, which means that P−1 = P . Thus a symmetric matrix can be diagonalized as A = PDP ,
  20. 20. Powers of diagonalizable matrices Remember if A is diagonalizable then Ak = (PDP−1 )k = (PDP−1 )(PDP−1 ) · · · (PDP−1 ) k −1 −1 −1 = PD(P P)D(P P) · · · D(P P)DP−1 = PDk P−1
  21. 21. Another way to look at it If v is an eigenvector corresponding to eigenvalue λ, then Ak v =
  22. 22. Another way to look at it If v is an eigenvector corresponding to eigenvalue λ, then Ak v = λ k v
  23. 23. Another way to look at it If v is an eigenvector corresponding to eigenvalue λ, then Ak v = λ k v If v1 , . . . vn are eigenvectors with eigenvalues λ1 , . . . , λn , then Ak (c1 v1 + · · · + cn vn )
  24. 24. Another way to look at it If v is an eigenvector corresponding to eigenvalue λ, then Ak v = λ k v If v1 , . . . vn are eigenvectors with eigenvalues λ1 , . . . , λn , then Ak (c1 v1 + · · · + cn vn ) = c1 λk v1 + · · · + cn λk vn 1 n
  25. 25. Another way to look at it If v is an eigenvector corresponding to eigenvalue λ, then Ak v = λ k v If v1 , . . . vn are eigenvectors with eigenvalues λ1 , . . . , λn , then Ak (c1 v1 + · · · + cn vn ) = c1 λk v1 + · · · + cn λk vn 1 n If A is diagonalizable, there are n linearly independent eigenvectors, so any v can be written as a linear combination of them.
  26. 26. Outline Hatsumon Concept Review Eigenbusiness Diagonalization The Spectral Theorem The split case The symmetric case Iterations Applications Back to Fibonacci Markov chains
  27. 27. Setting up the Fibonacci sequence Recall the Fibonacci sequence defined by f (k + 2) = f (k) + f (k + 1), f (0) = 1, f (1) = 1
  28. 28. Setting up the Fibonacci sequence Recall the Fibonacci sequence defined by f (k + 2) = f (k) + f (k + 1), f (0) = 1, f (1) = 1 Let’s let g (k) = f (k + 1). Then g (k + 1) = f (k + 2) = f (k) + f (k + 1) = f (k) + g (k).
  29. 29. Setting up the Fibonacci sequence Recall the Fibonacci sequence defined by f (k + 2) = f (k) + f (k + 1), f (0) = 1, f (1) = 1 Let’s let g (k) = f (k + 1). Then g (k + 1) = f (k + 2) = f (k) + f (k + 1) = f (k) + g (k). f (k) So if y(k) = , we have g (k) f (k + 1) g (k) 0 1 y(k + 1) = = = y(k) g (k + 1) f (k) + g (k) 1 1
  30. 30. Setting up the Fibonacci sequence Recall the Fibonacci sequence defined by f (k + 2) = f (k) + f (k + 1), f (0) = 1, f (1) = 1 Let’s let g (k) = f (k + 1). Then g (k + 1) = f (k + 2) = f (k) + f (k + 1) = f (k) + g (k). f (k) So if y(k) = , we have g (k) f (k + 1) g (k) 0 1 y(k + 1) = = = y(k) g (k + 1) f (k) + g (k) 1 1 So if A is this matrix, then y(k) =
  31. 31. Setting up the Fibonacci sequence Recall the Fibonacci sequence defined by f (k + 2) = f (k) + f (k + 1), f (0) = 1, f (1) = 1 Let’s let g (k) = f (k + 1). Then g (k + 1) = f (k + 2) = f (k) + f (k + 1) = f (k) + g (k). f (k) So if y(k) = , we have g (k) f (k + 1) g (k) 0 1 y(k + 1) = = = y(k) g (k + 1) f (k) + g (k) 1 1 So if A is this matrix, then y(k) = Ak y(0).
  32. 32. Diagonalize 0 1 The eigenvalues of A = are found by solving 1 1 −λ 1 0= = (−λ)(1 − λ) − 1 1 1−λ = λ2 − λ − 1
  33. 33. Diagonalize 0 1 The eigenvalues of A = are found by solving 1 1 −λ 1 0= = (−λ)(1 − λ) − 1 1 1−λ = λ2 − λ − 1 The roots are √ √ 1+ 5 1− 5 ϕ= ϕ= ¯ 2 2
  34. 34. Diagonalize 0 1 The eigenvalues of A = are found by solving 1 1 −λ 1 0= = (−λ)(1 − λ) − 1 1 1−λ = λ2 − λ − 1 The roots are √ √ 1+ 5 1− 5 ϕ= ϕ= ¯ 2 2 Notice that ϕ + ϕ = 1, ¯ ϕϕ = −1 ¯ (These facts make later calculations simpler.)
  35. 35. Eigenvectors We row reduce to find the eigenvectors: −ϕ 1 −ϕ 1 −ϕ ¯ −ϕ 1 A − ϕI = = 1 1−ϕ 1 ϕ ←+ ¯ − 0 0 1 So is an eigenvector for A corresponding to the eigenvalue ϕ. ϕ
  36. 36. Eigenvectors We row reduce to find the eigenvectors: −ϕ 1 −ϕ 1 −ϕ ¯ −ϕ 1 A − ϕI = = 1 1−ϕ 1 ϕ ←+ ¯ − 0 0 1 So is an eigenvector for A corresponding to the eigenvalue ϕ. ϕ 1 Similarly, is an eigenvector for A corresponding to the ϕ¯ eigenvalue ϕ. ¯
  37. 37. Eigenvectors We row reduce to find the eigenvectors: −ϕ 1 −ϕ 1 −ϕ ¯ −ϕ 1 A − ϕI = = 1 1−ϕ 1 ϕ ←+ ¯ − 0 0 1 So is an eigenvector for A corresponding to the eigenvalue ϕ. ϕ 1 Similarly, is an eigenvector for A corresponding to the ϕ¯ eigenvalue ϕ. So now we know that ¯ 1 1 y(k) = c1 ϕk + c2 ϕk ¯ ϕ ϕ ¯
  38. 38. What are the constants? To find c1 and c2 , we solve 1 1 1 1 1 c1 = c1 + c2 = 1 ϕ ¯ ϕ ϕ ϕ ¯ c2 −1 c1 1 1 1 1 ϕ −1 ¯ =⇒ = = c2 ϕ ϕ ¯ 1 ϕ−ϕ ¯ −ϕ 1 1 ϕ−1 ¯ 1 = ϕ−ϕ ϕ+1 ¯ 1 1 ϕ =√ 5 −ϕ ¯
  39. 39. Finally Putting this all together we have ϕ 1 ϕ ¯ 1 y(k) = √ ϕk − √ ϕk¯ 5 ϕ 5 ϕ ¯ f (k) 1 ϕk+1 − ϕk+1 ¯ =√ k+2 − ϕk+2 g (k) 5 ϕ ¯
  40. 40. Finally Putting this all together we have ϕ 1 ϕ ¯ 1 y(k) = √ ϕk − √ ϕk¯ 5 ϕ 5 ϕ ¯ f (k) 1 ϕk+1 − ϕk+1 ¯ =√ k+2 − ϕk+2 g (k) 5 ϕ ¯ So √ √   k+1 k+1 1 1+ 5 1− 5 f (k) = √  −  5 2 2
  41. 41. Markov Chains Recall the setup: T is a transition matrix giving the probabilities of switching from any state to any of the other states.
  42. 42. Markov Chains Recall the setup: T is a transition matrix giving the probabilities of switching from any state to any of the other states. We seek a steady-state vector, i.e., a probability vector u such that Tu = u.
  43. 43. Markov Chains Recall the setup: T is a transition matrix giving the probabilities of switching from any state to any of the other states. We seek a steady-state vector, i.e., a probability vector u such that Tu = u. This is nothing more than an eigenvector of eigenvalue 1!
  44. 44. Theorem If T is a regular doubly-stochastic matrix, then 1 is an eigenvalue for T all other eigenvalues of T have absolute value less than 1.
  45. 45. Let u be an eigenvector of eigenvalue 1, scaled so it’s a probability vector. Let v2 , . . . , vn be eigenvectors corresponding to the other eigenvalues λ2 , . . . , λn . Then for any initial state x(0), we have x(k) = Ak x(0) = Ak (c1 u + c2 λ2 v2 + · · · + cn λn vn ) = c1 u + c2 λk v2 + · · · + cn λk vn 2 n So x(k) → c1 u Since each x(k) is a probability vector, c1 = 1. Hence x(k) → c1 u

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