Upcoming SlideShare
×

# Lesson 16 The Spectral Theorem and Applications

6,782
-1

Published on

1 Like
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• @waggonerc Should be equation (1)

Are you sure you want to  Yes  No
• slide 15, your \ref{} didn't compile successfully

Are you sure you want to  Yes  No
• Oh I see, thanks a lot for the clarification!

Are you sure you want to  Yes  No
• The 'v' in the first bullet point is any eigenvector but the 'v' in the third bullet point is any vector in R^n. Sorry about the confusing reuse of the same variable. :-)

Are you sure you want to  Yes  No
• slide 32: 'If A is diagonalizable, there are n linearly independent eigenvectors, so any v can be written as a linear combination of them' From my understanding v is just any eigenvector. This means that eigenvectors are linearly independent and can be written as a linear combination of each other.

And as I understand it the definition of linearly independent vectors are that they can NOT be written as a linear combination of each other.

So somewhere I've made a mistake or the slide is wrong. Please tell me where the problem is.

Are you sure you want to  Yes  No
Views
Total Views
6,782
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
153
5
Likes
1
Embeds 0
No embeds

No notes for slide

### 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
1. #### A particular slide catching your eye?

Clipping is a handy way to collect important slides you want to go back to later.