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Day 2 Eigenvectors

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Day 2 Eigenvectors

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  1. 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 was 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.

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

  3. x1 x 2 …xn λ1 x1,λ2x2… λnxn x1 x2 …xn S-1AS =Λ Capital 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-1AS =Λ λ 1 0 … 0 0 λ2 …0 … 0,0…0 λn Eigenvalue matrix, Λ

  4. S-1AS =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 n is 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. Diagonalization problem 1 Diagonalize the matrix if possible

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

  8. Diagonalize the matrix if possible

  9. When can we diagonalize a matrix? If a matrix has n independent eigenvectorsthen 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

  10. What are the Eigenvalues and eigenvectors of A2 , A3 If Ax = λx multiply both sides of the equation by A A2x = Aλx = λAx = λ2x This tells us that the eigenvectors of A2x = λ2x. If we square A then we square the eigenvalues of a A but the eigenvectors are the same as the eigenvectors of A.

  11. We can get the same information from the formula S-1AS =Λ S-1AS = Λ AS = S Λ A = S Λ S-1 Therefore A2 = S Λ S-1 S Λ S-1 A2 = S Λ2 S-1 Note this is telling us the same information that we found before.

  12. 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.

  13. Raise matrix to a power In a previous slide we were able to diagonalize the matrix. Find A4

  14. Note: we found the eigenvalues and eigenvectors on slide 7. Ak = S Λk S-1 To Find A4 Multiply S D4 S-1 We usually do not use this equation in this form. However, this demonstrates that diagonalization tells us about powers of a matrix.

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

  16. 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.

  17. 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?

  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? 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.

  19. How does this usually get applied? uk+1 = A uk With starting condition u0 u1 = Au0, u2 = A2 u0, u3 = A3 u0, uk = Aku0 To solve Write u0 as a combination of eigenvectors. u0 = c1x1 + c2x2 + … + cnxn Multiply both sides of the equation by A (Note: the x’s are eigenvectors) Au0 = c1λ1x1 + c2 λ2x2 + … + cn λnxn Aku0 = c1λ1kx1 + c2 λ2kx2 + … + cn λnkxn

  20. 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.

  21. 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 termsEquation: y(k+2) = y(k+1) + y(k) How fast is this sequence growing? How could we approximate the 100th term of the sequence?

  22. 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

  23. Find the determinant of A - λI 1- λ 1 = λ2 – λ – 1 by the quadratic formula • - λλ1 = ½ (1 + √5) λ2 = ½ (1 - √5) λ1 ≈ 1.618 λ 2 ≈ -0.618 What do the eigenvalues tell us about the growth of the Fibonacci Sequence?

  24. Fibonacci solution λ2 – λ – 1 = 0 Find the eigenvectors: Recall: λ2 – λ – 1 = 0 for these two values

  25. Fibonacci solution To find eigenvectors Recall: λ2 – λ – 1 = 0 for these two values

  26. Fibonacci solution Use our key equation Plug in λ1 ≈ 1.618 λ 2 ≈ -0.618 And the corresponding eigenvectors for x1 and x2. Then use the initial conditions to find the c1 and c2.

  27. Fibonacci solution A100u0 = c1½ (1 + √5)100 x1 + c2 (1 -√5)100 x2 x1 and x2 are the eigenvectors by inspections the eigenvectors are Recall: λ2 – λ – 1 = 0 for these two values.

  28. Fibonacci solution Ak u0 = c1(λ1)kλ1 + c2 (λ2)k λ2 1 1 λ1 = ½ (1 + √5) λ2 = ½ (1 - √5) Now use the initial conditions to find c1 and c2 u0 = 1 u0 = c1x1 + c2x2 0 [ ] [ ] [ ] λ1c1 + λ2c2 = 1 1 c1 + 1 c2 = 0 c1 = √5/5 c2 = -√5/5

  29. 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."

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

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