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ENGG2013 Unit 18 The characteristic polynomial

ENGG2013 Unit 18 The characteristic polynomial. Mar, 2011. Linear Discrete-time dynamical system. Three objects are required to specifie a linear discrete-time dynamical system. State vector u (t): a vector of length n, which summarizes the status of the system at time t.

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ENGG2013 Unit 18 The characteristic polynomial

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  1. ENGG2013 Unit 18The characteristic polynomial Mar, 2011.

  2. Linear Discrete-time dynamical system Three objects are required to specifie a linear discrete-time dynamical system. • State vector u(t): a vector of length n, which summarizes the status of the system at time t. • Transitional matrix A: how to obtain the state vector u(t+1) at time t+1 from the state vector u(t) at time t. u(t+1) = Au(t). • Initial state u(0): the starting point of the system. ENGG2013

  3. Example • The unemployment rate problem in midterm • u(t) is the unemployment rate in the t-th month and e(t) is 1-u(t). ENGG2013

  4. Last time • Given a square matrix A, a non-zero vector v is called an eigenvector of A, if we an find a real number (which may be zero), such that • This number  is called the eigenvalue of A corresponding to the eigenvector v. Matrix-vector product Scalar product of a vector ENGG2013

  5. Another geometric picture • Take for example. • Define a recursion by u(t+1) = A u(t) • The initial vector is u(0) = [0 0.5]T. u(9) u(8) u(7) u(6) u(0) ENGG2013

  6. Dependency on initial condition • Define u(t+1) = A u(t) • The initial vector is u(0) = [1 -3]T. u(1) u(5) u(0) u(6) u(7) u(8) u(9) ENGG2013

  7. Eigenvector • An eigenvector of is a nonzero vector v such that if we start from u(t) = v,we will stay on the linewith direction v. ENGG2013

  8. A recipe for calculating eigenvalue  is an eigenvalue of A • A x =  x for some nonzero vector x • A x =  Ix for some nonzero vectorx • (A –  I ) x = 0 has a nonzero solution • A –  I is not invertible • det ( A –  I ) = 0 ENGG2013

  9. Characteristic polynomial • Given a square matrix A, if we expand the determinant the result is a polynomial in variable , and is called characteristic polynomial of A. • The roots of the characteristic polynomial are precisely the eigenvalues of A. ENGG2013

  10. First eigenvalue of • Eigenvalue = 1.5, the corresponding eigenvector is where k is anynonzero constant. • The initial point u(0)is somewhere on theline y = x. u(0) ENGG2013

  11. Another eigenvalue of • Eigenvalue = 0.5, the corresponding eigenvector is where k is anynonzero constant. u(0)=(-10,10) u(1) u(2) ENGG2013

  12. The direction [-1 1]T is not stable • In this example, if we start from a point very close to the line y= –x, for example,if the initial point isu(0)=(-9.9, 10), it will diverge. u(0) u(1) u(2) ENGG2013

  13. The direction [-1 1]T is not stable • If we start from another point very close to the line y= –x, u(0) = say (-10, 9.9), it will also diverge. u(0) u(1) u(2) ENGG2013

  14. Fibonacci sequence • F1 = 1, F2 = 1, and for n > 2, Fn = Fn–1+Fn–2. • The Fibonacci numbers are 1,1,3,5,8,13,21,34,55,89,144,… • Define a vector • The recurrence relation in matrix form ENGG2013

  15. How to find F1000 without going through the recursion? • F1000 also counts the number of binary strings of length 1000 with no consecutive ones. • We need a closed-form formula for the Fibonacci numbers. ENGG2013

  16. Closed-form formula • An expressions involving finitely many + –  , and some well-known functions. • http://en.wikipedia.org/wiki/Closed-form_expression • Integral, infinite series etc. (anything which involves the concept of limit in calculus) are not allowed. • For example, the roots of x2+x+1= 0 can be written in closed-form expression, namely ENGG2013

  17. Closed-form formula (cont’d) • By the theory of Abel and Galois, a polynomial in degree 5 or higher in general has no closed-form formula. • The function has no closed-form formula • Geometric series 1+x+x2+x3+… has closed-form formula 1/(1– x) if |x|<1. ENGG2013

  18. Niels Henrik Abel http://en.wikipedia.org/wiki/Niels_Henrik_Abel • 5 August 1802 – 6 April 1829 • Norwegian mathematician • Gave the first rigorous proofthat quintic equation in generalcannot be solved using radical. ENGG2013

  19. Évariste Galois • October 25, 1811 – May 31, 1832 • French mathematician • Tell us precisely under what condition aa polynomial is solvable using radical. • Galois theory of equations. http://en.wikipedia.org/wiki/Galois ENGG2013

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