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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. 1. State vector u(t): a vector of length n, which summarizes the status of the system at time t. 2. 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) = A u(t). 3. Initial state u(0): the starting point of the system. kshum ENGG2013 2 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). kshum ENGG2013 3 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 Matrix-vector product Scalar product of a vector • This number is called the eigenvalue of A corresponding to the eigenvector v. kshum ENGG2013 4 Another geometric picture • Take for example. u(9) 10 • Define a recursion by u(t+1) = A u(t) 8 u(8) 7 6 y – The initial vector is u(0) = [0 0.5]T. 9 u(7) 5 4 u(6) 3 2 1 u(0) 0 0 2 4 6 8 10 x kshum ENGG2013 5 Dependency on initial condition u(1) -2 • Define u(t+1) = A u(t) – The initial vector is u(0) = [1 -3]T. u(5) u(0) -4 u(6) -6 u(7) -8 y -10 u(8) -12 -14 -16 -18 u(9) -18 -16 -14 -12 -10 -8 -6 -4 -2 0 2 x kshum ENGG2013 6 Eigenvector • An eigenvector of vector v such that if we start from u(t) = v, we will stay on the line with direction v. kshum ENGG2013 is a nonzero 7 A recipe for calculating eigenvalue is an eigenvalue of A A x = x for some nonzero vector x A x = I x for some nonzero vectorx (A – I ) x = 0 has a nonzero solution A – I is not invertible det ( A – I ) = 0 kshum ENGG2013 8 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. kshum ENGG2013 9 First eigenvalue of • Eigenvalue = 1.5, the corresponding eigenvector is 10 9 8 where k is any nonzero constant. 7 y 6 5 4 3 • The initial point u(0) is somewhere on the line y = x. kshum 2 1 u(0) 0 0 ENGG2013 2 4 6 8 10 x 10 Another eigenvalue of • Eigenvalue = 0.5, the corresponding eigenvector is 10 u(0)=(-10,10) 8 where k is any nonzero constant. 6 u(1) 4 u(2) y 2 0 -2 -4 -6 -8 -10 -10 kshum ENGG2013 -5 0 x 5 10 11 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 is u(0) u(0)=(-9.9, 10), u(1) it will diverge. u(2) 10 8 6 4 y 2 0 -2 -4 -6 -8 -10 -10 kshum ENGG2013 -5 0 x 5 10 12 The direction [-1 1]T is not stable • If we start from another point very close to the line y= –x, u(0) u(0) = say (-10, 9.9), u(1) it will also diverge. 10 8 6 4 u(2) y 2 0 -2 -4 -6 -8 -10 -10 kshum ENGG2013 -5 0 x 5 10 13 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 kshum ENGG2013 14 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. kshum ENGG2013 15 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 kshum ENGG2013 16 Closed-form formula (cont’d) • By the theory of Abel and Galois, a polynomial in degree 5 or higher in general has no closedform formula. • The function has no closed-form formula • Geometric series 1+x+x2+x3+… has closedform formula 1/(1– x) if |x|<1. kshum ENGG2013 17 http://en.wikipedia.org/wiki/Niels_Henrik_Abel Niels Henrik Abel • 5 August 1802 – 6 April 1829 • Norwegian mathematician • Gave the first rigorous proof that quintic equation in general cannot be solved using radical. kshum ENGG2013 18 Évariste Galois kshum ENGG2013 http://en.wikipedia.org/wiki/Galois • October 25, 1811 – May 31, 1832 • French mathematician • Tell us precisely under what condition a a polynomial is solvable using radical. • Galois theory of equations. 19