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. kshumENGG20132

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). kshumENGG20133

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. kshumENGG20134 Matrix-vector productScalar product of a vector

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. kshumENGG20135 u(0) u(9) u(8) u(7) u(6)

Dependency on initial condition Define u(t+1) = A u(t) – The initial vector is u(0) = [1 -3] T. kshumENGG20136 u(0) u(9) u(8) u(7) u(6) u(1) u(5)

Eigenvector An eigenvector of is a nonzero vector v such that if we start from u(t) = v, we will stay on the line with direction v. kshumENGG20137

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 kshumENGG20138

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. kshumENGG20139

First eigenvalue of Eigenvalue = 1.5, the corresponding eigenvector is where k is any nonzero constant. The initial point u(0) is somewhere on the line y = x. kshumENGG u(0)

Another eigenvalue of Eigenvalue = 0.5, the corresponding eigenvector is where k is any nonzero constant. kshumENGG u(0)=(-10,10) u(1) u(2)

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)=(-9.9, 10), it will diverge. kshumENGG u(0) u(1) u(2)

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. kshumENGG u(0) u(1) u(2)

Fibonacci sequence F 1 = 1, F 2 = 1, and for n > 2, F n = F n–1 +F n–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 kshumENGG201314

How to find F 1000 without going through the recursion? F 1000 also counts the number of binary strings of length 1000 with no consecutive ones. We need a closed-form formula for the Fibonacci numbers. kshumENGG201315

Closed-form formula An expressions involving finitely many + –  , and some well-known functions. – Integral, infinite series etc. (anything which involves the concept of limit in calculus) are not allowed. For example, the roots of x 2 +x+1= 0 can be written in closed-form expression, namely kshumENGG201316

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+x 2 +x 3 +… has closed- form formula 1/(1– x) if |x|<1. kshumENGG201317

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. kshumENGG

Évariste 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. kshumENGG