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ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011. EXAMPLE 1 kshum ENGG2013 2 Q6 in midterm • u(t): unemployment rate in the t-th month. • e(t)= 1-u(t) • The unemployment rate in the next month is given by a matrix multiplication • Equilibrium: Solve Unemployment rate at equilibrium = 0.2 kshum ENGG2013 3 Equilibrium Unstable kshum Stable ENGG2013 4 If stable, how fast does it converge to the equilibrium point? Slow convergence 0.2 kshum Fast convergence 0.2 ENGG2013 5 Question • Suppose that the initial unemployment rate at the first month is x(1), (for example x(1)=0.25), and suppose that the unemployment evolves by matrix multiplication Find an analytic expression for x(t), for all t. kshum ENGG2013 6 EXAMPLE 2 kshum ENGG2013 7 How to count? • Count the number of binary strings of length n with no consecutive ones. kshum ENGG2013 8 SOLVING RECURRENCE RELATION kshum ENGG2013 9 • • • • F1 = 1 F2 = 1 For n > 2, Fn = Fn-1+Fn-2. The Fibonacci numbers are – 1,1,3,5,8,13,21,34,55,89,144 kshum ENGG2013 http://en.wikipedia.org/wiki/Fibonacci_number Fibonacci numbers 10 A matrix formulation • Define a vector • Initial vector • Find the recurrence relation in matrix form kshum ENGG2013 11 A general question • Given initial condition and for t 2 Find v(t) for all t. kshum ENGG2013 12 Matrix power • Need to raise a matrix to a very high power kshum ENGG2013 13 A trivial special case • Diagonal matrix • The solution is easy to find • Raising a diagonal matrix to the power t is easy. kshum ENGG2013 14 Decoupled equations • When the equation is diagonal, we have two separate equation, each in one variable kshum ENGG2013 15 DIAGONALIZATION kshum ENGG2013 16 Problem reduction • A square matrix M is called diagonalizable if we can find an invertible matrix, say P, such that the product P–1 M P is a diagonal matrix. • A diagonalizable matrix can be raised to a high power easily. – Suppose that P–1 M P = D, D diagonal. – M = P D P–1. – Mn = (P D P–1) (P D P–1) (P D P–1) … (P D P–1) = P Dn P–1. kshum ENGG2013 17 Example of diagonalizable matrix • Let • A is diagonalizable because we can find a matrix such that kshum ENGG2013 18 Now we know how fast it converges to 0.2 • The matrix can be diagonalized kshum ENGG2013 19 Convergence to equilibrium • The trajectory of the unemployment rate – the initial point is set to 0.1 0.2 0.19 0.18 Unemployment rate 0.17 0.16 0.15 0.14 0.13 0.12 0.11 0.1 kshum 1 2 3 4 5 6 month (t) ENGG2013 7 8 9 10 20 EIGENVECTOR AND EIGENVALUE kshum ENGG2013 21 How to diagonalize? • How to determine whether a matrix M is diagonalizable? • How to find a matrix P which diagonalizes a matrix M? kshum ENGG2013 22 From diagonalization to eigenvector • By definition a matrix M is diagonalizable if P–1 M P = D for some invertible matrix P, and diagonal matrix D. or equivalently, kshum ENGG2013 23 The columns of P are special • Suppose that kshum ENGG2013 24 Definition • 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 an eigenvalue of A, corresponding to the eigenvector v. kshum ENGG2013 25 Important notes • If v is an eigenvector of A with eigenvalue , then any non-zero scalar multiple of v also satisfies the definition of eigenvector. k0 kshum ENGG2013 26 Geometric meaning • A linear transformation L(x,y) given by: L(x,y) = (x+2y, 3x-4y) x x + 2y y 3x – 4y • If the input is x=1, y=2 for example, the output is x = 5, y = -5. kshum 27 Invariant direction • An Eigenvector points at a direction which is invariant under the linear transformation induced by the matrix. • The eigenvalue is interpreted as the magnification factor. • L(x,y) = (x+2y, 3x-4y) • If input is (2,1), output is magnified by a factor of 2, i.e., the eigenvalue is 2. kshum 28 Another invariant direction • • L(x,y) = (x+2y, 3x-4y) If input is (-1/3,1), output is (5/3,-5). The length is increased by a factor of 5, and the direction is reversed. The corresponding eigenvalue is -5. kshum 29 Eigenvalue and eigenvector of First eigenvalue = 2, with eigenvector where k is any nonzero real number. Second eigenvalue = -5, with eigenvector where k is any nonzero real number. kshum ENGG2013 30 Summary • Motivation: want to solve recurrence relations. • Formulation using matrix multiplication • Need to raise a matrix to an arbitrary power • Raising a matrix to some power can be easily done if the matrix is diagonalizable. • Diagonalization can be done by eigenvalue and eigenvector. kshum ENGG2013 31