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.

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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).

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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.

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Matrix-vector product Scalar 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.

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0 2 4 6 8 100

1

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9

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x

y

u(0)

u(9)

u(8)

u(7)

u(6)

-18 -16 -14 -12 -10 -8 -6 -4 -2 0 2

-18

-16

-14

-12

-10

-8

-6

-4

-2

x

y

Dependency on initial condition

• Define u(t+1) = A u(t)– The initial vector is

u(0) = [1 -3]T.

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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 linewith direction v.

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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

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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.

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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.

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0 2 4 6 8 100

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9

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x

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u(0)

Another eigenvalue of

• Eigenvalue = 0.5, the corresponding eigenvector is

where k is anynonzero constant.

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-10 -5 0 5 10-10

-8

-6

-4

-2

0

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10

x

y

u(0)=(-10,10)

u(1)

u(2)

-10 -5 0 5 10-10

-8

-6

-4

-2

0

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10

x

y

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.

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u(0)

u(1)

u(2)

-10 -5 0 5 10-10

-8

-6

-4

-2

0

2

4

6

8

10

x

y

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.

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u(0)

u(1)

u(2)

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

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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.

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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

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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.

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Niels Henrik Abel

• 5 August 1802 – 6 April 1829• Norwegian mathematician• Gave the first rigorous proof

that quintic equation in generalcannot be solved using radical.

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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.

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