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
2
3
4
5
6
7
8
9
10
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
1
2
3
4
5
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7
8
9
10
x
y
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
2
4
6
8
10
x
y
u(0)=(-10,10)
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
• 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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http
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