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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.
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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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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.
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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
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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
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Eigenvector
• An eigenvector of
vector v such that
if we start from u(t) = v,
we will stay on the line
with direction v.
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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
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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
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.
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2
1
u(0) 0 0
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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
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-5
0
x
5
10
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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
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-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
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-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
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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 closedform formula.
• The function
has no
closed-form formula
• Geometric series 1+x+x2+x3+… has closedform formula 1/(1– x) if |x|<1.
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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.
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Évariste Galois
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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.
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