1. Explain why N is an n-dimensional vector space. [Hints: First say why N is a vector space. Then refer to the discussion above about initial values determining a unique solution. Think of
   (y₀, y₁, ... , y_n-1)

   as an arbitrary element of Rⁿ, and define a linear transformation U: Rⁿ -> N by mapping a vector of initial values to the corresponding solution. Show that U is an isomorphism.]

Now suppose we do specify the sequence {z_k}. Then solving the equation

is the same as solving T({y_k}) = {z_k} for {y_k}. This is just like solving the matrix-vector equation Ax = b, where A and b are known. And we know that we can separate the solution process into two steps: First find the general solution x₀ of the homogeneous equation Ax = 0. Then find one particular solution x_p of Ax = b. The general solution of Ax = b is x₀+x_p.

2. Show that an exponential sequence, y_k = r^k, is a solution of the homogeneous equation
   y_k+n + a₁ y_k+n-1 + .... + a_n-1 y_k+1 + a_n y_k = 0

   if and only if

   rⁿ + a₁ r^n-1 + ... + a_n-1 r + a_n = 0.

The polynomial equation is Step 2 is called the auxiliary equation or characteristic equation. Its solutions r₁, r₂, ..., are called characteristic roots of the difference equation.

We saw in Part 3 that two exponential sequences based on different characteristic roots are linearly independent. This is true for any number of exponential sequences with distinct bases, but the general argument is much harder. If the auxiliary equation has n distinct roots, then these exponential sequences provide a basis for the null space N of T. But there is no guarantee that there will be n distinct roots -- and even if there are, they may not all be real numbers. In this module we will consider only situations in which the characteristic roots are real and distinct.

3. Suppose
   rⁿ + a₁ r^n-1 + ... + a_n-1 r + a_n = 0

   is the characteristic equation of an n-th order linear difference equation. Explain why r = 0 cannot be one of the characteristic roots.

Now we consider the linear independence of exponential sequences {r^k} belonging to distinct characteristic roots. Suppose r₁, r₂, ..., r_n are distinct roots of the characteristic equation

rⁿ + a₁ r^n-1 + ... + a_n-1 r + a_n = 0.

If the sequences {r₁^k}, {r₂^k}, ..., {r_n^k} are linearly dependent, then there exist scalars c₁, c₂, ..., c_n, not all of which are 0, such that

c₁ r₁^k + c₂ r₂^k + ... + c_n r_n^k = 0

for every k = 0, 1, 2, ... .

4. Explain why the vector c = [c₁,c₂,...,c_n]^T would have to be a solution of the matrix equation Rc = 0, where R is the matrix
   

   [Hint: Write out the equation preceding this step for the specific cases of k = 0, 1, 2, ..., n-1.] The matrix R is called a Vandermonde matrix of order n.

5. Experiment in your worksheet with determinants of Vandermonde matrices of orders 3, 4, and 5. Explain why such a determinant can never be 0 if the numbers r₁, r₂, ..., r_n are distinct. Then explain why the corresponding exponential sequences must be linearly independent.

In the next Part we will put this theory to work and solve a specific third-order linear difference equation.

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modules at math.duke.edu

Copyright CCP and the author(s), 1999