Background. The matrix A has as an eigenvalue, with the associated eigenvector provided , and the eigenvector V corresponding to is a non-zero solution to . Generalized eigenvectors will be of the form

Computer Lab Work.

Exercise 1. (a) Find the general solution to the system of D.E.'s X' = A X, where

(b) Find the solution in (a) that has the I.C.'s
,
and plot the solutions over the interval 0 <= t <= 3.

Solution. Find the eigenvectors of A. We want to familiarize ourselves with the NullSpace command, because it must be used to find the generalized eigenvectors.

Thus, the matrix has two length 1 chains and a length 2 chain (defect 1), i.e. we need to construct one generalized eigenvector for the first eigenvalue. This requires that we find V2 so that .

Any linear combination of the above vectors can be used for V2, for convenience we use the following vector.

Question. Are the four solutions linearly independent?
Check this out by finding the determinant. If it is not zero at at least one point then the functions are linearly independent and form a basis for the general solution.

Since the determinant is not identically zero, its o.k. to we proceed. The following linear combination is the general solution.

Extract the four coordinate functions from Y[t] and call them y1[t], y2[t], y3[t]], y4[t].

Plot the functions

Exercise 2. (a) Find the general solution to the system of D.E.'s X' = A X, where

(b) Find the solution in (a) that has the I.C.'s
,
and plot the solutions over the interval 0 <= t <= 3.

Solution. Find the eigenvectors of A. We want to familiarize ourselves with the NullSpace command, because it must be used to find the generalized eigenvectors.

Thus, the matrix has one length 1 chains and a length 3 chain (defect 2), i.e. we need to construct two generalized eigenvectors for the first eigenvalue. This requires that we find V3 so that .

Not all linear combinations of the above vectors can be used for V3, choose something so that you do not get the zero vector when you multiply it times M.

Question. Are the four solutions linearly independent?
Check this out by finding the determinant. If it is not zero at at least one point then the functions are linearly independent and form a basis for the general solution.

Since the determinant is not identically zero, its o.k. to we proceed. The following linear combination is the general solution.

Extract the four coordinate functions from Y[t] and call them y1[t], y2[t], y3[t], y4[t].

Plot the functions

Solutions.

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(c) John H. Mathews, 1998