Page 323 - Advanced engineering mathematics

10.2 Solution of X = AX for Constant A 303

To carry out this strategy we will focus on the special case that A is a real, constant matrix.
Taking a cue from the constant coefficient, second order differential equation, attempt solutions
λt
of the form X=Ee , with E an n ×1 matrix of numbers and λ a number. For this to be a solution,
we need
λt
λt
λt
(Ee ) = Eλe = AEe .
This will be true if
AE = λE,
which holds if λ is an eigenvalue of A with associated eigenvector E.

THEOREM 10.5

Let A be an n × n matrix of real numbers. If λ is an eigenvalue with associated eigenvector E,
then Ee is a solution of X = AX.
λt

We need n linearly independent solutions to write the general solution of X = AX. The next
theorem addresses this.

THEOREM 10.6

Let A be an n × n matrix of real numbers. Suppose A has eigenvalues λ 1 ,··· ,λ n , and sup-
pose there are n corresponding eigenvectors E 1 ,··· ,E n that are linearly independent. Then
λ 1 t
E 1 e ,··· ,E n e λ n t are linearly independent solutions.
When the eigenvalues are distinct, we can always find n linearly independent eigenvectors.
But even when the eigenvalues are not distinct, it may still be possible to find n linearly indepen-
dent eigenvectors, and in this case, we have n linearly independent solutions, hence the general
solution. We can also use these solutions as columns of a fundamental matrix.

EXAMPLE 10.6
We will solve the system

42

X = X.
33
A has eigenvalues of 1,6 with corresponding eigenvectors

1 1
E 1 = and E 2 = .
−3/2 1
These are linearly independent (the eigenvalues are distinct), so we have two linearly independent
solutions

1 t 1 6t
e and e .
−3/2 1
The general solution is

1 t 1 6t
X(t) = c 1 e + c 2 e .
−3/2 1

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October 14, 2010 20:32 THM/NEIL Page-303 27410_10_ch10_p295-342