Lecture Notes in Differential Equations - Bruce E. Shapiro

314 LESSON 31. LINEAR SYSTEMS
Let y 1 , ..., y n denote the column vectors of W. Then by (31.93)
Equating columns,
( y
′
1 · · · y ′ n)
= W
′
(31.94)
= AW (31.95)
= A ( )
y 1 · · · y n (31.96)
= ( )
Ay 1 · · · Ay n (31.97)
y ′ i = Ay i , i = 1, . . . , n (31.98)
hence each column of W is a solution of the differential equation.
Furthermore, by property (3) of of the Matrix Exponential, W = e At is
invertible. Since a matrix is invertible if and only if all of its column vectors
are linearly independent, this means that the columns of W form a linearly
independent set of solutions to the differential equation. To prove that they
are a fundamental set of solutions, suppose that y(t) is a solution of the
initial value problem with y(t 0 ) = y 0 .
We must show that it is a linear combination of the columns of W. Since
the matrix W is invertible, the numbers C 1 , C 2 , ..., C n , which are the components
of the vector
exist. But
C = [W(t 0 )] −1 y 0 (31.99)
Ψ = WC (31.100)
⎛ ⎞
= ( C 1
) ⎜
y 1 · · · y n ⎝
⎟
. ⎠ (31.101)
C n
= C 1 y 1 + · · · + C n y n (31.102)
is a solution of the differential equation, and by (31.99), Ψ(t 0 ) = W (t 0 )C =
y 0 , so that Ψ(t) also satisfies the initial value problem. By the uniqueness
theorem, y and Ψ(t) must be identical.
Hence every solution of y ′ = Ay is a linear combination of the column
vectors of W, because any solution can be considered a solution of some
initial value problem. Thus the column vectors form a fundamental set of
solutions, and hence W is a fundamental matrix.