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10.1 Linear Systems 299

for all t in I. This means that

1 (t) = c 2 2 (t) + ··· + c n n (t)
for all t in I, hence that 1 (t), 2 (t),··· , n (t) are linearly dependent on I. This completes the
proof.

Thus far, we know how to test n solutions of the homogeneous system for linear indepen-
dence on an open interval. We will now show that n linearly independent solutions are all that
are needed to specify all solutions.

THEOREM 10.3

Let A(t) =[a ij (t)] be an n × n matrix of functions that are continuous on an open interval I.
Then

1. The system X = AX has n linearly independent solutions on I.

2. Given n linearly independent solutions 1 (t),,··· , n (t) defined on I, every solution
on I is a linear combination of 1 (t),,··· , n (t).

Proof To prove that there are n linearly independent solutions, define the n × 1 constant
matrices
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 0 0
0
0
1
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
0 0
⎜ ⎟ ⎜ ⎟ ⎜.⎟
.
(1)
(2)
(n)
E = ⎜.⎟,E = ⎜.⎟,··· ,E = ⎜ ⎟.
⎜ ⎟
⎜ . ⎟
⎜ ⎟
0
.
.
⎜ . ⎟ ⎜ . ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
0 0 0
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
0 0 1
Choose any t 0 in I. We know that the initial value problem
X = AX;X(0) = E ( j)

has a unique solution j (t),for j = 1,2,··· ,n. These solutions are linearly independent,
because, the way the initial conditions were chosen, the n × n matrix whose columns are these
solutions evaluated at t 0 is I n , with determinant 1. This proves part (1).
To prove conclusion (2), suppose 1 ,··· , n are n linearly independent solutions on I.
Let be any solution. We want to show that is a linear combination of 1 ,··· , n .Pick
any t 0 in I. Form the n × n nonsingular matrix S having the linearly independent vectors
1 (t 0 ),··· , n (t 0 ) as its columns and consider the linear system of n algebraic equations in n
unknowns:
⎛ ⎞
c 1
c 2
⎜ ⎟
S⎜ . ⎟ = (t 0 ).
⎜ ⎟
.
⎝ . ⎠
c n
Because S is nonsingular, this algebraic system has a unique solution for numbers c 1 ,c 2 ,··· ,c n
such that
(t 0 ) = c 1 1 (t 0 ) + ··· + c n n (t 0 ).
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October 14, 2010 20:32 THM/NEIL Page-299 27410_10_ch10_p295-342