Lecture Notes in Differential Equations - Bruce E. Shapiro

317
By a similar argument, the second determinant is a 22 W (t), the third one
is a 33 W (t), ..., and the nth one is a nn W (t). Therefore
W ′ (t) = (a 11 + a 22 + · · · + a nn )W (t) = (traceA)W (t) (31.112)
Dividing by W (t) and integrating produces the desired result.
Theorem 31.9. Let y 1 , ..., y n : I → R n be solutions of the linear system
y ′ = Ay, where A is an n × n constant matrix, on some interval I ⊂ R,
and let W (t) denote their Wronskian. Then the following are equivalent:
1. W (t) ≠ 0 for all t ∈ I.
2. For some t 0 ∈ I, W (t 0 ) ≠ 0
3. The set of functions y 1 (t), ..., y n (t) are linearly independent.
4. The set of functions y 1 (t), ..., y n (t) form a fundamental set of solutions
to the system of differential equations on I.
Proof. (1) ⇒ (2). Suppose W (t) ≠ 0 for all t ∈ I (this is (1)). Then pick
any t 0 ∈ I. Then ∃t 0 ∈ I such that W (t 0 ) ≠ 0 (this is (2)).
(2) ⇒ (1). This follows immediately from Abel’s formula.
(1) ⇒ (3). Since the Wronskian is nonzero, the fundamental matrix is invertible.
But a matrix is invertible if and only if its column vectors are
linearly independent.
(3) ⇒ (1). Since the column vectors of the fundamental matrix are linearly
independent, this means the matrix is invertible, which in turn implies that
its determinant is nonzero.
(3) ⇒ (4). This was proven in a previous theorem.
(4) ⇒ (3). Since the functions form a fundamental set, they must be linearly
independent.
Since (4) ⇐⇒ (3) ⇐⇒ (1) ⇐⇒ (2) all four statements are equivalent.
Theorem 31.10. Let A be an n × n matrix with n linearly independent
eigenvectors v 1 , ..., v n with corresponding eigenvalues λ 1 , ..., λ n . Then
y 1 = v 1 e λ1t , . . . y n = v n e λnt (31.113)
form a fundamental set of solutions for y ′ = Ay.