R is the reals. xi : R → R are differentiable for i = 1, 2, ... , n and satisfy xi' = ai1x1 + ... + ainxn for some constants aij ≥ 0. Also limt→∞ xi(t) = 0. Can the functions xi be linearly independent?
Solution
Answer: no.
Fairly easy.
Let A be the n x n matrix (aij). Let AT be its transpose. We know that the trace of AT is non-negative (since all its elements are non-negative). So it must have an eigenvalue λ with Re(λ) ≥ 0. Let v = (vi) be a corresponding eigenvector and consider the linear combination y = ∑ vixi. We have y' = ∑ viaijxj = λ ∑ vjxj = λ y. So, integrating, y = k eλt for some constant k. But |eλt| does not tend to zero as t → ∞, since Re(λ) ≥ 0. We are given that each xi(t) and hence y(t) → 0, so k = 0. In other words we have found a (non-trivial) linear combination of the xi which is zero, so they are not linearly independent.
© John Scholes
jscholes@kalva.demon.co.uk
12 Dec 1998