Putnam 1995, Problem A-5

Let \( x_1, x_2, \ldots, x_n \) be differentiable (real-valued) functions of a single variable \( t \) which satisfy
\[
\begin{aligned}
\frac{dx_1}{dt} &= a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n \\
\frac{dx_2}{dt} &= a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n \\
&\vdots \\
\frac{dx_n}{dt} &= a_{n1}x_1 + a_{n2}x_2 + \cdots + a_{nn}x_n
\end{aligned}
\]
for some constants \( a_{ij} > 0 \). Suppose that for all \( i \), \( x_i(t) \to 0 \) as \( t \to \infty \). Are the functions \( x_1, x_2, \ldots, x_n \) necessarily linearly dependent?

Solution:
Everyone (presumably) knows that the set of solutions of a system of linear first-order differential equations with constant coefficients is \( n \)-dimensional, with basis vectors of the form \( f_i(t) \vec{v}_i \) (i.e. a function times a constant vector), where the \( \vec{v}_i \) are linearly independent. In particular, our solution \( \vec{x}(t) \) can be written as \( \sum_{i=1}^n c_i f_i(t) \vec{v}_i \).  
Choose a vector \( \vec{w} \) orthogonal to \( \vec{v}_2, \dots, \vec{v}_n \) but not to \( \vec{v}_1 \). Since \( \vec{x}(t) \to 0 \) as \( t \to \infty \), the same is true of \( \vec{w} \cdot \vec{x} \); but that is simply \( (\vec{w} \cdot \vec{v}_1) c_1 f_1(t) \). In other words, if \( c_1 \neq 0 \), then \( f_1(t) \) must also go to 0.  
However, it is easy to exhibit a solution which does not go to 0. The sum of the eigenvalues of the matrix \( A = (a_{ij}) \), also known as the trace of \( A \), being the sum of the diagonal entries of \( A \), is nonnegative, so \( A \) has an eigenvalue \( \lambda \) with nonnegative real part, and a corresponding eigenvector \( \vec{v} \). Then \( e^{\lambda t} \vec{v} \) is a solution that does not go to 0. (If \( \lambda \) is not real, add this solution to its complex conjugate to get a real solution, which still doesn't go to 0.)  
Hence one of the \( c_i \), say \( c_1 \), is zero, in which case \( \vec{x}(t) \cdot \vec{w} = 0 \) for all \( t \).