Are these functions necessarily linearly independent?

[Ackbach]

Here is this week's POTW:

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Let $x_1, x_2,\dots,x_n$ be differentiable (real-valued) functions of a single variable $t$ which satisfy
\begin{align*}
\d{x_1}{t}&=a_{11}x_1+a_{12}x_2+\cdots+a_{1n}x_n \\
\d{x_2}{t}&=a_{21}x_1+a_{22}x_2+\cdots+a_{2n}x_n \\
\vdots \\
\d{x_n}{t}&=a_{n1}x_1+a_{n2}x_2+\cdots+a_{nn}x_n
\end{align*}
for some constants $a_{ij}>0$. Suppose that for all $i, \; x_i(t)\to 0$ as $t\to\infty$. Are the functions $x_1,x_2,\dots,x_n$ necessarily linearly independent?

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[Ackbach]

Re: Problem Of The Week # 216 - May 17, 2016

This was Problem A-5 in the 1995 William Lowell Putnam Mathematical Competition.

No one answered this week's POTW. The solution, attributed to Kiran Kedlaya and his associates, follows:

Spoiler

It is known 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 $\sum_{i=1}^{n}
c_{i}f_{i}(t) \vec{v}_{1}$.

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 $c_{i} \neq 0$, then $f_{i}(t)$ 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 $t$.