TITLE: Solution of a homogeneous system is the null solution if it converges to $0$ for $x\to\pm\infty$
QUESTION [2 upvotes]: Let $A\in\mathbb R^{d\times d}$ be diagonizable with eigenvalues $\lambda_1,\dots,\lambda_d\in\mathbb R$.
Show that: If $y\in C^1(\mathbb R\to\mathbb R^d)$ is a solution of the system $y'=Ay$ satisifying $\lim_{x\to\pm\infty} y(x)=0$, then $y(x)\equiv 0$.
Thoughts: Let $v_1,\dots,v_d$ be eigenvectors of $A$ for the eigenvalues $\lambda_1,\dots,\lambda_d$. If $\lambda_i\neq\lambda_j$ for $i\neq j$, then $\{x\mapsto v_1\exp(\lambda_1x),\dots,x\mapsto v_d\exp(\lambda_dx)\}$ is a basis for the vector space of solutions. The condition $y(x)\to 0$ for $x\to\pm\infty$ implies that all components $y_i$ of $y$ must go to $0$ as well. These can be written as sums of exponential functions of the form $\exp(\lambda_i x)$ in each component, which implies that all coefficients have to be $0$ (using the linear independence of these functions) and thus $y(x)\equiv 0$. Why exactly is this the case, and how can one show this in the more general case where $\lambda_i=\lambda_j$ is allowed?

REPLY [3 votes]: Suppose $y(0) = \sum_k y_k v_k$ where $v_k$ are the eigenvectors.
Then $y(t) = \sum_k y_k  e^{\lambda_k t} v_k$.
Since $y(t) \to 0$ as $|t| \to \infty$ we must have $y_k e^{|\lambda_k t|} \to 0$ and since $e^{|\lambda_k t|} \ge 1$ we must have $y_k = 0$.
This is true whether or not the eigenvalues are distinct.
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