= Solution
The <independence of eigenvectors for distinct eigenvalues> ensures that such <eigenvectors> form a <linearly independent> set. For completeness, apply $\prod_{j\ne k}(M-\lambda_j I)$ to a proposed relation $\sum_i c_i\mathbf e_i=0$. It leaves $c_k\prod_{j\ne k}(\lambda_k-\lambda_j)\mathbf e_k=0$, so every $c_k=0$. Thus the three <eigenvectors> form a <basis> over the relevant field and $\mathbf x=\sum_i a_i(t)\mathbf e_i$. Substitution into the <linear system of ordinary differential equations> gives $\sum_i(\dot a_i-\lambda_i a_i)\mathbf e_i=0$; <linear independence> implies $\dot a_i=\lambda_i a_i$. Hence
$$
\boxed{\mathbf x(t)=\sum_{i=1}^3 A_i e^{\lambda_i t}\mathbf e_i.}
$$
A real matrix may have a complex conjugate pair of <eigenvalues>. In that case the calculation is over $\mathbb C$ and conjugate coefficients yield real solutions; equivalently use real and imaginary parts of the complex modes. No unstated assumption that all three <eigenvalues> are real is needed.
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