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A fundamental matrix of a system of n homogeneous linear ordinary differential equations

\begin{equation} \dot{\mathbf{x}}(t) = A(t) \mathbf{x}(t) > \end{equation}

is a matrix-valued function $\Psi(t)$ whose columns are linearly independent solutions of the system. -Wikipedia

Now, I also know that the system has solution $\mathbf{x}=e^{At}\mathbf{x}(0)$. The columns of this matrix are linearly independent. Does this mean one can use $e^{At}$ as a fundamental matrix?

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  • $\begingroup$ Yes, you are correct. $\endgroup$ – Artem Nov 7 '15 at 20:05
  • $\begingroup$ Even when you ommit the $\vec{x}(0)$? $\endgroup$ – Michael Angelo Nov 7 '15 at 20:12
  • $\begingroup$ If you do not omit $x(0)$ you will have a vector, not a matrix. $\endgroup$ – Artem Nov 7 '15 at 20:13
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    $\begingroup$ You are correct again, "it is right to say" $\endgroup$ – Artem Nov 7 '15 at 20:18
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    $\begingroup$ That you can generally solve $y'(t)=A(t)y(t)$ using the matrix exponential, as the question in combination of quote and follow-up stated. For that you need that $A(s)$ and $A(t)$ commute for all $s,t$, which is the case for $A(t)=A=const.$. $\endgroup$ – LutzL Nov 8 '15 at 9:18

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