# Is the matrix exponential of a system of linear differential equations the fundamental matrix of that system?

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

$$\dot{\mathbf{x}}(t) = A(t) \mathbf{x}(t) >$$

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?

• Yes, you are correct. – Artem Nov 7 '15 at 20:05
• Even when you ommit the $\vec{x}(0)$? – Michael Angelo Nov 7 '15 at 20:12
• If you do not omit $x(0)$ you will have a vector, not a matrix. – Artem Nov 7 '15 at 20:13
• You are correct again, "it is right to say" – Artem Nov 7 '15 at 20:18
• 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 $A(t)=A=const.$. – LutzL Nov 8 '15 at 9:18