Define the Fundamental Matrix of e, resulting in multiplication with Inverse Fundamental Matrix where t=0 Given A is a n×n matrix, Let $\phi(t)$ be a fundamental matrix for the homogeneous linear system $X' = Ax$.
Then the solution of the initial value problem $X' = AX,\ X(0)=X_0$ is given by $X(t) = \Phi(t)\Phi(0)^{-1}X_0$.
$e^{At}$ is defined as:
$$
e^{At} = I +At + \frac {A^2}2t^2 + ... +\frac {A^n}nt^n
$$
...which is an infinite expression. How can it be proven that
$$
e^{At} = \Phi(t)\Phi^{-1}(0)
$$
even though it is a finite definition?
 A: We have
$(e^{At})' = Ae^{At}, \tag 1$
and we also have
$(\Phi(t) \Phi^{-1}(0))' = \Phi'(t) \Phi^{-1}(0); \tag 2$
since $\Phi(t)$ is a fundamental matrix solution, 
$\Phi'(t) = A\Phi(t); \tag 3$
thus (2) yields
$(\Phi(t) \Phi^{-1}(0))' = A\Phi(t) \Phi^{-1}(0); \tag 4$
comparing (1) and (4), we see that both $e^{At}$ and $\Phi(t)\Phi(0)$ satisfy a matrix differential equation of the form
$Z'(t) = AZ(t). \tag 5$
We next observe that at 
$t = 0 \tag 6$
we have
$e^{A(0)} = e^0 = I, \tag 7$
and
$\Phi(0)\Phi^{-1}(0) = I \tag 8$
as well; thus $e^{At}$ and $\Phi(t)\Phi^{-1}(0)$ each satisfy the same differential equation (5) with the same initial conditions (7)-(8); thus, by uniqueness of solutions,
$e^{At} = \Phi(t)\Phi^{-1}(t) \tag 9$
for all times $t$.
Another way to arrive at this result is to set
$W(t) = e^{-At}\Phi(t)\Phi^{-1}(0); \tag{10}$
then
$W'(t) = (e^{-At})' \Phi(t)\Phi^{-1}(0) + e^{-At}(\Phi(t)\Phi^{-1}(0))'$
$= -Ae^{-At} \Phi(t)\Phi^{-1}(0) + e^{-At} \Phi'(t)\Phi^{-1}(0); \tag{11}$
via (3) this becomes
e^
$W'(t) = -Ae^{-At} \Phi(t)\Phi^{-1}(0) + e^{-At} A\Phi(t)\Phi^{-1}(0)$
$= -Ae^{-At} \Phi(t)\Phi^{-1}(0) + Ae^{-At} \Phi(t)\Phi^{-1}(0) = 0; \tag{12}$
thus
$W(t) = \text{constant} = W(0)$
$= e^{-A(0)}\Phi(0)\Phi^{-1}(0) = e^0\Phi(0)\Phi^{-1}(0) = I; \tag{13}$
combining (10) and (13) yields
$ e^{-At}\Phi(t)\Phi^{-1}(0) = I, \tag{14}$
that is,
$e^{At} = \Phi(t)\Phi^{-1}(0). \tag{15}$
