Let $A$ be an $n \times n$ matrix. I need to show that if $\Phi$ is a fundamental matrix of the system $$Y^{\prime}=AY $$ then there exists an invertible matrix B such that $\Phi = e^{At}B$, where $e^{At}$ is the exponential of the matrix $A$.
I started out supposing I have $n$ linearly independent solutions $Y_{1},Y_{2},\cdots , Y_{n}$ to the system $Y^{\prime}=AY$. Then, $\Phi$ a fundamental matrix implies that $$\Phi = \begin{pmatrix} Y_{1} & Y_{2} & \cdots & Y_{n} \end{pmatrix}$$
But, I don't know where to go from here.
Could somebody please help me with this proof?
Thank you ahead of time for your time and patience.
Since
$\Phi = (Y_1 \; Y_2 \; \cdots \; Y_n), \tag 1$
that is, the columns of $\Phi$ are $n$ linearly independent solutions of
$Y' = AY, \tag 2$
we have
$\Phi' = (Y_1' \; Y_2' \; \cdots \; Y_n') = (AY_1 \; AY_2 \; \cdots \; AY_n) = A\Phi, \tag 3$
or
$\Phi' - A\Phi = 0; \tag 4$
now consider the matrix
$e^{-At} \Phi; \tag 5$
we have
$(e^{-At} \Phi)' = e^{-At} \Phi' - Ae^{-At} \Phi = e^{-At} \Phi' - e^{-At} A\Phi = e^{-At}(\Phi' - A \Phi) = 0; \tag 6$
it follows that $e^{-At}\Phi$ is a constant matrix $B$:
$ e^{-At} \Phi = B; \tag 7$
this may be written
$\Phi = e^{At}B; \tag 8$
note that both $e^{-At}$ and $\Phi$ are invertible; the inverse of $e^{-At}$ is $e^{At}$:
$e^{At}e^{-At} = e^{At - At} = e^0 = I, \tag 9$
and $\Phi$ is invertible since it's columns are linearly independent. So $B$ is invertible as well. And we are done.
Note: If follows from (7) that in fact
$B = \Phi(0) = (Y_1(0) \; Y_2(0) \; \cdots \; Y_n(0)). \tag{10}$