To linearize a system, in one of the steps I am required to find the fundamental matrix $\Phi$(t) of a system such that $\Phi$(0)=I. The example system my professor used:
$\dot{x} = x - y - x^3 - xy^2$
$\dot{y} = x + y - x^2y - y^3$
$\dot{z} = \lambda z$
Then $\vec\gamma(t) = (\cos{t}, \sin{t}, 0)^T $ is a periodic orbit of the system.
We compute $D\vec f(\vec x)= \left( \begin{array}{ccc} 1-3x^2-y^2 & -1-2xy & 0 \\ 1-2xy & 1-x^2-3y^2 & 0 \\ 0 & 0 & \lambda \end{array} \right) $
Then the linearization is given by $\frac{d}{dt}\vec x=A(t)\vec x $ where
$A(t) = D\vec f(\vec\gamma(t)) = \left( \begin{array}{ccc} -2\cos^2{t} & -1-\sin^2{t} & 0 \\ 1-\sin^2{t} & -2\sin^2{t} & 0 \\ 0 & 0 & \lambda \end{array} \right)$
This all makes sense until he says "Clearly, this is its fundamental matrix:"
$ \Phi(t) = \left( \begin{array}{ccc} e^{-2t}\cos{t} & -\sin{t} & 0 \\ e^{-2t}\sin{t} & \cos{t} & 0 \\ 0 & 0 & e^{\lambda t} \end{array} \right)$
Am I missing something? How does one get from the periodic matrix A(t) to the fundamental matrix?