Show that System $(I)$ is stable iff $X(t)$ is bounded. I have a theorem:
For a linear homogeneous system:
$$\dfrac{dx}{dt}=A(t)x \tag{I}$$
Where $A(t)=(a_{ij}(t))_{n \times n} \in C(\mathbb{R}^+,\mathbb{R}^{n \times n})$
Suppose that $X(t)$ be the fundamental matrix solution of the following reference system $(I)$.
Let $K(t,s)=X(t)X^{-1}(s)$ be the Cauchy matrix of the following reference system $(I)$.
Prove that:


*

*a/ System $(I)$ is stable iff $X(t)$ is bounded, it means $\exists M>0$ such that $$\|X(t)\| \le M, \forall t \ge 0 $$

*b/ System $(I)$ is asymptotically stable iff  $$\lim_{t \to +\infty}X(t)=0$$
=================================================================
I have stuck when I try to show this theorem.
I have tried using the definition of stable, asymptotically stable. But I have no solution. 
=================================================================
Ps: (Or) if somebody knows/reads this theorem (book/pdf/djvu...) then you can post it. 
Can anyone help me!
Any help will be appreciated! Thanks/
 A: Here's my solution for Question a: 
a/ A linear homogeneous system $\dot{x}=A(t)x$ is stable iff $X(t)$ is bounded.
Without loss of generality we can assume that $X(t_0)=E$, then $x(t)=X(t)x_0$.


*

*If $X(t)$ is bounded. Whence $$\left \| x(t) \right \|=\left \| X(t)x_0 \right \| \le \left \| X(t) \right \| \left \| x_0 \right \| \le M\left \| x_0 \right \|<\epsilon, \text{for}\left \| x_0 \right \|<\delta:=\frac{\epsilon}{M}$$
Hence, $x \equiv 0$ is stable.

*If $x \equiv 0$ is stable. Whence 
$$\forall \varepsilon >0, \exists \delta >0 \ \text{such that}\  \left \| x_0 \right \| \le \delta \implies \left \| x(t) \right \| = \left \| X(t)x_0 \right \| < \epsilon, \forall t \ge t_0$$
Thus $$\left \| X(t)\alpha  \right \|=\left \| X(t)\alpha  \cdot \dfrac{\delta}{\left \| \alpha \right \|}\right \|\cdot \dfrac{\left \| \alpha \right \|}{\delta}<\dfrac{\epsilon}{\delta}\left \| \alpha \right \|, \forall \alpha \in \mathbb{R}^n$$
Hence, $\left \| X(t)  \right \| \le M:=\dfrac{\epsilon}{\delta} \blacksquare $.
====================================================
My solution for Question b:


*

*If $\lim_{t \to \infty}X(t)=0$ then since $x(t)=X(t)X^{-1}(t_0)x_0$ we have:
$$\left \| x(t) \right \| \le \left \| X(t) \right \|\cdot  \left \| X^{-1}(t_0)\right \|\cdot \left \| x_0 \right \|$$
Thus,  $\lim_{t \to \infty}\left \| x(t;t_0,x_0) \right \| =0$, forall $x(t)$ of $(I)$.


Hence, $(I)$ is asymptotically stable.


*

*If $(I)$ is asymptotically stable. How we can show that $\lim_{t \to \infty}X(t)=0$?

A: The key step is to note that $x(t)=K(t,s)x_0$ solves the initial value problem
$$
\dot x(t)=A(t)x(t)\\
x(s)=x_0.
$$
Now just take the definitions of stability and use $x(t)=K(t,s)x_0=X(t)X^{-1}(s)x_0$. Furthermore, note that $X^{-1}(s)x_0$ is a constant vector so it can be absorbed in some other constant when you take the norm.
Assume the equilibrium point is at the origin (you can always do it).
Stable $\iff$ $||x(t)||< M $ $\iff$ $|| X(t)X^{-1}(s)x_0 ||=||X(t)||\cdot||X^{-1}(s)x_0||<M$ $\iff$ $ || X(t)||<\frac{M}{||X^{-1}(s)x_0||}=P $. 
The division is well defined, so $P$ is another constant.  
Exponential Stability $\iff$ $||x(t)||< Me^{-ct}, \, c>0, \, t>s $ $\iff$ $|| X(t)X^{-1}(s)x_0 ||=||X(t)||\cdot||X^{-1}(s)x_0||<Me^{-ct}$ $\iff$ $ || X(t)||<\frac{Me^{-ct}}{||X^{-1}(s)x_0||}=Pe^{-ct} $. Next, $\displaystyle\lim_{t\to\infty}Pe^{-ct}=0$. 
