Determine the values ​of the $m$ parameter so that the origin is stable in the sense of Liapunov. Determine the values ​​of the $m$ parameter so that the origin is stable in the sense of Liapunov.
$$\dot{x}=\begin{pmatrix}-(m+1)& 0 & 2(m+1)\\0 & -2 & 0\\-(m+1) & 0 & 2(m+1)\end{pmatrix}x$$
The first thing I have tried is to verify that zero is a point of equilibrium and clearly it is, now what follows is to see that the matrix above has all the eigenvalues ​​with a negative real part, this is equivalent to the definition of stability in the sense of Liapunov? Could someone help me understand this and tell me if I'm confused? Thank you very much.
Let $\phi_t$ denote the flow of the differential equation $\dot{x}=f(x)$ defined
for all $t\in \mathbb{R}$. An equilibrium point $x_o$ of $\dot{x}=f(x)$ is stable if for all $\epsilon > 0$ there exists a $\delta > 0$ such that for all $x\in N_{\delta}(x_0)$ and $t > 0$ we have $\phi_t (x)\in N_{\epsilon}(x_0)$
 A: This system is a LTI system $\dot x=Ax$,
$$
A=\left(\begin{array}{ccc}
-(m+1)&0&2(m+1)\\
0&-2&0\\
-(m+1)&0&2(m+1)\\
\end{array}\right).
$$
A LTI system is 


*

*asymptotically stable if and only if 
all the eigenvalues of $A$ have negative real part;

*stable (but not asymptotically stable) if and only if 
all the eigenvalues have non positive real part and the eigenvalues with 
zero real part have corresponding Jordan blocks of size $1$;

*unstable if and only if there exists at least one eigenvalue with 
positive real part or a Jordan block of size $>1$ corresponding to an eigenvalue with 
zero real part 


The eigenvalues of $A$ are $-2$, $0$ and $m+1$, thus, if $m>-1$ then the system is unstable. If $m<-1$ then $A$ is diagonalizable (as it has distinct eigenvalues), therefore, all of its Jordan blocks are of size 1 and the system is stable (but not asymptotically stable).  In the case when $m=-1$
$$
A=\left(\begin{array}{rrr}
0&0&0\\
0&-2&0\\
0&0&0\\
\end{array}\right);
$$
it is already in Jordan normal form and  all of its Jordan blocks are also of size 1.
