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Given a matrix $A \in \Bbb R^{3 \times 3}$ , which is non-constant of the form $$y'(t)=A(t)y(t)$$ I want to check the stability of the trivial solution.

My problem is the following:

If $$y(t)=\begin{pmatrix}y_1(t) \\ y_2(t) \\ y_3(t)\end{pmatrix}$$ My matrix consists not of $t$ but $y_1,y_2,y_3$ and I wanted to know how do I show this ? I mean I can't calculate the eigenvalues because of the $y_i's$, this method is only for constant matrices as far as I know.

Edit: My matrix looks like: $$\begin{pmatrix}2(1-y_2-y_1) & 1 & 0 \\ -y_2 & -1 & -2 \\ 1&0 & -1\end{pmatrix}$$

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    You need to be more specific about what is $A(t)$. – 2017-01-27
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    @JohnB: Put it in Edit, Thanks – 2017-01-28
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    You should try to find a positive definite function $V(y)$, e.g., $V(y)=\sum_i y^2_i$ s.t. $\frac{d}{dt}V(y)<0$. If you manage to find such a function, the system is stable. It seems, however, that a mere sum of squares does not work. – 2017-01-28

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The linearization of the system at the origin gives the equation $$ \begin{pmatrix}y_1'\\y_2'\\y_3'\end{pmatrix}= \begin{pmatrix}2 & 1 & 0 \\ 0 & -1 & -2 \\ 1&0 & -1\end{pmatrix} \begin{pmatrix}y_1\\y_2\\y_3\end{pmatrix}. $$ The eigenvalues of the matrix are $0$ and $\pm\sqrt3$. Since one of the eigenvalues is positive (or if you prefer, since one of the eigenvalues has positive real part), the origin is unstable. So the original system has an unstable manifold and the origin is again unstable.

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    @ John B: Thanks haven't thought of just plugging in 0 :D – 2017-01-30