Differential equation system changing the eigen-values The following is a problem from a past exam,  that I couldn't answer (or at least I didn't knew how to answer).
Consider $\Omega:=\begin{pmatrix} \alpha & 1 \\ 0 & \alpha \end{pmatrix}$,
$P,A\in M_{2x2}(\Bbb R)$, where $P$ is an invertible matrix and $A=P\,\Omega P^{-1}$. 
Let $\Phi$ be the matrix given by: $\Phi(t)=\begin{pmatrix} \Phi_{11}(t) & \Phi_{12}(t) \\ \Phi_{21}(t) & \Phi_{22}(t) \end{pmatrix}=P\begin{pmatrix} e^{\alpha t} & te^{\alpha t} \\ 0 & e^{\alpha t}\end{pmatrix}P^{-1}$.
They asked to find the values of $\alpha$ such that: (a) the flow of the equation expands the volume*, (b) all the solutions thend to infinite when $t\to \infty$, (c) the equation has nonzero bounded solutions.
I'm not sure I understood what they asked, or what I was suppose to say. In (a), I just wrote "When $\alpha>0$", but I don't understand what it means with expands the volume. We know that the phase portrait should look something like this:
 
(In the picture, $\alpha>0$).
In (b), I also wrote "When $\alpha>0$", because if $\alpha<0$ then the phase prortrait should look like: 

And finally for (c), I just said when $\alpha=0$, because I thought that's when the phase portrait is bounded.
It's not hard to belive that I might have gotten everything wrong, I want to know how this is suppose to be.
*For (a), I don't know if that's the correct interpretation of what they asked in spanish: el flujo de la ecuación expande el volumen.
 A: (a) Think of a small region in the $xy$ plane.  As time goes on each point moves, according to the system of differential equations, thus $\pmatrix{x\cr y\cr}$ evolves to $\pmatrix{\widetilde{x}\cr \widetilde{y}} = \Phi(t) \pmatrix{x\cr y\cr}$.  The matrix $\Phi(t)$ is the Jacobian matrix for this transformation, and its determinant gives you the factor by which volume is multiplied.  In this case
$\det \Phi(t) = \det \pmatrix{e^{\alpha t} & t e^{\alpha t}\cr 0 & e^{\alpha t}\cr} = e^{2\alpha t}$, so you were correct: the volume expands if $\alpha > 0$.
It stays constant if $\alpha = 0$ and it contracts if $\alpha < 0$.
(b): Actually the answer is that there are no such $\alpha$, because of the constant solution $(0,0)$.
But that's probably not what was meant.  If you ask for what $\alpha$ all
nonzero solutions go to $\infty$ as $t \to +\infty$, the answer is again $\alpha > 0$.  Note that in this case for $\alpha = 0$ most solutions do go
to $\infty$, but those on the line through the eigenvector $P \pmatrix{1\cr 0\cr}$ stay constant.
(c): Yes, $\alpha = 0$.  Again you got the right answer, but for the wrong reason.
For $\alpha < 0$ all solutions go to $\infty$ as $t \to -\infty$, for $\alpha > 0$ they go to $\infty$ as $t \to +\infty$.  For $\alpha = 0$, most solutions go to $\infty$ (both for $t \to +\infty$ and $t \to -\infty$), but as I mentioned above there is a line of constant solutions.
