Given the following system of ODEs, it is in my understanding that to sketch trajectories in the $SI$-plane, one has to first find the critical point(s), which is only the point $(0, 0)$ in this case, then linearize the system, that is derive its Jacobian matrix at $(0, 0)$, and then find the eigenvalues of the new system. However, the result is that the eigenvalues are $\lambda = -\beta$ and $\lambda = 0$, so the trajectories appear to be just straight vertical lines, which seems dubious to me.
$S'=-\alpha S I,\\ I'=\alpha S I - \beta I,\\ R'=\beta I$
$S(0)=S_0,\\ I(0)=I_0,\\ R(0)=R_0$
In this system, $S(t), I(t)$, and $R(t)$ represent various kinds of populations, and $\alpha, \beta >0$ are constants.
I would appreciate some advice.
Another approach:
The lines $S=0$ and $I=0$ are invariant. The second is composed of fixed points and on the first you have $I'=-\beta I$ showing that all goes to the origin. Finally, the line $S=\beta/\alpha$ determines almost all that is missing, since $I'=(\alpha S-\beta)I$. You can sketch the phase portrait considering the signs of $S'$ and $I'$ in the six regions determined by the 3 lines.
Notice that all but one of the invariant curves has the line $S=0$ as an asymptote (I write on purpose "invariant curves", not "trajectories").