Stability of these fixed points Say we have the set of nonlinear equations, where $\alpha>0$:
$$\begin{matrix}
\frac{dx}{dt}=x[1-\alpha x-y]\\ 
\frac{dy}{dt}=y[1-x-\alpha y]
\end{matrix}$$
I have determined that the fixed points of this system when $\alpha \neq 1$ are $$(0,0), (0,\frac{1}{\alpha}), (\frac{1}{\alpha},0), (\frac{\alpha (1-\alpha)}{1-\alpha ^{2}},\frac{1-\alpha}{1-\alpha^{2}})$$
And the Jacobian is $$J= \begin{pmatrix}
1-2\alpha x-y& -x\\ 
-y & 1-x-2\alpha y
\end{pmatrix}$$
So, the fixed point $(0,0)$ has eigenvalues $\lambda=1,1$, so this fixed point is unstable.
But what about the stability of the other fixed points? For example, for the fixed point $(0,\frac{1}{\alpha})$ the Jacobian becomes 
$$J= \begin{pmatrix}
\frac{\alpha -1}{\alpha}& 0\\ 
-\frac{1}{\alpha} & -1
\end{pmatrix}$$
which gives eigenvalues of $\lambda=\frac{\alpha -1}{\alpha},-1?$ But I am struggling at this point because the first eigenvalue could be positive or negative depending on the value of $\alpha$, correct? Have I made a mistake somewhere or am I missing something?
Thanks.
 A: There is no problem with your answer. Just do it on a case by case basis depending on $\alpha$. For example:
$$\bigg(0,\frac 1\alpha\bigg) \text{ is a }\begin{cases} \text{saddle} & \alpha<0 \\ \text{stable node} & 0<\alpha <\frac 12 \\ \text{stable inflected node } & \alpha = \frac 12 \\ \text{stable node} & \frac 12<\alpha < 1 \\ \text{saddle} & \alpha > 1\end{cases}$$
where of course, a saddle is unstable by definition.
When $\alpha = 1$, you get a zero eigenvalue, so you will have to expand out the center manifold in order to determine the stability in this case.

EDIT:
Have you been taught how to find the eigenvalues for a general matrix?
For any matrix $A$, $\lambda$ is an eigenvalue of $A$ if and only if $\lambda$ satisfies $\det(A-\lambda I)=0$
In particular, the determinant of a $2 \times 2$ matrix is
$$ \det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad-bc$$
Here, the Jacobian is
$$J = \begin{pmatrix}
-\frac{\alpha}{\alpha+1} & -\frac{1}{\alpha+1} \\
-\frac{1}{\alpha+1} & -\frac{\alpha}{\alpha+1}
\end{pmatrix}
$$
So we want to find $\lambda$ such that $\det(J-\lambda I)=0$.
\begin{align}
0 & =\det(J-\lambda I) \\
& = \det \begin{pmatrix} -\frac{\alpha}{\alpha+1}-\lambda & -\frac{1}{\alpha+1} \\
-\frac{1}{\alpha+1} & -\frac{\alpha}{\alpha+1}-\lambda
\end{pmatrix} \\
& = \bigg(-\frac{\alpha}{\alpha+1}-\lambda\bigg)\bigg(-\frac{\alpha}{\alpha+1}-\lambda\bigg) - \bigg(-\frac{1}{\alpha+1}\bigg)\bigg(-\frac{1}{\alpha+1}\bigg) \\
& = \bigg(\lambda + \frac{\alpha}{\alpha+1}\bigg)^2 - \bigg(\frac{1}{\alpha+1}\bigg)^2 \\
\end{align}
Solving for $\lambda$, we find that
$$\lambda = \color{red}{-}1 \qquad \text{ or } \qquad \lambda = \color{red}{-}\frac{\alpha-1}{\alpha+1}$$
I leave it up to you to do the analysis.
