For which values of $\alpha$ is the solution stable? If we have the following differential equation
$X'=\begin{bmatrix} 1 & 1 \\ -1 & \alpha  \end{bmatrix}X$
for which values of $\alpha$ is the zero solution stable?
My attempt: For stability we need the real part of all eigenvalues to be negative and geometric and algebraic multiplicity to be the same.I tried computing the eigenvalues and see the different cases but it is an incredible amount of work and I got stuck.
Would there be any simpler and nicer way to do this(with knowledge of a first course in ODE)?
Thanks
 A: This question is essentially about linear algebra, namely, analyzing the eigenvalues of the matrix, call it $A$, as a function of the (presumptively real) parameter $\alpha$.
The characteristic polynomial of the matrix, whose roots are the eigenvalues $\lambda, \mu$ of $A$, is
$$p(t) := \det(t I - A) = t^2 - (\alpha + 1) t + (\alpha + 1) .$$ Thus, $\lambda + \mu = \lambda \mu = \alpha + 1$. But if both $\lambda$ and $\mu$ have negative real part, then $\alpha + 1 = \lambda + \mu < 0 < \lambda \mu = \alpha + 1$, a contradiction.
Remark The claim is in particular valid even when the roots $\lambda, \mu$ are not real. In that case---which occurs precisely when the discriminant $\alpha^2 - 2 \alpha - 3 = (\alpha + 1)(\alpha - 3)$ of $p(t)$ is negative, i.e., when $\alpha \in (-1, 3)$---the roots are complex conjugates, say, $\beta \pm \gamma i$, where $\beta, \gamma \in \Bbb R$. If the common real part, $\beta$, of the roots is negative, then their sum, $\alpha + 1 = 2 \beta$, is negative but their product, $\alpha + 1 = \beta^2 + \gamma^2$, is positive, a contradiction.
