I have came to certain problem in a differential question, but I fail to solve it. I look for sufficient conditions to make $q$ to converge to zero.
\begin{equation} \alpha \dot{q} + \lambda q = 0 \end{equation}
$\alpha$, $\lambda$ $\in \mathbb{R}^{m \times n}$. I have already accomplished to conjecturate a necessary condition: if $T = \{t \in \mathbb{R}^+ \, \mid \,\dot{q} = e^{Ct} q(0)\}$, than, for $\theta = \begin{bmatrix} \alpha & \lambda \end{bmatrix}$ and $G = \begin{bmatrix} C \\ I \end{bmatrix}$, hence $S = \{ \alpha, \lambda \in \mathbb{R}^{m \times n} \, | \, \theta \, G = 0 \}$. It is a $\mathbf{necessary}$ condition since, though $T \Rightarrow S$, the opposite argument is not true ($S \nRightarrow T$). I thank sincerely for the help.
Best regards, Bruno
Let me change a little your notation: you are asking for conditions on a system of differential equations $\alpha \dot{q}(t)+\lambda q(t)=0$, where $\alpha=\begin{bmatrix} \alpha_a &\mathbb{1}\\ \mathbb{1} &\alpha_u \end{bmatrix}$ and $\lambda=\begin{bmatrix} \lambda_a &\mathbb{1}\\ \mathbb{1} &\lambda_u \end{bmatrix}$, such that $q(t)\xrightarrow{t\to\infty} 0$.
Well, supposing $\alpha$ is invertible, the original equations are equivalent to $\dot{q}(t)=\left(\alpha^{-1}\lambda\right) q(t)$, which has as solution $q(t)=q(0) e^{\left(\alpha^{-1}\lambda\right) t}$. Hence, your question would be when $ e^{\left(\alpha^{-1}\lambda\right) t}\xrightarrow{t\to\infty} 0$. One simple necessary and sufficient contition that warranties that $ e^{\left(\alpha^{-1}\lambda\right) t}\xrightarrow{t\to\infty} 0$ is that all the real parts of the eigenvalues of $\alpha^{-1}\lambda$ are negative.