Let $A=A\left(t_0\right)$ for some fixed $t_0$. Let $\delta \in(0, \pi)$ denote the angle between the eigenvector corresponding to $\lambda_1$ and that corresponding to $\lambda_2$. Show that there exists an invertible matrix $Q$ such that $$ Q A Q^{-1}=B:=\left[\begin{array}{cc} \lambda_1 & \left(\lambda_2-\lambda_1\right) \cot (\delta) \\ 0 & \lambda_2 \end{array}\right] $$ This question is a part of the journey to prove: $$ \|x(t)\| \rightarrow \infty \text { as } t \rightarrow \infty, $$ even though $A(t)$ has negative eigenvalues for each fixed $t$. Assume that $A(t)$ is a $2 \times 2$ matrix with real, distinct eigenvalues, $\lambda_1(t)<\lambda_2(t)<0$ for each $t$.
My try: Let $\mathcal{B}$ denote the set of $2 \times 2$ matrices with eigenvalues of negative real part such that $x \cdot B x>0$ for some $x \in \mathbb{R}^2$. I have proved that if $(1)$ occurs, then $A(t) \in \mathcal{B}$ for some $t>0$.
Now suppose the eigenvector corresponding to $\lambda_1$ is $u_1$ and that corresponding to $\lambda_2$ is $u_2$, then $\cos(\delta)=\frac{u_1^Tu_2}{||u_1||||u_2||}.$ Now I am really stuck about how to pull this off to the $A_{12}$ place with $\cot(\delta)$.