Given an ODE $$\ddot{\mathbf{x}}=\mathbf{Ax}$$ where $\ddot{\mathbf{x}},\dot{\mathbf{x}}\in\mathbb{R}^n$ and $\mathbf{A}\in\mathbb{R}^{n\times n}$ is the symmetric matrix $$\begin{pmatrix}1&-0.5&0&0&\ldots&-0.5\\-0.5&1&-0.5&0&\ldots&0\\0&-0.5&1&-0.5&\ldots&0\\\vdots&0&\vdots&\vdots&\ddots&\vdots\\0&0&0&-0.5&1&-0.5\\-0.5&0&0&0&-0.5&1\end{pmatrix}$$
How does one determine the stability of the equilibria?
Does it depend on the eigenvalues of $\mathbf{A}$, as it would do if the ODE was $\dot{\mathbf{x}}=\mathbf{Ax}$? If so, how would we compute the eigenvalues of the above $n\times n$ matrix?
Building on advice from the comments section: Let $$\mathbf{u} = \dot{\mathbf{x}}$$ $$\mathbf{v} = \mathbf{x}$$ Then the ODE is reduced to a first order system: $$\dot{\mathbf{u}} = \mathbf{Av}$$ $$\dot{\mathbf{v}} = \mathbf{u}$$
Or, in matrix form:
$$\pmatrix{\dot{\mathbf{u}}\\ \dot{\mathbf{v}}}=\pmatrix{\mathbf{0}&\mathbf{A}\\ \mathbf{I}&\mathbf{0}}\pmatrix{\mathbf{u}\\ \mathbf{v}}$$ From the article https://en.wikipedia.org/wiki/Matrix_differential_equation#Stability_of_the_two-state-variable_case we see that the system is stable iff all the eigenvalues of this new matrix have a negative real part. So we wish to solve $$\det\pmatrix{-\lambda\mathbf{I}&\mathbf{A}\\ \mathbf{I}&-\lambda\mathbf{I}}=0$$ Where do I go from here to find the stabilities?