It is well known the differential equation related to a single degree of freedom mass-spring-damper system,
$$ m \ddot{x} + d \dot{x} + k x = 0 , $$
where, $m > 0$ , $d > 0$ and $k > 0$ , can be written as a first-order matrix differential equation by choosing $\xi = [x \quad \dot{x}]^T$:
$$ \dot{\xi} = \underbrace{\begin{bmatrix} 0 & 1 \\ -\frac{k}{m} & -\frac{d}{m} \end{bmatrix}}_{\mathbf{A}} \xi. $$
The matrix $\mathbf{A}$ is diagonalizable if $k \ne \frac{d^2}{4 m}$ since its eigenvectors become linearly independent.
Now, given the following second-order matrix differential equation,
$$ \mathbf{M} \ddot{\mathbf{x}} + \mathbf{D} \dot{\mathbf{x}} + \mathbf{K} \mathbf{x} = \mathbf{0}, $$
with $\mathbf{x} \in {\rm I\!R}^n$ , $\mathbf{M} \in {\rm I\!R}^{n \times n}$ , $\mathbf{D} \in {\rm I\!R}^{n \times n}$ and $\mathbf{K} \in {\rm I\!R}^{n \times n}$, the same can be done by letting $\boldsymbol{\xi} = [\mathbf{x} \quad \dot{\mathbf{x}}]^T$:
$$ \dot{\boldsymbol{\xi}} = \underbrace{\begin{bmatrix} \mathbf{O} & \mathbf{I} \\ -\mathbf{M}^{-1} \mathbf{K} & -\mathbf{M}^{-1} \mathbf{D} \end{bmatrix}}_{\tilde{\mathbf{A}}} \boldsymbol{\xi}. $$
Assuming the matrices $\mathbf{M}$ , $\mathbf{D}$ and $\mathbf{K}$ to be positive-definite, what can be said about the diagonalizability of the matrix $\tilde{\mathbf{A}}$? Am I missing something that is easy to notice?
Thank you so much in advance.