Modal analysis

Remarks

Such a basis exists by simultaneous diagonalisation of the symmetric matrices $M$ and $K$, with $M$ positive definite. The decoupled equations follow by substituting $x$ into the system and multiplying on the left by $v_k^T$. If $K$ is positive semidefinite, $\lambda_k = \omega_k^2 \ge 0$ with natural frequencies $\omega_k$. With a damping matrix $C$, the equations decouple in the same way when $v_j^T C v_k = 0$ for $j \ne k$, for example for proportional damping $C = \alpha M + \beta K$.

Example: $M = I_2$, $K = \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}$, $v_1 = \frac{1}{\sqrt 2}(1, 1)$, $v_2 = \frac{1}{\sqrt 2}(1, -1)$, $\lambda_1 = 1$, $\lambda_2 = 3$. The modal coordinates $\eta_1 = (x_1 + x_2)/\sqrt 2$ and $\eta_2 = (x_1 - x_2)/\sqrt 2$ satisfy $\ddot \eta_1 + \eta_1 = (f_1 + f_2)/\sqrt 2$ and $\ddot \eta_2 + 3 \eta_2 = (f_1 - f_2)/\sqrt 2$.

Non-example: the standard basis $e_1, e_2$ in the same system is not a basis of mode shapes, since $K e_1 = (2, -1)^T$ is not a multiple of $e_1$. In these coordinates the equations stay coupled.