Consider a basic mechanical system driven by the second order ODEs $$ \mathbf{M\ddot{x}+D\dot{x}+Kx=0}$$ where the mass matrix $\mathbf M$ is positive definite; no specific assumptions other than constant coefficients are made on $\mathbf D$ and $\mathbf K$. The above equation can be recast in the first-order form $$ \mathbf{\dot{y}=Ay}$$ with $$\mathbf A=\begin{bmatrix}\mathbf 0 & \mathbf I \\ -\mathbf{M}^{-1}\mathbf K & -\mathbf{M}^{-1}\mathbf D \end{bmatrix}$$ Are you aware of a mechanical system for which $\mathbf A$ cannot be diagonalized in $\mathbb C$?

  • $\begingroup$ Note that in mechanics, $K$ is always symmetric positive-definite (Maxwell-Betti theorem). Also, assuming $M,K,C$ are real, any real matrix is diagonalisable in $\mathbb{C}$, so I don't think you will find any example. $\endgroup$ – anderstood Jun 23 '16 at 21:13
  • $\begingroup$ If I remember my linear algebra right, any matrix with det $\neq 0$ can be diagonalised over the complex numbers ... $\endgroup$ – Sanya Jun 23 '16 at 21:16
  • $\begingroup$ @Sanya no. See en.wikipedia.org/wiki/Diagonalizable_matrix You might always find the required number of eigenvalues, but the dimension of the associated eigenspace might be smaller, in which case you cannot diagonalize. $\endgroup$ – pluton Jun 23 '16 at 23:17
  • $\begingroup$ @anderstood. If you consider a rotating shaft in a viscous oil film, the linearized dynamics (including the linearized dynamics of the oil film) is such that $\mathbf K$ and $\mathbf D$ have no specific mathematical structure. $\endgroup$ – pluton Jun 23 '16 at 23:19
  • $\begingroup$ @pluton I'm not quite sure how this is possible. What assumption of the Maxwell-Betti's theorem is not satisfied in that case? Maybe it no longer holds for _visco_elasticity, but I am not sure. $\endgroup$ – anderstood Jun 24 '16 at 1:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.