$$
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Modal Analysis
Consider a conservative \(n\)-dimensional linear mechanical system with mass
matrix \(M > 0\) and stiffness matrix \(K \geq 0\):
\[M\ddot{x} + Kx = 0\]
Then \(\inv{M} K\) is self-adjoint for the inner-product \(M\), which means
there exists an \(M\)-orthogonal basis \(B\) of eigenvectors:
\[\inv{M}K = BS\inv{B}\]
for some diagonal matrix \(S\). By definition of \(B\), the mass matrix in this
basis is the identity:
\[B^T M B = I\]
In this basis, the stiffness matrix becomes diagonal:
\[B^T K B = \underbrace{B^T M B}_I S \underbrace{\inv{B} B}_I = S\]
and the linear system may be understood as a sum of mechanically independent
(that is, \(M\)-orthogonal) one-dimensional subsystems:
\[\ddot{y} + S y = 0\]
where \(y = Bx\).
TODO eigenfrequencies
TODO mass-spring/mesh laplacian
Computation
From a Cholesky factorization \(M = LL^T\) and the \(M\)-orthogonality of \(B\)
one can check that \(L^TB = U\) for some orthogonal matrix \(U \in O(n)\). To
compute either \(B\) or \(U\) and \(S\) there are several options, ordered by
decrasing order of efficiency/numerical stability:
- solve the generalized eigenvalue problem \(M, K\) to obtain \(B, S\) and then \(U\)
- solve the regular eigenvalue problem on \(M^{-1}K\)
- notice that \(USU^T = UB^TKBU^T = L^{-1} K L^{-T}\) since \(B^T = U^TL^{-1}\)
and obtain \(U, S\) by diagonalization of \(L^{-1} K L^{-T}\)
Notes
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