Comparison to linear algebra
In linear algebra, it is more common to see the standard form of an eigensystem which is
expressed as:
![{\displaystyle [A][x]=[x]\lambda }](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6cc02b192ec5e73eceb5cfd676486440debe07a)
Both equations can be seen as the same because if the general equation is
multiplied through by the inverse of the mass,
,
![{\displaystyle \mu ={\frac {1}{\lambda }}=\sum _{i=1}^{k}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/449609e8aa68b2c385154ad0e2e6d006dc04508a)
![{\displaystyle \chi ^{2}=\sum _{i=1}^{k}{\frac {(X_{i}-np_{i})^{2}}{np_{i}}}=\sum _{i=1}^{k}{\frac {(X_{i}-b)^{2}}{b}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e0d0a7d657b615072a5b9d59b2ab07c083282cf)
![{\displaystyle \chi ^{2}=\sum _{i=1}^{k}{\frac {(X_{i}-np_{i})^{2}}{np_{i}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f913da525ce2eab906ae72cf9c39c4f92d9a3c6f)
![{\displaystyle p_{i}={\frac {e^{-(x-\mu )^{2}/(2\sigma ^{2})}}{\sqrt {2\pi \sigma ^{2}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/285c99c931b1dcf8c391320f7821f1d85c8e63f0)
![{\displaystyle P(x)={\frac {1}{2}}{\Big [}1+\operatorname {erf} {\Big (}{\frac {x-\mu }{\sqrt {2\sigma ^{2}}}}{\Big )}{\Big ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ec95f9f58c6b3e86b5fcae7b4e08e1ea160391b)
![{\displaystyle Probabilty(V>observed)=Q_{KP}([{\sqrt {N}}+0.155+0.24/{\sqrt {N}}]D)=Q_{KP}(\lambda )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d827d0e12ccccbafe4f7d7fc8fff40d7bd189189)
![{\displaystyle Q_{KP}(\lambda )=\sum _{j=1}^{\infty }(4j^{2}{\lambda }^{2}-1)e^{-2j^{2}{\lambda }^{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/97bb0966be4089c071987698dedc2a73e678c481)
![{\displaystyle V=D_{+}+D_{-}~=\max _{-\infty <x<\infty }[S_{N}(x)-P(x)]+\max _{-\infty <x<\infty }[P(x)-S_{N}(x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe32e94b6f0d2a496490ef6dc9042947a7cdb55f)
![{\displaystyle {\begin{bmatrix}-13.73&0.85&0.15&0.50&-0.10&0.13&-0.99\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c510659d24a4d243dee418ba7315481e5fe28ce6)
![{\displaystyle {\begin{bmatrix}1&-2&-3&-4&0&0\\0&3&2&1&1&0\\0&2&5&3&0&1\end{bmatrix}}{\begin{bmatrix}Z\\x\\y\\z\\s\\t\end{bmatrix}}={\begin{bmatrix}0\\10\\15\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9d121d7cf434b3f4c5295a8718f10009f68d6bb6)
but the eigenvectors are the same.
Methods of solution
For linear elastic problems that are properly set up (no rigid body rotation or translation),
the stiffness and mass matrices and the system in general are positive definite.
These are the easiest matrices to deal with because the numerical methods commonly
applied are guaranteed to converge to a solution. When all the qualities of the system are
considered:
- Only the smallest eigenvalues and eigenvectors of the lowest modes are desired
- The mass and stiffness matrices are sparse and highly banded
- The system is positive definite
a typical prescription of solution is first to tridiagonalize the system using the
Lanczos algorithm. Next, use the QR algorithm to find the eigenvectors and eigenvalues of
this tridiagonal system. If inverse iteration is used, the new eigenvalues will
relate to the old by
, while the eigenvectors of the original can
be calculated from those of the tridiagonalized matrix by:
![{\displaystyle [r^{n}]=[Q][v^{n}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/84fdda88501ac45f289796e7177c1453e55d808c)
where
is a Ritz vector approximately equal to
the eigenvector of the original system,
is the matrix
of Lanczos vectors, and
is the
eigenvector
of the tridiagonal matrix.