Module 11 · Lesson 11.1
The general eigenproblem
From two degrees of freedom to any number: what changes is only that the determinant can no longer be expanded by hand.
Why this matters
Module 9 solved a 2DOF system by expanding a determinant into a quadratic. At three degrees of freedom that becomes a cubic, at four a quartic, and beyond that there is no formula at all — and even where a formula exists, extracting roots from a high-degree polynomial is numerically hopeless.
So the practical method is entirely different from the hand method, while the CONCEPT is unchanged. This lesson is about the concept, the properties every correct set of modes must have, and how to tell whether the set your software has returned is credible.
By the end of this lesson you should be able to
- State the general eigenproblem and what its solutions are
- Explain why the polynomial route fails beyond a few degrees of freedom
- List the properties a correct set of modes must satisfy
- Interpret rigid-body modes, repeated frequencies and local modes
What you should already know
- The 2DOF eigenvalue problem (Module 9)
- Assembly of M and K (Module 10)
- Matrix multiplication and the idea of a determinant
The problem, unchanged
For any number of degrees of freedom, free undamped vibration gives
(K − ω²M)φ = 0
and a non-trivial solution requires
det(K − ω²M) = 0
For an n-degree-of-freedom system that determinant expands into a polynomial of degree n in ω², so there are exactly n natural frequencies and n mode shapes.
Why nobody expands the determinant
Two reasons, and both are fatal rather than inconvenient.
There is no formula beyond degree four. Abel proved in 1824 that the general quintic has no solution in radicals. A five-storey shear building is already past the point where a closed form exists.
Polynomial roots are numerically appalling. Even where a formula exists, the roots of a high-degree polynomial are extraordinarily sensitive to its coefficients. Perturb a coefficient in the twentieth significant figure and a root can move in the third. For a fifty-degree-of-freedom system the polynomial route is not merely slow — it produces answers with no correct digits.
So real eigensolvers never form the polynomial. They work directly on the matrices, by iteration: subspace iteration, Lanczos, or — for the small systems in this course — Jacobi rotations. This course's library uses a Cholesky reduction followed by cyclic Jacobi, which is unconditionally stable and returns orthogonal shapes even when frequencies repeat.
What you should take from this: the eigenproblem is a matrix problem, not a polynomial problem. The polynomial is a way of understanding why there are n answers, and it is not how they are found.
Four properties every correct set of modes has
Real and positive eigenvalues. For symmetric positive-definite M and K, every ω² is real and positive. A complex or negative ω² means an error in the matrices — most often an unsymmetric K or a zero on the diagonal of M.
Orthogonality. φᵢᵀMφⱼ = 0 and φᵢᵀKφⱼ = 0 for i ≠ j. This is derived in the next lesson and it is the property everything else depends on.
Completeness. The n mode shapes span the space, so ANY possible deformation of the model can be written as a combination of them. There is no motion the modes cannot describe.
Predictable shape pattern. For a shear building, mode n has exactly n − 1 sign changes down the height. It is a free check that costs nothing to apply.
Three things that look like errors and are not
A rigid-body mode, with ω = 0. The structure can move without storing strain energy. Legitimate for an unrestrained model — a satellite, an aircraft, a bridge deck before its bearings engage. A grounded building that reports one has lost a support, and that IS an error.
Repeated frequencies. A symmetric building has two lateral modes at the same frequency. Any linear combination of two modes with the same frequency is also a mode, so the individual shapes the solver returns are not unique — but the plane they span is. A solver that produces non-orthogonal shapes at a repeated frequency is a poor solver.
A local mode, where one part of the structure moves and the rest hardly does. Common in detailed models: a slender parapet, a plant item on a flexible support, a long unbraced beam. Legitimate, and usually carrying almost no effective mass — so it matters for the local element and not for the global response.
And one thing that IS an error
A mode with a very long period AND almost no effective mass, in a model that should be fully connected. That is a mechanism: a degree of freedom with mass and essentially no stiffness. The mode shape identifies which node it is — everything will be near zero except one entry.
Try it
Every mode this building has
Select a mode and watch it. Then change how it is normalised and watch every number move while the physics does not.
Normalisation
- Mode
- 1
- ωn
- 12.4 rad/s
- fn
- 1.98 Hz
- Tn
- 0.506 s
- Sign changes in the shape
- 0
- Shape as shown
- 0.253, 0.503, 0.728, 0.901, 1.000
- Modal mass with this scaling
- 8.8e5
- Worst orthogonality residual
- 5.6e-16
Mode n always has n − 1 of them for a shear building.
Change the normalisation. Every number in the shape row changes and the ratios between them do not — the roof still moves 3.96 times as far as the first floor whichever scaling you choose. That is why a mode-shape ORDINATE from software means nothing on its own, and why comparing two ordinates from the same mode means everything.
What this shows: A mode shape has an arbitrary magnitude and a meaningful shape. Any multiple of it satisfies the eigenproblem equally, so only the RATIOS between its entries carry information.
Worked example
The three modes of a uniform three-storey building
Given
- Three storeys, each of mass 400 tonnes
- Each storey stiffness 600 MN/m
- Uniform, so the classical closed-form solution applies as a check
Find
All three frequencies and shapes, and every check that can be applied to them.
Assumptions
- Shear building, undamped, lumped mass
Predict first
A modal analysis of a fully restrained ten-storey building returns a first mode with a period of 62 s and an effective mass of 0.05%. What is it?
Practice
A model has 8 degrees of freedom. How many natural frequencies does it have?
Practice
A uniform three-storey shear building has m = 500 tonnes per floor and k = 750 MN/m per storey. What is its first natural frequency, in rad/s? Use the ratio ω₁ = 0.4450√(k/m).
Practice
For that building, what is the third natural frequency in rad/s? Use ω₃ = 1.8019√(k/m).
Check yourself
Why do real eigensolvers not find the roots of det(K − ω²M) = 0 directly?
Worked example
Why the eigenproblem is generalised, not standard
Given
- An MDOF system with mass matrix M and stiffness matrix K
- A proposal to solve K φ = λ φ instead of K φ = ω² M φ
Find
What the shortcut would get wrong
Summary
- (K − ω²M)φ = 0 gives exactly n frequencies and n shapes for n degrees of freedom
- The determinant explains WHY there are n; it is not how they are found
- No closed form exists beyond four degrees of freedom, and polynomial roots are unstable
- Real solvers iterate on the matrices directly — here, Cholesky reduction plus Jacobi
- Every ω² is real and positive for symmetric positive-definite M and K
- Rigid-body modes, repeated frequencies and local modes are all legitimate in context
- A long period WITH negligible effective mass in a restrained model is a mechanism
This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint