Math for ME · Chapter 10 of 19 · Intermediate

Eigenvalues, Eigenvectors, and Modes

Every structure has preferred ways to vibrate. Eigenvalues are their frequencies; eigenvectors are their shapes. This chapter matters as much as any in the course.

The thread: Inside a matrix hide special directions it only stretches, never turns. Those directions are a structure's vibration modes and the key to whether it is stable.

01

Readiness check

From Linear Systems. Tick only what you can do closed-notes.

  • Multiply a matrix by a vector quickly and correctly.
  • Compute a 2×2 determinant.
  • Solve a quadratic equation.
  • Explain what a homogeneous system Ax = 0 asks.
  • Recognize a symmetric matrix on sight.
0 or 1 weak itemsContinue with this chapter.
2 weak itemsRedo the Linear Systems ladder; eigen-analysis is built from those moves.
3 or more weak itemsStep back to Linear Systems and complete it fully.
02

The core idea

An eigenvector is a direction the matrix does not turn, only stretches. The stretch factor is the eigenvalue.

Av = λvdet(A − λI) = 0

The characteristic equation (right) finds the λ values; substituting each back finds its vector. For vibrating systems, the eigenvalues of the mass-stiffness problem are ω² (squared natural frequencies) and the eigenvectors are the mode shapes.

The skill works when: the system is linear: small vibrations, elastic structures, linearized controls. Symmetric matrices (the usual case) give real eigenvalues and perpendicular modes.
The skill breaks down when: amplitudes grow large (nonlinearity) or damping couples the modes; then eigen-analysis is a starting point, not the answer.
The concept. Most directions get rotated by a matrix. Eigen-directions only scale. Structures vibrate naturally along exactly those directions.
03

The skills, taught in order

10.1 The eigenproblem

For a square matrix A, an eigenvector v is a direction that A only stretches, never turns, and the eigenvalue λ is the stretch factor:

Av = λv

Rewriting this as (A − λI)v = 0 shows that a nonzero v can exist only when A − λI is singular, which is exactly how the eigenvalues are found.

10.2 The procedure

Three steps, the same every time:

StepDo
1. Eigenvaluessolve det(A − λI) = 0, the characteristic equation
2. Eigenvectorsfor each λ, solve (A − λI)v = 0 for the direction v
3. Verifymultiply back and check Av = λv

An eigenvector is a direction, so any nonzero multiple of it is the same eigenvector; pick a convenient scaling.

10.3 Symmetric and positive-definite matrices

Structural stiffness and mass matrices are symmetric, and symmetry is a gift: it guarantees real eigenvalues and mutually perpendicular eigenvectors. If, in addition, every eigenvalue is positive, the matrix is positive definite, the mathematical signature of a stable elastic structure that stores energy for any deflection.

10.4 Natural frequencies and mode shapes

Free vibration of a mass-stiffness system reduces to the eigenproblem Kv = ω²mv. Each eigenvalue is a squared natural frequency and each eigenvector is a mode shape:

λ = ω²

Take the square root for the angular frequency, then divide by 2π for hertz. This is exactly what a modal analysis in any FEA package reports.

10.5 Eigenvalues decide stability

For a linear dynamic system, solutions behave like eλt, so the sign of the real part of each eigenvalue is the verdict:

EigenvalueBehaviour
negative real partdecays: stable
positive real partgrows: unstable
imaginary part presentoscillates at that frequency

Engineering connection: Vibrations, Controls, FEA, structural dynamics.

04

Worked example: two masses, two modes

Two 1 kg masses are connected wall-spring-mass-spring-mass-spring-wall, every spring k = 100 N/m. Find the natural frequencies and mode shapes.

Figure 1. The governing model and its two modes: in-phase (slow) and out-of-phase (fast).
  1. ProblemFind ω₁, ω₂ and the mode shapes for the chain in Figure 1.
  2. Given / findm = 1 kg, k = 100 N/m, stiffness matrix K = [2k, −k; −k, 2k]. Find eigenvalues and eigenvectors of K/m.
  3. AssumptionsSmall horizontal motion, no damping, linear springs.
  4. ModelFree vibration x = v sin(ωt) turns the equations of motion into the eigenproblem Kv = ω²mv: eigenvalues of K/m are ω².
  5. Equationsdet(K − ω²mI) = 0
  6. SolveWith a = ω²m/k: det[2−a, −1; −1, 2−a] = (2−a)² − 1 = 0, so a = 1 or a = 3. Then ω² = 100 or 300: ω₁ = 10 rad/s (1.59 Hz) with v₁ = (1, 1), and ω₂ = 17.3 rad/s (2.76 Hz) with v₂ = (1, −1).
  7. CheckMultiply back: K(1,1) = (k, k) = 100·(1,1): λ = 100 with m = 1. Mode logic: moving together never stretches the middle spring (lower frequency); moving apart works all three springs (higher frequency). The symmetric K gave perpendicular modes: (1,1)·(1,−1) = 0.
  8. ConclusionIf this machine is excited near 1.59 Hz or 2.76 Hz it will resonate in exactly these shapes. Modal analysis in FEA reports precisely this pair of lists: frequencies and shapes.
Result. ω₁ = 10 rad/s, mode (1, 1); ω₂ = 17.3 rad/s, mode (1, −1).
04b

Worked example 2: is this system stable?

A linearised system evolves as ẋ = Ax with A = [0, 1; −2, −3] (a damped oscillator written in state form). Find its eigenvalues and decide whether the system is stable.

  1. Given / findA = [0, 1; −2, −3]. Find the eigenvalues and the stability verdict.
  2. ModelSolutions grow or decay like eλt, so stability is read from the signs of the eigenvalues' real parts.
  3. Characteristic equationdet(A − λI) = det[−λ, 1; −2, −3−λ] = (−λ)(−3−λ) − (1)(−2) = λ² + 3λ + 2.
  4. Solveλ² + 3λ + 2 = (λ + 1)(λ + 2) = 0, so λ = −1 and λ = −2.
  5. VerdictBoth eigenvalues are real and negative, so every solution decays: the system is stable.
  6. CheckThe two decay rates are e−t and e−2t; the slower one, e−t, sets the settling time. No positive or purely imaginary eigenvalue appears, so there is neither growth nor undamped oscillation.
  7. ConclusionThe same det(A − λI) = 0 that gives vibration frequencies also gives stability. Controls reads pole locations this way to decide whether a feedback design settles or runs away.
Result. λ = −1, −2: both negative, so the system is stable and settles without oscillating.
05

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Eigenvector "solved" as a unique vectorConfusion when (2, 2) also works"What does Av = λv say about scaled v?"Eigenvectors are directions: any nonzero multiple is the same mode. Normalize by convention.
Expecting unturned directions from every matrix to be obviousGuessing modes instead of solving"Did I actually solve det(A − λI) = 0?"Intuition guesses; the characteristic equation decides. Verify by multiplying back.
Frequency confused with eigenvalueHz reported where ω² was computed"Is my λ a frequency, or a frequency squared?"For Kv = ω²mv: λ = ω². Take the root, then divide by 2π for Hz.
Sign of eigenvalue ignored in stabilityA growing solution called stable"Does eλt grow or decay for my λ?"Positive real part grows (unstable); negative decays. The sign is the verdict.
06

Practice ladder

Level 1 · Direct skill

Find the eigenvalues of A = [3, 1; 1, 3].

Show answer

det(A − λI) = (3−λ)² − 1 = 0, so λ = 2 and λ = 4. Symmetric matrix, real eigenvalues.

Then find the eigenvalues of the triangular matrix B = [5, 2; 0, 3].

Show answer

For a triangular matrix the eigenvalues are the diagonal entries: λ = 5 and λ = 3, since (5−λ)(3−λ) = 0.

Level 2 · Mixed concept

For the same A, find both eigenvectors and verify they are perpendicular.

Show answer

λ = 2: (A − 2I)v = 0 gives v = (1, −1). λ = 4: v = (1, 1). Dot product: 1 − 1 = 0. Perpendicular, as symmetry guarantees.

A system has eigenvalues λ = −2 and λ = +1. Is it stable, and which mode dominates the long-term behaviour?

Show answer

Unstable: the λ = +1 mode grows like et and eventually dominates, while the λ = −2 mode decays. A single positive eigenvalue is enough to make a system unstable.

Level 3 · Independent problem

Re-solve the worked example with the middle spring doubled (k, 2k, k pattern: K = [3k, −2k; −2k, 3k], m = 1, k = 100). What happened to each mode and why?

Show answer

(3−a)² − 4 = 0 gives a = 1 and a = 5: ω₁ = 10 rad/s unchanged, ω₂ = √500 = 22.4 rad/s. Mode 1 (together) never works the middle spring, so stiffening it changes nothing; mode 2 (opposite) works it hard, so it speeds up. Mode shapes explain hardware changes.

Level 4 · Transfer to real engineering

Film a real two-mass system (two phones on a ruler bridge, two masses on elastic bands) and excite it gently. Identify the in-phase and out-of-phase motions, estimate both frequencies from the video, and compare their ratio to the ideal √3.

What good work looks like

Frequency estimates from frame counting, the measured ratio against √3 ≈ 1.73, and a sentence on why reality deviates (unequal masses, damping, support flex).

07

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Here are my eigenvalues and vectors. I will verify by multiplying Av; check only whether my verification logic is sound."
"Describe a physical change to a two-mass system; I will predict which mode's frequency moves before computing."
"Find the eigenvalues." The characteristic-equation routine must be automatic before Vibrations.
"What are the modes of this structure?" Predict from physics first; compute second; software third.

Portfolio task

Write a one-page modal study of the worked example: the eigenproblem solved by hand, both modes sketched, the Level 3 modification analyzed, and a closing paragraph on what a designer does with mode shapes (move resonances away from excitation frequencies).

Must include: the multiply-back verification for both modes and the orthogonality check.
08

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Define eigenvalue and eigenvector in one sentence each.

An eigenvector is a direction a matrix maps onto itself; the eigenvalue is the scale factor applied along it.

2. How do you find eigenvalues, mechanically?

Solve det(A − λI) = 0, the characteristic equation; one λ per root.

3. What two gifts do symmetric matrices give?

Real eigenvalues and mutually perpendicular eigenvectors: the reason structural modes are clean.

4. Connect eigenvalues to natural frequency.

For Kv = ω²mv, each eigenvalue is a squared natural frequency and its vector is the mode shape.

5. How do eigenvalues decide stability?

Solutions behave like eλt: eigenvalues with positive real part grow (unstable), negative decay (stable).

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-solve the two-mass example, including verification.
+3 daysThe Level 3 stiffened system from a blank page.
+7 daysMixed set: an eigenproblem plus a Linear Systems elimination.
+30 daysWatch for ω = √(k/m) reappearing in the ODE oscillator.
09

Textbook mapping

ItemMapping
Main sourceKreyszig, Advanced Engineering Mathematics, Ch 8 (matrix eigenvalue problems)
Core topics10.1 Eigenvalue idea · 10.2 Eigenvector idea · 10.3 Characteristic equation · 10.4 Diagonalization (intro) · 10.5 Symmetric matrices · 10.6 Positive definiteness · 10.7 Frequencies and modes · 10.8 Stability · 10.9 Mass-spring examples
Engineering connectionVibrations, Controls, FEA modal analysis, structural dynamics, buckling.
Skip on first passComplex eigenvalues beyond the stability reading, Jordan forms, repeated-eigenvalue pathology.
Read nextComplex Numbers for Engineers.