Lecture 10: What Survives?

Eigenvalues and eigenvectors — the directions that transformation preserves.

The Hook

Apply this matrix to many vectors:

$$A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$$

Most vectors change direction. But watch these two:

These vectors only get scaled — their direction survives!

v₁ is scaled by 3. v₂ is scaled by 1 (unchanged).

These special directions are eigenvectors. The scale factors are eigenvalues.

Eigenvector Visualizer

2
1
1
2
Trace (a + d)
4
Det (ad - bc)
3
λ₁
3
λ₂
1

Recognition: The SAMENESS Primitive

SAMENESS: "What doesn't change?"

Under a linear transformation, most directions rotate. But some directions are special — they only stretch or compress, never rotate.

Finding these invariant directions reveals the essential structure of the transformation.

When to Use Eigenvalues/Eigenvectors

SituationEigenanalysis gives you
"What directions does this matrix preserve?"Eigenvectors
"By how much are they scaled?"Eigenvalues
"Simplify this matrix?"Diagonalization
"Solve coupled differential equations?"Decouple via eigenvectors
"Find normal modes of vibration?"Eigenvectors of Hessian
"Find molecular orbitals?"Eigenvectors of Hamiltonian
"Long-term behavior of iterated process?"Dominant eigenvalue
"Is this matrix stable?"Check if all |λ| < 1
"Principal components?"Eigenvectors of covariance

Definition

Eigenvector and Eigenvalue

Definition: Let A be an n × n matrix. A nonzero vector v is an eigenvector of A if: $$A\mathbf{v} = \lambda \mathbf{v}$$ for some scalar λ. The scalar λ is the eigenvalue corresponding to v.

Interpretation:

Terminology:

The Eigenspace

Definition: For eigenvalue λ, the eigenspace Eλ is: $$E_\lambda = \{\mathbf{v} : A\mathbf{v} = \lambda\mathbf{v}\} = \ker(A - \lambda I)$$

This is a subspace containing all eigenvectors for λ (plus the zero vector).

Finding Eigenvalues

The Characteristic Equation

Key insight: Av = λv can be rewritten:

$$A\mathbf{v} - \lambda\mathbf{v} = \mathbf{0}$$ $$(A - \lambda I)\mathbf{v} = \mathbf{0}$$

For a nonzero solution v to exist, (A - λI) must be singular:

$$\det(A - \lambda I) = 0$$

This is the characteristic equation.

The Characteristic Polynomial

Definition: The characteristic polynomial of A is: $$p(\lambda) = \det(A - \lambda I)$$

The eigenvalues are the roots of p(λ) = 0.

2×2 Case

For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$:

$$A - \lambda I = \begin{pmatrix} a - \lambda & b \\ c & d - \lambda \end{pmatrix}$$
$$p(\lambda) = (a - \lambda)(d - \lambda) - bc = \lambda^2 - (a+d)\lambda + (ad - bc)$$
Key formula: $$p(\lambda) = \lambda^2 - \text{tr}(A)\lambda + \det(A)$$

The eigenvalues satisfy:

Quadratic formula:

$$\lambda = \frac{\text{tr}(A) \pm \sqrt{\text{tr}(A)^2 - 4\det(A)}}{2}$$
Example: $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$

tr(A) = 4, det(A) = 4 - 1 = 3

$$\lambda = \frac{4 \pm \sqrt{16 - 12}}{2} = \frac{4 \pm 2}{2}$$

λ₁ = 3, λ₂ = 1

Check: 3 + 1 = 4 = tr(A) ✓, 3 × 1 = 3 = det(A) ✓

Special Cases

Triangular matrices: Eigenvalues are the diagonal entries.

$$\begin{pmatrix} a & * & * \\ 0 & b & * \\ 0 & 0 & c \end{pmatrix} \Rightarrow \lambda_1 = a, \lambda_2 = b, \lambda_3 = c$$

Finding Eigenvectors

Once you have eigenvalue λ, find eigenvectors by solving:

$$(A - \lambda I)\mathbf{v} = \mathbf{0}$$

This is a homogeneous system — find the null space of (A - λI).

Example continued: A = [2 1; 1 2], λ₁ = 3
$$A - 3I = \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}$$

Solve: -v₁ + v₂ = 0, so v₁ = v₂.

Eigenvector: v₁ = (1, 1) (or any scalar multiple)


For λ₂ = 1:

$$A - I = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$$

Solve: v₁ + v₂ = 0, so v₂ = -v₁.

Eigenvector: v₂ = (1, -1)

Step-by-Step Eigenvalue Calculator

Properties of Eigenvalues

Algebraic and Geometric Multiplicity

Key Properties

  1. A matrix and its transpose have the same eigenvalues.
  2. Similar matrices have the same eigenvalues: if B = P⁻¹AP, then B and A have the same spectrum.
  3. tr(A) = sum of eigenvalues (counting multiplicity).
  4. det(A) = product of eigenvalues.
  5. A is invertible ⟺ all eigenvalues are nonzero.
  6. Eigenvalues of A⁻¹ are 1/λᵢ (with same eigenvectors).
  7. Eigenvalues of Aᵏ are λᵢᵏ (with same eigenvectors).

Special Matrix Types

Symmetric matrices (A = Aᵀ):

Physical matrices (Hamiltonians, Hessians, covariance matrices) are typically symmetric — the spectral theorem guarantees nice behavior.

Orthogonal matrices (AᵀA = I):

Positive definite matrices:

Complex Eigenvalues

When They Occur

Real matrices can have complex eigenvalues. This happens when:

$$\text{tr}(A)^2 - 4\det(A) < 0$$

The eigenvalues are complex conjugates: λ = α ± βi.

Example: Rotation by θ ≠ 0°, 180°
$$R(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$

tr(R) = 2cos θ, det(R) = 1

$$\lambda = \frac{2\cos\theta \pm \sqrt{4\cos^2\theta - 4}}{2} = \cos\theta \pm i\sin\theta = e^{\pm i\theta}$$

The eigenvalues are complex! (No real direction is preserved by a rotation.)

Complex Eigenvalues: Rotation

60°
λ₁
e^(i60°)
λ₂
e^(-i60°)
|λ|
1

Interpretation

Complex eigenvalues indicate rotation within a plane.

Diagonalization

The Goal

Diagonalization means finding P such that:

$$P^{-1}AP = D = \begin{pmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{pmatrix}$$

The Method

If A has n linearly independent eigenvectors v₁, ..., vₙ:

  1. Form P with eigenvectors as columns: P = [v₁ | v₂ | ... | vₙ]
  2. Then P⁻¹AP = D where D has eigenvalues on diagonal
Why it works:
$$AP = A[\mathbf{v}_1 | \cdots | \mathbf{v}_n] = [A\mathbf{v}_1 | \cdots | A\mathbf{v}_n] = [\lambda_1\mathbf{v}_1 | \cdots | \lambda_n\mathbf{v}_n]$$
$$= [\mathbf{v}_1 | \cdots | \mathbf{v}_n] \begin{pmatrix} \lambda_1 & & \\ & \ddots & \\ & & \lambda_n \end{pmatrix} = PD$$

Therefore: AP = PD ⟹ P⁻¹AP = D

When Diagonalization Fails

A matrix is not diagonalizable if it doesn't have n linearly independent eigenvectors.

This happens when geometric multiplicity < algebraic multiplicity for some eigenvalue.

Example of non-diagonalizable matrix:
$$A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$

Characteristic polynomial: (1-λ)² = 0, so λ = 1 (algebraic multiplicity 2).

Eigenspace: A - I = [0 1; 0 0]. Null space is span{(1, 0)}. Geometric multiplicity = 1.

Only one independent eigenvector — not diagonalizable.

Why Diagonalization Matters

In the eigenvector basis, the transformation is just scaling along axes:

$$D = P^{-1}AP \Rightarrow A = PDP^{-1}$$

Powers: $A^k = PD^kP^{-1}$ (and Dᵏ is trivial — just raise diagonal entries to kth power)

Exponentials: $e^{At} = Pe^{Dt}P^{-1}$ (useful for differential equations)

Chemistry Connection: Molecular Orbitals

The Secular Equation

In Hückel theory, molecular orbital energies are eigenvalues of the Hamiltonian matrix:

$$\mathbf{H}\mathbf{c} = E\mathbf{c}$$

where:

Hückel Molecular Orbital Calculator

Example: Ethylene
$$\mathbf{H} = \begin{pmatrix} \alpha & \beta \\ \beta & \alpha \end{pmatrix}$$

Setting x = (α - E)/β:

$$\det\begin{pmatrix} \alpha - E & \beta \\ \beta & \alpha - E \end{pmatrix} = (\alpha - E)^2 - \beta^2 = 0$$

(α - E) = ±β

E₁ = α + β (bonding), E₂ = α - β (antibonding)

Since β < 0, E₁ < E₂.


Eigenvectors (MO coefficients):

Chemistry Connection: Normal Modes

Vibrational Analysis

The Hessian matrix of second derivatives:

$$H_{ij} = \frac{\partial^2 V}{\partial q_i \partial q_j}$$

Eigenvalue problem: Hq = ω²q

Chemistry Connection: Quantum Mechanics

The Eigenvalue Problem

The time-independent Schrödinger equation:

$$\hat{H}\psi = E\psi$$

This IS an eigenvalue problem:

Allowed energies are eigenvalues. Stationary states are eigenvectors.

Summary: Eigenvalue Toolkit

StepMethod
1. Find eigenvaluesSolve det(A - λI) = 0
2. For each λ, find eigenvectorsSolve (A - λI)v = 0
3. CheckVerify Av = λv

Quick Reference

PropertyFormula
Characteristic polynomialp(λ) = det(A - λI)
Sum of eigenvaluestr(A)
Product of eigenvaluesdet(A)
Eigenvalues of Aᵏλᵢᵏ
Eigenvalues of A⁻¹1/λᵢ
Symmetric ⟹Real eigenvalues, orthogonal eigenvectors

Exercises

Exercise 1: 2×2 Eigenvalues

Find the eigenvalues and eigenvectors of:

(a) $\begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}$

(b) $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$

(c) $\begin{pmatrix} 3 & 1 \\ 0 & 3 \end{pmatrix}$

Show Solution

(a) tr = 7, det = 12 - 2 = 10. λ = (7 ± √(49-40))/2 = (7 ± 3)/2. λ₁ = 5, λ₂ = 2.

For λ = 5: (A - 5I)v = [-1 2; 1 -2]v = 0 → v = (2, 1)

For λ = 2: (A - 2I)v = [2 2; 1 1]v = 0 → v = (1, -1)


(b) tr = 0, det = 1. λ = (0 ± √(-4))/2 = ±i. Complex eigenvalues λ = ±i.

This is rotation by 90° — no real eigenvectors.


(c) tr = 6, det = 9. λ = (6 ± √(36-36))/2 = 3. λ = 3 (double root).

For λ = 3: (A - 3I)v = [0 1; 0 0]v = 0 → v = (1, 0) only.

Only one eigenvector — not diagonalizable.

Exercise 2: Verification

For $A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$:

(a) Find eigenvalues using tr and det.

(b) Find eigenvectors.

(c) Verify that eigenvectors are orthogonal (why should they be?).

(d) Construct P and verify P⁻¹AP = D.

Show Solution

(a) tr = 2, det = 1 - 4 = -3. λ = (2 ± √(4+12))/2 = (2 ± 4)/2. λ₁ = 3, λ₂ = -1.

(b) For λ = 3: [-2 2; 2 -2]v = 0 → v₁ = (1, 1)

For λ = -1: [2 2; 2 2]v = 0 → v₂ = (1, -1)

(c) v₁ · v₂ = 1 - 1 = 0 ✓ Orthogonal because A is symmetric.

(d) P = [1 1; 1 -1], P⁻¹ = [1/2 1/2; 1/2 -1/2]

P⁻¹AP = [3 0; 0 -1] = D ✓

Exercise 3: Properties

For a matrix with eigenvalues 2, 3, 5:

(a) What is the trace?

(b) What is the determinant?

(c) What are the eigenvalues of A²?

(d) What are the eigenvalues of A⁻¹?

Show Solution

(a) tr = 2 + 3 + 5 = 10

(b) det = 2 × 3 × 5 = 30

(c) Eigenvalues of A²: 4, 9, 25 (square each)

(d) Eigenvalues of A⁻¹: 1/2, 1/3, 1/5

Exercise 4: Complex Eigenvalues

Find the eigenvalues of the rotation matrix:

$$R(60°) = \begin{pmatrix} 1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & 1/2 \end{pmatrix}$$

Express them in the form e^(iθ). What does this tell you geometrically?

Show Solution

tr = 1, det = 1/4 + 3/4 = 1.

λ = (1 ± √(1-4))/2 = (1 ± √(-3))/2 = 1/2 ± i√3/2

In polar form: λ = cos(60°) ± i sin(60°) = e^(±iπ/3)

Geometrically: rotation by 60° has eigenvalues on unit circle at angles ±60°.

Exercise 5: Hückel Theory

For the allyl system (3 carbons in a row), the Hückel matrix is:

$$\mathbf{H} = \begin{pmatrix} \alpha & \beta & 0 \\ \beta & \alpha & \beta \\ 0 & \beta & \alpha \end{pmatrix}$$

(a) Setting α = 0 and β = -1, find the eigenvalues.

(b) Find the eigenvectors (MO coefficients).

(c) Which MO is bonding? Nonbonding? Antibonding?

Show Solution

(a) With α = 0, β = -1: H = [0 -1 0; -1 0 -1; 0 -1 0]

Characteristic polynomial: -λ³ + 2λ = -λ(λ² - 2) = 0

Eigenvalues: λ = 0, √2, -√2

(Energies: E = α + √2|β|, α, α - √2|β|)


(b) For λ = √2: c ∝ (1, √2, 1) (bonding)

For λ = 0: c ∝ (1, 0, -1) (nonbonding — node at center)

For λ = -√2: c ∝ (1, -√2, 1) (antibonding)


(c) Bonding: E = α + √2|β| (lowest energy).

Nonbonding: E = α.

Antibonding: E = α - √2|β| (highest energy).

Next

Lecture 11: Symmetry

The eigenvectors of symmetric matrices are orthogonal. Molecular symmetry operations form groups. The mathematics of invariance.

Chemical Thinking: The Grammar of Reality

Lecture 10 — What Survives?