Eigenvalues and Eigenvectors
Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors reveal the fundamental behaviour of linear transformations. An eigenvector of a matrix is a non-zero vector whose direction is unchanged (or reversed) by the transformation — it is only scaled by a factor called the eigenvalue.
Definition
For a square matrix A, a non-zero vector v is an eigenvector with eigenvalue λ if:
Av = λv
This means that applying the transformation A to v simply multiplies v by the scalar λ:
- If λ > 1: v is stretched
- If 0 < λ < 1: v is compressed
- If λ < 0: v is reversed and scaled
- If λ = 1: v is unchanged (a fixed direction)
- If λ = 0: v is mapped to zero (the matrix is singular)
Finding Eigenvalues
Rearranging Av = λv:
Av − λv = 0 → (A − λI)v = 0
For non-zero solutions, the matrix (A − λI) must be singular, so:
det(A − λI) = 0 — the characteristic equation
2×2 Case
For A = [[a, b], [c, d]]:
det(A − λI) = (a − λ)(d − λ) − bc = 0
This gives: λ² − (a + d)λ + (ad − bc) = 0
Which is: λ² − tr(A)λ + det(A) = 0
where tr(A) = a + d is the trace and det(A) = ad − bc.
Key fact: The sum of eigenvalues = tr(A) and the product of eigenvalues = det(A). Use this to check your answers.
3×3 Case
For a 3×3 matrix, the characteristic equation is a cubic. Expand det(A − λI) by cofactors along any row or column.
Example: A = [[2, 1, 0], [0, 3, 0], [1, 0, 1]]
det(A − λI) = (2−λ)[(3−λ)(1−λ)] − 1[0 − 0] + 0 = (2−λ)(3−λ)(1−λ)
So the eigenvalues are λ = 1, 2, 3.
Finding Eigenvectors
For each eigenvalue λ, solve (A − λI)v = 0.
This is a homogeneous system — there are infinitely many solutions (eigenvectors are not unique; any scalar multiple of an eigenvector is also an eigenvector).
Example: For A = [[4, 2], [1, 3]] with eigenvalue λ = 5:
(A − 5I)v = [[-1, 2], [1, -2]] [[x], [y]] = [[0], [0]]
−x + 2y = 0 → x = 2y. So v = t[[2], [1]] for any t ≠ 0.
The eigenvector (choosing t = 1) is v = [[2], [1]].
Properties of Eigenvalues
| Property | Result |
|---|---|
| A is singular | λ = 0 is an eigenvalue |
| Eigenvalues of A⁻¹ | 1/λ (same eigenvectors) |
| Eigenvalues of Aⁿ | λⁿ (same eigenvectors) |
| Eigenvalues of A + kI | λ + k (same eigenvectors) |
| Eigenvalues of kA | kλ (same eigenvectors) |
| Sum of eigenvalues | = trace of A |
| Product of eigenvalues | = determinant of A |
Diagonalisation
A matrix A is diagonalisable if it has n linearly independent eigenvectors (where A is n×n).
If P = [v₁ | v₂ | ... | vₙ] is the matrix whose columns are eigenvectors, and D = diag(λ₁, λ₂, ..., λₙ), then:
A = PDP⁻¹ and equivalently D = P⁻¹AP
Why is this useful? Computing Aⁿ becomes easy:
Aⁿ = PDⁿP⁻¹
Since D is diagonal, Dⁿ = diag(λ₁ⁿ, λ₂ⁿ, ..., λₙⁿ) — no matrix multiplication chain needed.
Worked Example: Full Diagonalisation
A = [[5, 4], [1, 2]]
Step 1: Characteristic equation: (5−λ)(2−λ) − 4 = λ² − 7λ + 6 = (λ−1)(λ−6) = 0
Eigenvalues: λ₁ = 1, λ₂ = 6.
Step 2: For λ₁ = 1: (A − I)v = [[4, 4], [1, 1]]v = 0 → x + y = 0 → v₁ = [[1], [−1]]
For λ₂ = 6: (A − 6I)v = [[−1, 4], [1, −4]]v = 0 → x = 4y → v₂ = [[4], [1]]
Step 3: P = [[1, 4], [−1, 1]], D = [[1, 0], [0, 6]]
P⁻¹ = (1/5)[[1, −4], [1, 1]]
Check: PDP⁻¹ = A ✓
Step 4: Aⁿ = P · [[1, 0], [0, 6ⁿ]] · P⁻¹
The Cayley-Hamilton Theorem
Every square matrix satisfies its own characteristic equation. If the characteristic equation is λ² − 7λ + 6 = 0, then:
A² − 7A + 6I = 0
This can be used to express higher powers of A in terms of A and I.
Geometric and Algebraic Multiplicity
- Algebraic multiplicity of λ: the number of times λ appears as a root of the characteristic equation
- Geometric multiplicity of λ: the dimension of the eigenspace (number of linearly independent eigenvectors for λ)
Geometric multiplicity ≤ algebraic multiplicity. A matrix is diagonalisable if and only if these are equal for every eigenvalue.
Exam Tips
- Always use tr(A) and det(A) checks: eigenvalue sum = trace, product = determinant
- For 3×3 characteristic cubics, try integer factors first (use the factor theorem)
- When finding eigenvectors, you need only one free parameter — don't try to find a unique solution
- Show your eigenvectors are correct by computing Av and checking it equals λv
- For diagonalisation, the order of columns in P must match the order of eigenvalues in D