What are Eigenvalues?
For a square matrix M, a scalar λ is an eigenvalue and a non-zero vector v is the corresponding eigenvector if:
Defining equation
Mv = λv
The matrix M maps the eigenvector v to a scalar multiple of itself. The vector changes length (by factor λ) but not direction — or it reverses direction if λ < 0.
Geometrically: most vectors are rotated and stretched by a transformation. Eigenvectors are special — they lie along invariant lines of the transformation. The matrix only stretches them.
λ > 0
Eigenvector is stretched or compressed in the same direction. λ = 1 means the vector is unchanged.
λ < 0
Eigenvector is reversed in direction and scaled. λ = −1 means the vector is reflected through the origin.
λ = 0
The matrix is singular. The eigenvector is mapped to the zero vector. det(M) = 0.
Why They Matter in Cambridge 9231
🎯 Exam Relevance
- Finding eigenvalues and eigenvectors for 2×2 and 3×3 matrices is directly tested — typically 6–10 marks.
- Diagonalisation of a matrix M = PDP⁻¹ allows rapid computation of Mⁿ — links back to induction from Lesson 2.2.
- Geometric interpretation is often required — invariant lines, area/volume scale factor via eigenvalues.
- Repeated eigenvalues and the question of diagonalisability are tested at A-level.
- Cayley-Hamilton theorem — M satisfies its own characteristic equation — may be invoked for matrix powers.
Deriving the Method
From Mv = λv we can write Mv − λIv = 0, i.e. (M − λI)v = 0. For a non-zero solution v to exist, the matrix (M − λI) must be singular:
Characteristic equation — the key formula
det(M − λI) = 0
This gives a polynomial in λ called the characteristic polynomial. Its roots are the eigenvalues. For an n×n matrix, this is a degree-n polynomial — so an n×n matrix has at most n eigenvalues (counted with multiplicity).
⚠️ Eigenvectors are Never Unique
If v is an eigenvector for λ, then so is kv for any non-zero scalar k. You may choose any convenient non-zero multiple. Cambridge accepts any correct scalar multiple as the eigenvector answer — but you must show the working for the null space of (M − λI).
Finding λ and v — The Method
There are always exactly two stages. You must complete both in every question.
2×2 Characteristic Equation
For M = [[a,b],[c,d]]
det(M − λI) = (a−λ)(d−λ) − bc = 0
⟹ λ² − (a+d)λ + (ad−bc) = 0
⟹ λ² − tr(M)·λ + det(M) = 0
tr(M) = a+d is the trace (sum of diagonal entries). This identity is worth memorising for speed.
3×3 Characteristic Equation
For a 3×3 matrix M
det(M − λI) = −λ³ + tr(M)λ² − (sum of 2×2 principal minors)λ + det(M) = 0
In practice, expand the 3×3 determinant directly — do not memorise the coefficient formula. Use the Factor Theorem: if λ=k gives det=0, then (λ−k) is a factor.
Key Properties
Trace and Determinant Relations
For any n×n matrix M with eigenvalues λ₁, λ₂, …, λₙ (counted with multiplicity):
Trace = sum of eigenvalues
tr(M) = λ₁ + λ₂ + ··· + λₙ
Use this to check your eigenvalues immediately.
Determinant = product of eigenvalues
det(M) = λ₁ · λ₂ · ··· · λₙ
A matrix is singular iff at least one eigenvalue is 0.
Further Properties
Eigenvalues of Mⁿ
λ₁ⁿ, λ₂ⁿ, …
Same eigenvectors, eigenvalues raised to power n.
Eigenvalues of M⁻¹
1/λ₁, 1/λ₂, …
Same eigenvectors. Requires all λᵢ ≠ 0.
Eigenvalues of kM
kλ₁, kλ₂, …
Scalar multiplication scales all eigenvalues.
Eigenvalues of Mᵀ
λ₁, λ₂, …
Same eigenvalues, but eigenvectors generally differ.
Distinct eigenvalues
⟹ Independent eigenvectors
Guaranteed linear independence — matrix is diagonalisable.
Symmetric matrices
Always real eigenvalues
Eigenvectors for distinct λ are orthogonal. M = Mᵀ.
Cayley-Hamilton Theorem
Statement
Every matrix satisfies its own characteristic equation.
If the characteristic polynomial is p(λ) = λ² − 7λ + 6, then p(M) = M² − 7M + 6I = 0. This can be used to express high powers of M in terms of lower powers.
E
Cayley-Hamilton — Express M² in terms of M and I
M = [[5,2],[2,2]] has characteristic equation λ² − 7λ + 6 = 0. Express M³ in terms of M and I.
Cayley-Hamilton
M² − 7M + 6I = 0 → M² = 7M − 6I
M³ = M·M²
M³ = M(7M − 6I) = 7M² − 6M
= 7(7M − 6I) − 6M
= 49M − 42I − 6M
= 7(7M − 6I) − 6M
= 49M − 42I − 6M
M³ = 43M − 42I
Repeated Eigenvalues
If the characteristic equation has a repeated root (algebraic multiplicity > 1), the matrix may or may not have enough independent eigenvectors to be diagonalised.
⛔ Repeated λ — Two Possible Outcomes
Diagonalisable case
The repeated eigenvalue has as many independent eigenvectors as its multiplicity. Example: λ=2 (multiplicity 2) has a 2D eigenspace.
The repeated eigenvalue has as many independent eigenvectors as its multiplicity. Example: λ=2 (multiplicity 2) has a 2D eigenspace.
Non-diagonalisable case
The eigenspace is smaller than the multiplicity. A defective matrix — cannot be diagonalised over ℝ. Cambridge will usually signal this situation explicitly.
The eigenspace is smaller than the multiplicity. A defective matrix — cannot be diagonalised over ℝ. Cambridge will usually signal this situation explicitly.
Diagonalisation
A matrix M is diagonalisable if it can be written as M = PDP⁻¹ where D is a diagonal matrix of eigenvalues and P is the matrix of corresponding eigenvectors.
⬡ The Diagonalisation Identity
M
=
P · D · P⁻¹
P
=
Matrix whose columns are eigenvectors v₁, v₂, (v₃)
D
=
Diagonal matrix of corresponding eigenvalues [[λ₁,0,…],[0,λ₂,…],…]
Mⁿ
=
P · Dⁿ · P⁻¹ where Dⁿ = [[λ₁ⁿ,0,…],[0,λ₂ⁿ,…],…]
💡 Why Mⁿ = PDⁿP⁻¹?
M² = (PDP⁻¹)(PDP⁻¹) = PD(P⁻¹P)DP⁻¹ = PD²P⁻¹. The inner P⁻¹P = I cancels. By induction, Mⁿ = PDⁿP⁻¹. Since D is diagonal, Dⁿ is trivial to compute — just raise each diagonal entry to the power n.
Conditions for Diagonalisability
| Condition | Diagonalisable? |
|---|---|
| n distinct eigenvalues | ✓ Always diagonalisable |
| Repeated eigenvalue, full eigenspace | ✓ Diagonalisable |
| Repeated eigenvalue, deficient eigenspace | ✗ Not diagonalisable over ℝ |
| Symmetric matrix (M = Mᵀ) | ✓ Always diagonalisable (orthogonally) |
| Complex eigenvalues (for real matrix) | ✗ Not diagonalisable over ℝ |
Worked Examples
1
3×3 Eigenvalues and Eigenvectors
Find the eigenvalues and eigenvectors of M = [[4,1,0],[0,3,1],[0,0,2]].
Characteristic eq
M is upper triangular — eigenvalues are the diagonal entries:
λ₁ = 4, λ₂ = 3, λ₃ = 2
(For triangular matrices, always use this shortcut — Cambridge rewards efficiency.)
Eigenvector λ=4
(M−4I)v = 0: [[0,1,0],[0,−1,1],[0,0,−2]]·v = 0
Row 3: −2z=0 → z=0. Row 2: −y+z=0 → y=0. x is free.
Row 3: −2z=0 → z=0. Row 2: −y+z=0 → y=0. x is free.
v₁ = [[1],[0],[0]]
Eigenvector λ=3
(M−3I)v = 0: [[1,1,0],[0,0,1],[0,0,−1]]·v = 0
Row 3: z=0. Row 1: x+y=0 → x=−y. y is free.
Row 3: z=0. Row 1: x+y=0 → x=−y. y is free.
v₂ = [[−1],[1],[0]]
Eigenvector λ=2
(M−2I)v = 0: [[2,1,0],[0,1,1],[0,0,0]]·v = 0
Row 2: y+z=0 → y=−z. Row 1: 2x+y=0 → 2x=z → x=z/2. Let z=2.
Row 2: y+z=0 → y=−z. Row 1: 2x+y=0 → 2x=z → x=z/2. Let z=2.
v₃ = [[1],[−2],[2]]
2
Repeated Eigenvalue
M = [[3,1],[0,3]]. Find eigenvalues and eigenvectors. Is M diagonalisable?
Characteristic eq
det(M−λI) = (3−λ)² = 0 → λ = 3 (repeated, multiplicity 2)
Eigenvectors λ=3
(M−3I)v = 0: [[0,1],[0,0]]·[[x],[y]] = 0 → y = 0, x free.
Only one independent eigenvector:
Only one independent eigenvector:
v = [[1],[0]]
Diagonalisable?
Algebraic multiplicity = 2, but geometric multiplicity (eigenspace dimension) = 1.
Not diagonalisable — deficient eigenspace.
3
Using Diagonalisation to find Mⁿ
M = [[3,2],[1,4]]. Find Mⁿ in terms of n.
Eigenvalues
det(M−λI) = (3−λ)(4−λ)−2 = λ²−7λ+10 = (λ−2)(λ−5) = 0
λ₁ = 2, λ₂ = 5
λ₁ = 2, λ₂ = 5
Eigenvectors
λ=2: (M−2I)v=0: [[1,2],[1,2]]v=0 → x=−2y → v₁=[[−2],[1]]
λ=5: (M−5I)v=0: [[−2,2],[1,−1]]v=0 → x=y → v₂=[[1],[1]]
λ=5: (M−5I)v=0: [[−2,2],[1,−1]]v=0 → x=y → v₂=[[1],[1]]
P and P⁻¹
P = [[−2,1],[1,1]] det(P) = −2−1 = −3
P⁻¹ = (−1/3)·[[1,−1],[−1,−2]] = [[(−1/3),(1/3)],[(1/3),(2/3)]]
P⁻¹ = (−1/3)·[[1,−1],[−1,−2]] = [[(−1/3),(1/3)],[(1/3),(2/3)]]
Mⁿ = PDⁿP⁻¹
Dⁿ = [[2ⁿ,0],[0,5ⁿ]]
Mⁿ = (1/3)·[[−2,1],[1,1]]·[[2ⁿ,0],[0,5ⁿ]]·[[−1,1],[1,2]]
Step 1: [[−2·2ⁿ, 5ⁿ],[2ⁿ, 5ⁿ]]
Step 2 × (1/3)·[[−1,1],[1,2]]:
(1,1): (−1/3)(−2·2ⁿ·(−1)+5ⁿ·1) — work entry by entry...
Mⁿ = (1/3)·[[−2,1],[1,1]]·[[2ⁿ,0],[0,5ⁿ]]·[[−1,1],[1,2]]
Step 1: [[−2·2ⁿ, 5ⁿ],[2ⁿ, 5ⁿ]]
Step 2 × (1/3)·[[−1,1],[1,2]]:
(1,1): (−1/3)(−2·2ⁿ·(−1)+5ⁿ·1) — work entry by entry...
Mⁿ = (1/3)·[[2·2ⁿ+5ⁿ, −2·2ⁿ+2·5ⁿ],[−2ⁿ+5ⁿ, 2ⁿ+2·5ⁿ]]
Check n=1: (1/3)[[4+5,−4+10],[−2+5,2+10]] = (1/3)[[9,6],[3,12]] = [[3,2],[1,4]] = M ✓
Practice Questions
Question 1 — 2×2 Eigenvalues and Eigenvectors
[6 marks]
Find the eigenvalues and corresponding eigenvectors of M = [[7, 4],[2, 5]].
Characteristic equation: λ² − tr(M)λ + det(M) = 0. Here tr = 12, det = 35−8 = 27. So λ²−12λ+27=0.
✓ Solution
det(M−λI) = (7−λ)(5−λ)−8 = λ²−12λ+27 = (λ−3)(λ−9) = 0λ₁ = 3, λ₂ = 9
λ=3: [[4,4],[2,2]]v=0 → x+y=0 → v₁ = [[1],[−1]]
λ=9: [[−2,4],[2,−4]]v=0 → −x+2y=0 → x=2y → v₂ = [[2],[1]]
Verify: M·v₁ = [[7−4],[2−5]] = [[3],[−3]] = 3v₁ ✓
Question 2 — 3×3 Eigenvalues
[7 marks]
Find all eigenvalues of M = [[2,1,1],[0,3,1],[0,0,1]] and find an eigenvector for each.
M is upper triangular — read the eigenvalues from the diagonal immediately. Then find eigenvectors for each.
✓ Solution
Upper triangular: λ₁=2, λ₂=3, λ₃=1λ=2: [[0,1,1],[0,1,1],[0,0,−1]]v=0 → z=0, y=0, x free → v₁=[[1],[0],[0]]
λ=3: [[−1,1,1],[0,0,1],[0,0,−2]]v=0 → z=0, then −x+y=0 → x=y → v₂=[[1],[1],[0]]
λ=1: [[1,1,1],[0,2,1],[0,0,0]]v=0 → 2y+z=0, z=−2y; x+y+z=0 → x=y → let y=2 → z=−4, x=2 → v₃=[[2],[2],[−4]] or [[1],[1],[−2]]
Question 3 — Diagonalisation and Mⁿ
[8 marks]
M = [[1,2],[3,2]]. Show M is diagonalisable, find P and D such that M = PDP⁻¹, and hence find M⁵.
Characteristic equation: λ²−3λ−4=0 → (λ−4)(λ+1)=0. Two distinct eigenvalues → diagonalisable. Build P from eigenvectors and compute M⁵ = PD⁵P⁻¹.
✓ Solution
det(M−λI) = (1−λ)(2−λ)−6 = λ²−3λ−4 = (λ−4)(λ+1) = 0λ₁=4, λ₂=−1 (distinct → diagonalisable ✓)
λ=4: [[−3,2],[3,−2]]v=0 → 3x=2y → v₁=[[2],[3]]
λ=−1: [[2,2],[3,3]]v=0 → x+y=0 → v₂=[[1],[−1]]
P=[[2,1],[3,−1]], D=[[4,0],[0,−1]]
det(P)=−2−3=−5, P⁻¹=(−1/5)[[−1,−1],[−3,2]]=[[1/5,1/5],[3/5,−2/5]]
D⁵=[[4⁵,0],[0,(−1)⁵]]=[[1024,0],[0,−1]]
M⁵=P·D⁵·P⁻¹=(1/5)[[2,1],[3,−1]]·[[1024,0],[0,−1]]·[[−1,−1],[−3,2]]
=(1/5)[[2048,−1],[3072,1]]·[[−1,−1],[−3,2]]
=(1/5)[[−2048−(−3),−2048+2(−1)],[−3072+3(−3),−3072+2]]...
[multiply entry by entry]
M⁵ = (1/5)[[2045, 2050],[3075, 3070]] — verify by checking M⁵·[[2],[3]]=4⁵·[[2],[3]]=[[2048],[3072]] ✓
Question 4 — Cayley-Hamilton
[4 marks]
M = [[7,4],[2,5]] (from Q1). Use the Cayley-Hamilton theorem to express M² and M³ in terms of M and I.
The characteristic equation is λ²−12λ+27=0. By Cayley-Hamilton, M²−12M+27I=0.
✓ Solution
Characteristic equation: λ²−12λ+27=0Cayley-Hamilton: M²−12M+27I = 0 → M² = 12M−27I
M³ = M·M² = M(12M−27I) = 12M²−27M = 12(12M−27I)−27M
= 144M−324I−27M
M³ = 117M − 324I
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Formula Sheet — Eigenvalues & Eigenvectors
Core Definition
Mv = λvDefining equation
det(M − λI) = 0Characteristic equation
(M − λI)v = 0Find eigenvectors
2×2 Shortcut
λ² − tr(M)λ + det(M) = 0—
tr(M) = a + dSum of diagonal
λ₁+λ₂ = tr(M)Check always
λ₁·λ₂ = det(M)Check always
Properties of Eigenvalues
Eigenvalues of Mⁿλ₁ⁿ, λ₂ⁿ, …
Eigenvalues of M⁻¹1/λ₁, 1/λ₂, …
Eigenvalues of kMkλ₁, kλ₂, …
Triangular matrixλᵢ = diagonal entries
Diagonalisation
M = PDP⁻¹—
Mⁿ = PDⁿP⁻¹—
Dⁿ diagonal[[λ₁ⁿ,0],[0,λ₂ⁿ]]
P columnsEigenvectors (in order)
Cayley-Hamilton
M satisfies p(M) = 0—
p(λ) = char. polynomial—
Use to reduce MⁿExpress via M, I
Diagonalisability
n distinct λAlways diagonalisable
Symmetric MAlways diagonalisable
Repeated λ, full spaceDiagonalisable
Deficient eigenspaceNot diagonalisable
📋 Cambridge Exam Strategy — Eigenvalues & Eigenvectors
- Always verify λ₁+λ₂ = tr(M) and λ₁·λ₂ = det(M) before moving to eigenvectors — one line of checking catches all arithmetic errors.
- For triangular matrices, read eigenvalues from the diagonal immediately — do not expand the determinant.
- For eigenvectors, write (M−λI)v=0 and row-reduce. Express the answer as a column vector with a free parameter, then set the parameter to 1 (or a value giving integer entries).
- For diagonalisation, state P and D explicitly and confirm the column ordering matches — λ₁ in column 1 of D must match v₁ in column 1 of P.
- For Mⁿ questions, use Mⁿ = PDⁿP⁻¹. Compute D first (trivial), then work left to right.
- Cayley-Hamilton: state "by Cayley-Hamilton" and quote the characteristic equation with M replacing λ. Then rearrange for M² before computing higher powers iteratively.