Eigenvalues and Eigenvectors — Summary
Eigenvalues/eigenvectors are a GATE EC favourite (nearly every year, 1–2 marks) and underpin state-space, stability, and system-mode analysis. Av = λv means v keeps its direction under A, scaled by λ.
Key results
- Characteristic equation: det(A − λI) = 0.
- Sum of eigenvalues = trace(A); product of eigenvalues = det(A).
- Eigenvalues of A⁻¹ are 1/λ; of Aᵏ are λᵏ; of A+kI are λ+k; A and Aᵀ share eigenvalues.
- Triangular/diagonal matrix: eigenvalues = diagonal entries.
- Symmetric (real) → real eigenvalues, orthogonal eigenvectors. Skew-symmetric → purely imaginary/zero. Orthogonal → |λ| = 1.
- Cayley–Hamilton: every matrix satisfies its own characteristic equation (use to compute Aⁿ, A⁻¹).
| Operation on A | Effect on eigenvalue λ | Eigenvector |
|---|---|---|
| A⁻¹ | 1/λ | same |
| Aᵏ | λᵏ | same |
| A + kI | λ + k | same |
| kA | kλ | same |
| Aᵀ | λ (same set) | may differ |
Exam Tricks & Tips
- 🎯 Don't expand the characteristic polynomial if trace and det pin down both eigenvalues of a 2×2 (λ₁+λ₂ = trace, λ₁λ₂ = det).
- 🎯 Idempotent matrix (A² = A) → eigenvalues 0 or 1; nilpotent (Aᵏ = 0) → all eigenvalues 0.
- 🎯 Cayley–Hamilton turns high powers into low-order polynomials — huge time saver for Aⁿ.
- 🎯 Repeated eigenvalue may lack a full set of independent eigenvectors (defective → not diagonalizable).
- 🎯 For a projection/rotation matrix, reason geometrically instead of computing.
- ❌ Common mistake: adding k to eigenvalues of kA — scaling A multiplies eigenvalues by k; adding kI adds k. Keep them separate.
Expected exam pattern
1-mark: eigenvalue of a 2×2/3×3, or of A⁻¹/A²/A+I given A's eigenvalues. 2-mark: eigenvector for a given λ, or apply Cayley–Hamilton to find A⁻¹ or a high power.
Quick recap
Solve det(A−λI)=0. Sum λ = trace, product λ = det. λ transforms predictably: 1/λ for inverse, λᵏ for power, λ+k for A+kI. Symmetric → real λ; Cayley–Hamilton reduces powers.
Eigenvalues and Eigenvectors — Flashcards
Cover the answer, recall, then check. 12 cards on eigenvalues and eigenvectors for GATE EC.
Q1. What equation gives the eigenvalues of A?
A1. The characteristic equation det(A − λI) = 0.
Q2. Relate sum and product of eigenvalues to A.
A2. Sum of eigenvalues = trace(A); product of eigenvalues = det(A).
Q3. If λ is an eigenvalue of A, what are the eigenvalues of A⁻¹, Aᵏ, and A + kI?
A3. 1/λ, λᵏ, and λ + k respectively — with the same eigenvector.
Q4. What are the eigenvalues of a triangular matrix?
A4. Its diagonal entries.
Q5. Eigenvalue nature: symmetric vs skew-symmetric vs orthogonal (real matrices)?
A5. Symmetric → real; skew-symmetric → purely imaginary or zero; orthogonal → magnitude 1.
Q6. State the Cayley–Hamilton theorem and one use.
A6. Every square matrix satisfies its own characteristic equation. Use it to express Aⁿ or A⁻¹ as a low-order polynomial in A.
Q7. Eigenvalues of an idempotent matrix (A² = A)?
A7. Only 0 and 1.
Q8. Eigenvalues of a nilpotent matrix (Aᵏ = 0)?
A8. All eigenvalues are 0.
Q9. Do A and Aᵀ have the same eigenvalues?
A9. Yes — same characteristic polynomial, so same eigenvalues (eigenvectors may differ).
Q10. When is a matrix diagonalizable?
A10. When it has n linearly independent eigenvectors (guaranteed if all eigenvalues are distinct, or if it is real symmetric).
Q11. For a 2×2 with trace 5 and det 6, find the eigenvalues.
A11. Roots of λ² − 5λ + 6 = 0 → λ = 2 and λ = 3.
Q12. What is a "defective" matrix?
A12. One with a repeated eigenvalue whose eigenvectors are fewer than its algebraic multiplicity → not diagonalizable.
Eigenvalues and Eigenvectors
Eigenvalues reveal the "natural axes" of a matrix — directions that a linear transformation only stretches, never rotates. They decide the stability of state-space systems, the modes of vibrating structures, and the diagonalisation that makes matrix powers trivial. GATE treats them as near-guaranteed marks, usually as a fast computation exploiting trace and determinant.
Core concept: For A·v = λ·v, λ is an eigenvalue and v its eigenvector: v points along an invariant direction that A merely scales by λ.
The theory
Beginner — the characteristic equation
Eigenvalues solve det(A − λI) = 0, a degree-n polynomial. For each λ, the eigenvector is found from (A − λI)v = 0 (the null space). Two invariants make computation fast:
- Sum of eigenvalues = trace(A) (sum of diagonal entries).
- Product of eigenvalues = det(A).
For a 2×2 matrix these two facts alone often give both eigenvalues by inspection.
Intermediate — properties
- Eigenvalues of a triangular/diagonal matrix are its diagonal entries.
- Symmetric real matrix → all eigenvalues real, eigenvectors orthogonal.
- Skew-symmetric → eigenvalues purely imaginary or zero.
- Orthogonal → eigenvalues have magnitude 1.
- Eigenvalues of Aⁿ are λⁿ; of A^(−1) are 1/λ; of A + kI are λ + k. Eigenvectors are unchanged by these operations.
- A singular matrix has at least one zero eigenvalue (det = 0 = Πλ).
Advanced — diagonalisation and Cayley-Hamilton
If A has n linearly independent eigenvectors, A = P·D·P^(−1) with D = diag(λi) and P their eigenvectors. Then Aᵏ = P·Dᵏ·P^(−1) — powers become trivial. Repeated eigenvalues may fail to give enough independent eigenvectors (defective matrix), needing Jordan form.
Cayley-Hamilton theorem: A satisfies its own characteristic polynomial, e.g. A² − (trace)A + det·I = 0 for 2×2 — a slick way to compute A^(−1) or reduce high powers.
Worked example
Find the eigenvalues of A = [4 1; 2 3].
trace = 4 + 3 = 7 = λ1 + λ2; det = 4·3 − 1·2 = 10 = λ1·λ2.
So λ² − 7λ + 10 = 0 → (λ−5)(λ−2) = 0 → λ = 5, 2.
Eigenvalues of A^(−1) would be 1/5 and 1/2; of A + 3I would be 8 and 5 — no fresh computation needed.
GATE relevance
Almost every GATE has an eigenvalue item: find eigenvalues/eigenvectors, use trace/det shortcuts, eigenvalues of A²/A^(−1)/A+kI, diagonalisability, or apply Cayley-Hamilton. Also underpins state-space stability (eigenvalues = poles) in Control Systems.
Exam tricks and shortcuts
- Use λ1+λ2 = trace, λ1·λ2 = det for 2×2 — usually no polynomial needed.
- Eigenvalues of A^(−1) = 1/λ, of Aⁿ = λⁿ, of A+kI = λ+k (same eigenvectors).
- Triangular matrix ⇒ eigenvalues sit on the diagonal.
- Mnemonic: "Trace sums, Det multiplies" the eigenvalues.
Thinking A + kI has different eigenvectors than A. Shifting by kI only shifts every eigenvalue to λ + k; the eigenvectors are identical. Similarly A^(−1) and Aⁿ keep A's eigenvectors — only the eigenvalues transform.
- ✓- Eigenvalues: det(A − λI) = 0; eigenvectors from (A − λI)v = 0.
- ✓- Σλ = trace(A); Πλ = det(A).
- ✓- Eig(Aⁿ)=λⁿ, Eig(A^(−1))=1/λ, Eig(A+kI)=λ+k; eigenvectors unchanged.
- ✓- Symmetric real ⇒ real λ, orthogonal eigenvectors.
- ✓- Diagonalisable ⇒ Aᵏ = P Dᵏ P^(−1); Cayley-Hamilton: A satisfies its char. eqn.
- ✓Eigenvalues are the scale factors along a matrix's invariant directions, tied to trace and determinant. Use those two invariants for speed, remember the transform rules for Aⁿ, A^(−1), A+kI, and reach for diagonalisation or Cayley-Hamilton when powers or inverses are asked.
Eigenvalues and Eigenvectors — Formula Sheet
Key formulas
- Characteristic equation: |A − λI| = 0.
- Eigenvector: (A − λI)x = 0, x ≠ 0.
- Sum of eigenvalues = trace(A); product = |A|.
- Properties: eigenvalues of Aⁿ are λⁿ; of A⁻¹ are 1/λ; of kA are kλ.
- Symmetric matrix: real eigenvalues, orthogonal eigenvectors.
- Diagonalisation: A = PDP⁻¹ (D = diag of eigenvalues).
- ✓- |A − λI| = 0 gives eigenvalues.
- ✓- Σλ = trace, Πλ = |A|.
- ✓- Eigenvalues of A⁻¹ are 1/λ; of Aⁿ are λⁿ.
Eigenvalues characterise a matrix's scaling directions; symmetric matrices are always diagonalisable.
Eigenvalues and Eigenvectors — Worked Example
Worked Example
Problem: For A = [[4, 1], [2, 3]], find the eigenvalues and the eigenvector associated with the larger eigenvalue.
Solution:
The characteristic equation is det(A − λI) = 0:
|[[4−λ, 1], [2, 3−λ]]| = (4−λ)(3−λ) − (1)(2) = 0.
Expand: λ² − 7λ + 12 − 2 = λ² − 7λ + 10 = 0.
Factor: (λ − 5)(λ − 2) = 0, so λ₁ = 5 and λ₂ = 2.
Sanity checks:
trace(A) = 4 + 3 = 7 = λ₁ + λ₂ ✓, and det(A) = 12 − 2 = 10 = λ₁λ₂ ✓.
Eigenvector for λ = 5: solve (A − 5I)v = 0.
A − 5I = [[−1, 1], [2, −2]].
Row 1 gives −x + y = 0 ⇒ y = x. (Row 2 is just −2 times row 1, so it is redundant.)
Choosing x = 1 gives the eigenvector v = [1, 1]ᵀ.
Verify: A[1, 1]ᵀ = [4+1, 2+3]ᵀ = [5, 5]ᵀ = 5[1, 1]ᵀ ✓.
Answer: Eigenvalues λ = 5 and λ = 2; eigenvector for λ = 5 is [1, 1]ᵀ (any nonzero scalar multiple).
- ✓- Eigenvalues are the roots of det(A − λI) = 0; sum of eigenvalues = trace, product = determinant — quick verification.
- ✓- An eigenvector is any nonzero vector in the null space of (A − λI); one row becomes redundant, so a free parameter always remains.
- ✓- Eigenvectors are defined only up to scale — normalise or set a convenient component to 1.