Eigenvalues and eigenvectors

Most matrix operations — multiplication, inversion, raising to a power — tell you nothing intuitively about what a matrix does geometrically; eigenvalues and eigenvectors are the rare exception that reveal the matrix's inner structure directly, and they underpin everything from Google's PageRank to quantum mechanics to principal component analysis.

Definition: For a square matrix A, a non-zero vector v is an eigenvector with corresponding eigenvalue λ if:

Av = λv

The matrix A maps the vector v to a scalar multiple of itself — it only scales v, it does not rotate or shear it. The scaling factor is λ.

Definition: The characteristic polynomial of A is det(A − λI) = 0. Its roots are the eigenvalues.

Definition: The eigenspace for eigenvalue λ is the null space of (A − λI) — the set of all eigenvectors corresponding to λ, plus the zero vector.

Finding eigenvalues — the characteristic equation

Step 1: Form the matrix (A − λI) by subtracting λ from each diagonal entry.

Step 2: Set det(A − λI) = 0. This gives a polynomial of degree n (for an n×n matrix).

Step 3: Solve the polynomial for λ. The roots are the eigenvalues.

For a 2×2 matrix: det(A − λI) = λ² − (tr A)λ + det A = 0

This is the fastest route for 2×2 problems in GATE and engineering entrances:

• Sum of eigenvalues = trace(A) (sum of diagonal entries)
• Product of eigenvalues = det(A)

Finding eigenvectors — the null space

For each eigenvalue λ, solve the homogeneous system (A − λI)v = 0 by row reduction. The solution space is the eigenspace.

There is always an infinite family of solutions (any scalar multiple of an eigenvector is also an eigenvector). Usually we express the eigenvector as the simplest non-zero representative.

Fully worked example

Question: Find eigenvalues and eigenvectors of A = [[2, 1], [1, 2]].

Solution:

Step 1: Characteristic equation = det(A − λI) = (2−λ)² − 1 = λ² − 4λ + 3 = 0

Step 2: Factorise: (λ−1)(λ−3) = 0 → λ₁ = 1, λ₂ = 3

Step 3: Verify via trace and det:

• Trace = 2+2 = 4 = 1+3 ✓
• Det = 4−1 = 3 = 1×3 ✓

Step 4: Eigenvector for λ₁ = 1:
(A − I) = [[1,1],[1,1]]. Row reduce → [1,1,0]. Solution: v₁ = s·(1, −1) for any scalar s.
Choose v₁ = (1, −1).

Step 5: Eigenvector for λ₂ = 3:
(A − 3I) = [[−1,1],[1,−1]]. Row reduce → [1,−1,0]. Solution: v₂ = s·(1, 1).
Choose v₂ = (1, 1).

Conclusion: Eigenvalues 1 and 3 with eigenvectors (1,−1) and (1,1). Geometrically, this symmetric matrix stretches along the diagonal direction (1,1) by factor 3 and compresses along the anti-diagonal (1,−1) by factor 1 (unchanged).

Key properties of eigenvalues — exam shortcuts

Property	Statement
Trace property	Σλᵢ = tr(A) (sum of diagonal entries)
Determinant property	Πλᵢ = det(A)
Powers	Eigenvalues of Aᵏ are λᵏ (same eigenvectors)
Inverse	Eigenvalues of A⁻¹ are 1/λ (same eigenvectors); requires λ ≠ 0
Triangular matrix	Eigenvalues are the diagonal entries (read directly)
Symmetric matrix	All eigenvalues are real
Orthogonal matrix
Identity matrix	Only eigenvalue is 1 (multiplicity n)
Nilpotent matrix	All eigenvalues are 0

Why the triangular shortcut matters: For a triangular matrix (upper or lower), no computation is needed — the eigenvalues are literally the diagonal. Always check for triangular structure first.

Diagonalization — the payoff

If A has n linearly independent eigenvectors (columns of P), then:

A = PDP⁻¹

where D is diagonal with eigenvalues on the diagonal. Then:

Aᵏ = PDᵏP⁻¹

Raising D to the k-th power is trivial: just raise each diagonal entry to the k-th power. This is the key computational advantage of diagonalization.

When is A diagonalizable? A is diagonalizable if and only if for every eigenvalue, its geometric multiplicity (dimension of eigenspace) equals its algebraic multiplicity (multiplicity as a root of the characteristic polynomial). A matrix with all distinct eigenvalues is always diagonalizable.

Real-world example (India-relevant): A Markov chain modelling student migration between three exam coaching centres can be diagonalized. The eigenvector corresponding to eigenvalue 1 gives the long-run steady-state distribution — how many students each centre will eventually attract regardless of the starting distribution. State-level coaching institute planners use exactly this analysis.

Cayley-Hamilton theorem

Every square matrix satisfies its own characteristic equation.

For a 2×2 matrix with characteristic polynomial λ² − (tr A)λ + (det A) = 0, we have:

A² − (tr A)·A + (det A)·I = 0

Useful application — finding A⁻¹:

Rearranging: (det A)·I = (tr A)·A − A²

→ A⁻¹ = [(tr A)·I − A] / det A (when det A ≠ 0)

This avoids row reduction entirely for 2×2 matrices, and is also valid for higher orders with the appropriate polynomial.

Question: Use Cayley-Hamilton to find A⁻¹ for A = [[2,1],[1,2]].

Solution:

Step 1: tr A = 4, det A = 3. Characteristic equation: A² − 4A + 3I = 0.

Step 2: 3I = 4A − A² → I = (4A − A²)/3 → A⁻¹ = (4I − A)/3.

Step 3: 4I − A = [[4,0],[0,4]] − [[2,1],[1,2]] = [[2,−1],[−1,2]].

Step 4: A⁻¹ = (1/3)[[2,−1],[−1,2]].

Conclusion: A⁻¹ = [[2/3, −1/3], [−1/3, 2/3]]. (Verify: A·A⁻¹ = I ✓)

Applications — why eigenvalues matter beyond the exam

1. Principal Component Analysis (PCA): The eigenvectors of the data covariance matrix are the principal components — the directions of maximum variance. Eigenvalues tell you how much variance each direction explains. Used in face recognition, data compression, genetics.

2. Google PageRank: The importance of a webpage is defined as the leading eigenvector (eigenvalue = 1) of the web's link-transition matrix. The entire search engine ranking is an eigenvalue problem.

3. Structural engineering (vibration analysis): Natural frequencies of a structure (bridge, building) are the square roots of eigenvalues of the mass-normalised stiffness matrix. Engineers must ensure no eigenfrequency matches the frequency of expected loads (earthquake, wind, traffic) to prevent resonance.

4. Quantum mechanics: Observable quantities (energy, momentum, position) are eigenvalues of Hermitian operators (Hamiltonians). The famous time-independent Schrödinger equation is Hψ = Eψ — an eigenvalue equation where E (energy levels) are eigenvalues.

5. Markov chains: The stationary probability distribution is the eigenvector of the transition matrix corresponding to eigenvalue 1 — directly applicable to queuing theory, population genetics, and financial modelling.

Cover the answer, recall, then check. 12 cards on the definitions and properties GATE tests every year.

Q1. Define an eigenvalue and eigenvector of a square matrix A.
A1. A scalar λ and a nonzero vector x with Ax = λx. λ is the eigenvalue, x the corresponding eigenvector. The zero vector is never an eigenvector.

Q2. What is the characteristic equation, and how do you get eigenvalues from it?
A2. det(A − λI) = 0. Its roots are the eigenvalues. For an n×n matrix it is a degree-n polynomial, so there are n eigenvalues (counting multiplicity, possibly complex).

Q3. What does the sum of all eigenvalues equal?
A3. The trace of A (sum of diagonal entries): Σλᵢ = trace(A). Great sanity check.

Q4. What does the product of all eigenvalues equal?
A4. det(A) = Πλᵢ. So A is singular ⇔ at least one eigenvalue is 0.

Q5. What are the eigenvalues of a triangular (or diagonal) matrix?
A5. Exactly its diagonal entries — no computation needed.

Q6. How do the eigenvalues of A and Aᵀ compare?
A6. They are identical (same characteristic polynomial). The eigenvectors, however, generally differ.

Q7. Distinguish algebraic multiplicity (AM) and geometric multiplicity (GM).
A7. AM = multiplicity of λ as a root of det(A − λI). GM = dim of eigenspace = n − rank(A − λI) = number of independent eigenvectors for λ. Always 1 ≤ GM ≤ AM.

Q8. If λ is an eigenvalue of A with eigenvector x, what is an eigenvalue of A + kI?
A8. λ + k, with the same eigenvector x. (Shifting adds k to every eigenvalue.)

Q9. For an idempotent matrix (A² = A), what can the eigenvalues be?
A9. Only 0 or 1. (Involutory A²=I → ±1; nilpotent Aᵏ=0 → all eigenvalues 0.)

Q10. Are eigenvectors for distinct eigenvalues independent?
A10. Yes — eigenvectors corresponding to distinct eigenvalues are always linearly independent.

Q11. A 3×3 matrix has trace 6 and two known eigenvalues 1 and 2. Find the third.
A11. Sum = trace ⇒ 1 + 2 + λ = 6 ⇒ λ = 3.

Q12. If λ is an eigenvalue of a rank-deficient (singular) matrix, what value must appear?
A12. 0 must be an eigenvalue, because det(A) = product of eigenvalues = 0.

Eigenvalues are the single most reliably tested idea in GATE CSE Engineering Mathematics — nearly every year there is a 1- or 2-mark question that reduces to "find/use an eigenvalue." The reason they matter beyond marks: they capture how a linear map stretches space, and they power PageRank, PCA, stability analysis, and Markov chains — all recurring in later CS subjects.

The core idea

For a square matrix A, an eigenvector x is a direction that A only scales, not rotates: Ax = λx. The scale factor λ is the eigenvalue. You find λ from the characteristic equation det(A − λI) = 0, then solve (A − λI)x = 0 for each eigenvector.

Key results (memorise cold)

Property	Result
Sum of eigenvalues	trace(A)
Product of eigenvalues	det(A)
Triangular/diagonal A	eigenvalues = diagonal entries
Eigenvalues of Aᵀ	same as A
Eigenvalue count	n (with multiplicity)
Multiplicities	1 ≤ GM ≤ AM
A + kI	eigenvalues shift by k

Exam Tricks & Tips

• 🎯 Trace and determinant first. For 2×2/3×3, use Σλ = trace and Πλ = det to guess or verify eigenvalues without expanding the full polynomial.
• 🎯 Special-matrix shortcuts: idempotent → {0,1}; involutory → {±1}; nilpotent → all 0; orthogonal → |λ| = 1; symmetric → all real.
• 🎯 Missing-eigenvalue trick: given some eigenvalues and the trace, back out the rest by subtraction.
• 🎯 Singular ⇔ 0 is an eigenvalue. If det = 0 the answer set must contain 0.
• 🎯 Same eigenvector under shifts: A, A+kI, A², A⁻¹ all share eigenvectors — only the eigenvalue transforms.
• ❌ Common mistake: treating the zero vector as an eigenvector, or forgetting GM can be strictly less than AM (that is exactly when A is not diagonalizable).

Expected exam pattern

Usually one direct MCQ/NAT: compute an eigenvalue of a 2×2/3×3 matrix, or apply "sum = trace / product = det," or identify eigenvalues of a special matrix. Occasionally combined with rank to test geometric multiplicity.

Quick recap

Solve det(A − λI) = 0. Sum of eigenvalues = trace, product = det. Diagonal/triangular give eigenvalues free. GM ≤ AM, and GM < AM means not diagonalizable. Shifts and powers keep eigenvectors fixed while transforming eigenvalues predictably.

Key formulas

• Characteristic equation: det(A − λI) = 0; roots are the eigenvalues λ.
• Eigenvector: (A − λI)x = 0, x ≠ 0.
• Trace: sum of eigenvalues = trace(A) = Σ aᵢᵢ.
• Product of eigenvalues = det(A).
• Eigenvalues of Aⁿ are λⁿ; of A⁻¹ are 1/λ; of A+kI are λ+k.
• A is diagonalizable if it has n linearly independent eigenvectors; then A = PDP⁻¹.
• Symmetric real matrix ⇒ all eigenvalues real, eigenvectors orthogonal.
• Cayley–Hamilton: every matrix satisfies its own characteristic equation.
• Algebraic multiplicity ≥ geometric multiplicity (= n − rank(A − λI)).

Usage: use trace and determinant as quick checks on computed eigenvalues before finding eigenvectors.

Worked Example

Problem: Find the eigenvalues and corresponding eigenvectors of the matrix
A = [[5, 4], [1, 2]].

Solution:
Eigenvalues satisfy det(A − λI) = 0:
det([[5 − λ, 4], [1, 2 − λ]]) = (5 − λ)(2 − λ) − (4)(1) = 0.
Expand: λ² − 7λ + 10 − 4 = λ² − 7λ + 6 = 0.
Factor: (λ − 1)(λ − 6) = 0 ⇒ λ₁ = 1, λ₂ = 6.

Verify with invariants: trace = 5 + 2 = 7 = 1 + 6 ✓; det = 10 − 4 = 6 = 1 × 6 ✓.

Eigenvector for λ = 6: solve (A − 6I)v = 0.
A − 6I = [[−1, 4], [1, −4]]. Row 1 gives −x + 4y = 0 ⇒ x = 4y ⇒ v = [4, 1]ᵀ.

Eigenvector for λ = 1: solve (A − I)v = 0.
A − I = [[4, 4], [1, 1]]. Row gives x + y = 0 ⇒ x = −y ⇒ v = [1, −1]ᵀ.

Answer: λ = 1 with eigenvector [1, −1]ᵀ, and λ = 6 with eigenvector [4, 1]ᵀ.