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Linear Algebra Unit 6: Eigenvalues & eigenvectors
Eigenvectors are directions a matrix only stretches. Eigenvalues are the stretch factors. They unlock diagonalization and long-run behavior.
Covers: Characteristic polynomial, diagonalization, applications (Markov chains, DEs)
Study guide
An eigenvector of a square matrix A is a nonzero vector v with Av = λv: A only stretches it by the factor λ, the eigenvalue. To find eigenvalues, solve the characteristic equation det(A − λI) = 0, a polynomial of degree n. For each eigenvalue, find eigenvectors by solving (A − λI)v = 0; the solutions form the eigenspace. The eigenvalues of a triangular matrix are its diagonal entries, their sum is the trace, and their product is the determinant. A is invertible exactly when 0 isn't an eigenvalue.
A matrix is diagonalizable if A = PDP⁻¹, where D is diagonal with the eigenvalues and P has the matching eigenvectors as columns. This requires n linearly independent eigenvectors. Distinct eigenvalues guarantee it; repeated eigenvalues need eigenspaces of full dimension. Diagonalization makes powers easy: Aᵏ = PDᵏP⁻¹, and Dᵏ just raises each diagonal entry to the kth power. Complex eigenvalues correspond to rotation-like behavior.
Applications: in a Markov chain, a stochastic matrix moves a probability vector from step to step, and the long-run steady state is the eigenvector for eigenvalue 1. In systems of differential equations x′ = Ax, each eigenpair gives a solution e^(λt)v, and the signs of the eigenvalues tell whether solutions grow, decay or oscillate. Dynamical systems xₖ₊₁ = Axₖ behave similarly, with |λ| > 1 meaning growth.
If Av = λv, then the same v is an eigenvector of A² (eigenvalue λ²), Aᵏ (λᵏ), A⁻¹ (1/λ) and A + cI (λ + c). So a matrix with A² = cI can only have eigenvalues ±√c, and the trace tells you how many of each. To compute Aᵏx₀, write x₀ = c₁v₁ + … + cₙvₙ in eigenvectors. Then Aᵏx₀ = c₁λ₁ᵏv₁ + … + cₙλₙᵏvₙ, and terms with |λ| < 1 die out. For a Markov matrix, the limit is the steady-state vector scaled so its entries add up to the same total as x₀'s.
An eigenvalue's algebraic multiplicity is its multiplicity as a root of the characteristic polynomial, and its geometric multiplicity is the dimension of its eigenspace, which is never larger. A matrix whose geometric multiplicity falls short for some eigenvalue is called defective and can't be diagonalized. An example is [2 1; 0 2].
Key ideas
- Av = λv with v ≠ 0.
- Find eigenvalues by solving det(A − λI) = 0.
- Find eigenvectors by solving (A − λI)v = 0.
- A is diagonalizable if it has n independent eigenvectors: A = PDP⁻¹.
- Diagonalization makes powers easy: Aᵏ = PDᵏP⁻¹.
- Trace = sum of eigenvalues; determinant = product of eigenvalues.
Formulas and rules
Common mistakes
- Accepting v = 0 as an eigenvector.
- Assuming every matrix is diagonalizable.
Words to know
- Definition of an eigenvector
- A nonzero v with Av = λv.
- Characteristic equation
- det(A − λI) = 0
- Diagonalizable matrix
- An n×n matrix with n linearly independent eigenvectors.
- Powers of a diagonalizable matrix
- If A = PDP⁻¹, then Aᵏ = PDᵏP⁻¹.
- Trace
- Sum of the diagonal entries = sum of the eigenvalues.
- Distinct eigenvalues
- Guarantee independent eigenvectors, so the matrix is diagonalizable.
- Eigenvalue
- A scalar λ with Av = λv for some nonzero v.
- Eigenspace
- All solutions of (A − λI)v = 0.
- Eigenvalues of a triangular matrix
- Its diagonal entries.
- Product of the eigenvalues
- det A
- Zero eigenvalue
- A is not invertible.
- Markov chain steady state
- The eigenvector for eigenvalue 1 of a stochastic matrix.
- Complex eigenvalues
- Indicate rotation-like behavior.
Sample practice questions
21 more questions like these, from level 1 to 5, are in the practice set.
Eigenvalues of [3 0; 0 5].
- 3 and 5
- 8
- 0
- 15
Show the answer and steps
Answer: A) 3 and 5
- A diagonal matrix's eigenvalues are its diagonal entries.
- 3 and 5.
Find the eigenvalues of [−2 0; 0 −5].
- 2 and 5
- −7
- 10
- −5 and −2
Show the answer and steps
Answer: D) −5 and −2
- A diagonal matrix's eigenvalues are its diagonal entries.
- −2 and −5.
An eigenvector of [2 1; 1 2] for λ = 3.
- ⟨1, −1⟩
- ⟨1, 0⟩
- ⟨2, 1⟩
- ⟨1, 1⟩
Show the answer and steps
Answer: D) ⟨1, 1⟩
- Solve (A − 3I)v = 0: [−1 1; 1 −1]v = 0.
- −v₁ + v₂ = 0 → v₁ = v₂.
- ⟨1, 1⟩.