Eigenvalues and eigenvectors
Eigenvalues and eigenvectors of 2 × 2 and 3 × 3 matrices through the characteristic polynomial, and at the top level the stationary distribution of a Markov chain.
How to solve it
An eigenvector is a direction the map does not turn but only stretches $\lambda$ times: $A\mathbf v = \lambda\mathbf v$. That can only happen if the matrix $A - \lambda I$ collapses a non-zero vector, that is, its determinant is zero.
Step by step
- Form the characteristic polynomial $\det(A - \lambda I)$: subtract $\lambda$ from the diagonal and compute the determinant. For $2 \times 2$: $\lambda^2 - (\text{trace})\lambda + \det A$.
- Its roots are the eigenvalues. For a triangular matrix they sit on the diagonal.
- An eigenvector for $\lambda$: solve $(A - \lambda I)\mathbf v = \mathbf 0$. Its equations are proportional — keep one and pick a non-zero solution.
- A Markov chain: the stationary distribution is the eigenvector for $\lambda = 1$, scaled so that its coordinates add up to $1$.
Common mistakes
- Subtracting $\lambda$ from every entry instead of only the diagonal.
- Getting $\mathbf v = \mathbf 0$ as an eigenvector — it does not count: a non-zero one is needed.
- Expecting a single eigenvector. Any non-zero multiple is one too.
- Forgetting the normalization in a Markov chain: the probabilities must add up to $1$.