Eigenvalues and eigenvectors
Av = λv means v is an eigenvector with eigenvalue λ. Solve det(A − λI) = 0.
Find eigenvalues from the characteristic equation det(A − λI) = 0. Then for each λ, solve (A − λI)v = 0 for the eigenvector.
Worked examples
A = (4 1; 2 3); det(A − λI) = (4-λ)(3-λ) − 2 = λ² − 7λ + 10. Eigenvalues 2, 5.
Eigenvector for λ = 2: (A − 2I)v = 0 ⇒ v = (1; −2).
Diagonalisation A = PDP−1 where D has eigenvalues on the diagonal.
Frequently asked questions
Why eigenvalues?
They reveal axes of stretch and compression; used in stability analysis, principal component analysis.
Always real?
Symmetric matrices: yes. General: may be complex.