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An eigenvalue λ satisfies Av = λv for some non-zero eigenvector v. Eigenvalues reveal the characteristic stretching/compressing directions of a linear transformation.
Trinn-for-trinn guide
- 1Solve det(A − λI) = 0 (characteristic equation)
- 2For 2×2: λ² − trace(A)λ + det(A) = 0
- 3Find eigenvectors by solving (A − λI)v = 0
Løste eksempler
Inndata
Matrix [[3,1],[1,3]]
Resultat
λ = 4 and λ = 2
Characteristic eq: λ²−6λ+8=0 → (λ−4)(λ−2)=0
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