learn.howToCalculate
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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.
ଷ୍ଟେପ୍-ଷ୍ଟେପ୍ ଗାଇଡ୍ |
- 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
ସମାଧାନ ହୋଇଥିବା ଉଦାହରଣ
ଇନପୁଟ୍
Matrix [[3,1],[1,3]]
ଫଳ
λ = 4 and λ = 2
Characteristic eq: λ²−6λ+8=0 → (λ−4)(λ−2)=0
learn.ctaText
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