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Mathematics I

Eigenvalues and Eigenvectors

Business I 242 words Free to read

The Directions a Matrix Cannot Bend

Apply a matrix to most vectors and they rotate and stretch. But special eigenvectors only get scaled along their line. Their scaling factor is the eigenvalue λ\lambda:

Av=λv,v0Av = \lambda v, \qquad v \ne 0

Finding them: Rewrite as (AλI)v=0(A - \lambda I)v = 0. A nonzero solution exists only if AλIA - \lambda I is singular, meaning its determinant is zero:

det(AλI)=0\det(A - \lambda I) = 0

This characteristic polynomial yields the eigenvalues as its roots. For each root, solving (AλI)v=0(A-\lambda I)v = 0 gives its eigenspace.

For a triangular matrix, eigenvalues sit right on the diagonal. Always use quick checksums to catch errors before moving forward.

The characteristic polynomial graphed against lambda, with the

The Eigen-Recipe & Dynamics

Step / CheckEquationYieldsScope
1. Polynomialdet(AλI)=0\det(A - \lambda I) = 0Eigenvalues λ\lambdaRoots
2. Nullspace(AλI)v=0(A-\lambda I)v = 0Eigenspace vvVectors
Check: Traceλi=tr(A)\sum \lambda_i = tr(A)Sum testErrors
Check: Detλi=detA\prod \lambda_i = \det AProduct testErrors

Why dynamics care: Repeated multiplication aligns states with the dominant eigenvector (largest λ|\lambda|).

λ>1:growthλ<1:decayλ=1:steady|\lambda| > 1: \text{growth} \quad |\lambda| < 1: \text{decay} \quad \lambda = 1: \text{steady}

Markov matrices have λ=1\lambda = 1; their eigenvector gives the long-run market share or Google's PageRank.

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Mathematics I