Eigenvectors B₁, B₂, ..., B_N are linearly independent and form a basis for R^N. Thus, any vector V can be writen as a linear combination:

Then

Similarly,
A^k V= b₁ &lambda₁^k B₁ + b₂ &lambda₂^k B₂ + ... + b_N &lambda_N^k B_N

Dividing by &lambda₁^k gives:
(1/&lambda₁^k) A^k V= b₁ B₁ + b₂ (&lambda₂^k /&lambda₁^k) B₂ + ... + b_N (&lambda_N^k /&lambda₁^k )B_N

Since &lambda₁ is the largest eigenvalue, the expressions (&lambda_j^k /&lambda₁^k) converge to 0 as k increases.

Thus, if we start with any vector V that is not in the subspace spanned by the eigenvectors B₂ through B_N, then (1/λ₁^k)A^k V converges to an eigenvector B₁.

Without [just a little] more analysis, we do not know &lambda₁. Instead, we scale the result of each successive power A^k V, to make it a unit vector. With just modest assumptions on the nature of the eigenvalues, such a process of taking powers should converge to a principal eigenvector for the dominent eigenvalue &lambda₁

created: 10 February 2007 last revised: 15 February 2007

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