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I'm trying to show that if $ \lambda_1 >0 \Rightarrow $ the vectors $ x^{(k)} $ converges to an eigenvector. Where k is the number of iteration, and k $ \rightarrow \infty $ I started by writing $ x^{(k)}$ $$ x^{(k)}= \frac{A x^{(k-1)} }{ ||A x^{(k-1)}||} $$ $$ x^{(k)}= \frac{A^{k-1} x^{(0)} }{ ||A^{k-1} x^{(0)}||} $$

In the assigment i was told, that i could use 1. there is a single eigenvalue of max modulus 2. there is a linearly independent set of n eigenvalues

I just can't see that this can help me. I tried to write out the coordinates, but that didn't help

Viktor Jeppesen
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Using fact (2), write your initial seed vector $x^{(0)}$ as a linear combination of all the eigenvectors, say: $$x^{(0)} = a_1 x_1 + a_2 x_2 + \cdots + a_n x_n,$$ where (by fact (1)) $x_1$ is the eigenvector with largest eigenvalue . Then, $$A^k x^{(0)} = a_1 \lambda_1^k x_1 + a_2 \lambda_2^k x_2 + \cdots + a_n \lambda_n^k x_n.$$ When you normalize, I think you should find that the denominator grows like $\lambda_1^k$ so that, in the limit, you're left with $x_1$.

Mark McClure
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