Barro35 schreef op 23 september 2015 23:16:
[...]
Alsjeblieft, veel succes er mee.
We will assume that the eigenvalues of S are $ \lambda_j $ and that
\[ 1 = \lambda_1 > |\lambda_2| \geq |\lambda_3| \geq \ldots \geq |\lambda_n| \]
We will also assume that there is a basis vj of eigenvectors for S with corresponding eigenvalues $ \lambda_j $ . This assumption is not necessarily true, but with it we may more easily illustrate how the power method works. We may write our initial vector I 0 as
\[ I^0 = c_1v_1+c_2v_2 + \ldots + c_nv_n \]
Then
\begin{eqnarray*} I^1={\bf S}I^0 &=&c_1v_1+c_2\lambda_2v_2 + \ldots + c_n\lambda_nv_n \\ I^2={\bf S}I^1 &=&c_1v_1+c_2\lambda_2^2v_2 + \ldots + c_n\lambda_n^2v_n \\ \vdots & & \vdots \\ I^{k}={\bf S}I^{k-1} &=&c_1v_1+c_2\lambda_2^kv_2 + \ldots + c_n\lambda_n^kv_n \\ \end{eqnarray*}
Since the eigenvalues $ \lambda_j $ with $ j\geq2 $ have magnitude smaller than one, it follows that $ \lambda_j^k\to0 $ if $ j\geq2 $ and therefore $ I^k\to I=c_1v_1 $ , an eigenvector corresponding to the eigenvalue 1.
It is important to note here that the rate at which $ I^k\to I $ is determined by $ |\lambda_2| $ . When $ |\lambda_2| $ is relatively close to 0, then $ \lambda_2^k\to0 $ relatively quickly. For instance, consider the matrix
\[ {\bf S} = \left[\begin{array}{cc}0.65 & 0.35 \\ 0.35 & 0.65 \end{array}\right]. \]