If $A$ is nonsingular $n \times n$ complex matrix with n distinct eigenvalues.
If $|\lambda_1| > |\lambda_i| , i=2,..,n$
Show the sequence $\vec{x_k}= \lambda_1^{-k}A^k\vec{x}$ converges to eigenvector $\vec{y}$ with eigenvalue $\lambda_1$, for most vectors $\vec{x}$, and what restrictions must be put on $\vec{x}$ .
First I will give my proof. My question is really related to the bold part above I am quite uncertain about how good my answer/proof is.
Proof:
$\vec{x_k}= \lambda_1^{-k}A^k\vec{x}$ .
A has $n$ distinct eigenvalues therefore:
$D = B^{-1}AB$ is diagonal with B the matrix of eigenvectors of A.
$\therefore \vec{x_k}= \lambda_1^{-k}BD^kB^{-1}\vec{x}= BD_0^kB^{-1}\vec{x}$ .
With $D_0 = diag(\frac{\lambda_i}{\lambda_1}) , i=1,..,n$ .
$\implies D_0^k \rightarrow diag(1,0,..,0)$ as $k \rightarrow +\infty$ .
$\implies \vec{x_k}= \vec{v}B^{-1}\vec{x} = \vec{y}$
Where $\vec{v}$ is the first column of B and the eigenvector of A corresponding to eigenvalue $\lambda_1$.
This is where I run into trouble. This can be shown to be an eigenvalue as long as:
(a): $x$ is not the zero vector.
(b): ......
And I cant figure out what other restrictions would have to be put on x.
If $(\lambda_i, v_i)$ is a eigenvalue, eigenvector pair of matrix $A$, then $(\lambda_i^k, v_i)$ is a pair for $A^k$. Since $A$ is diagonalizable, any vector $x$ can be written as $$x = \sum_{i=1}^n \alpha_iv_i.$$ $$\lambda_1^{-k}A^kx = \sum_{i=1}^n\alpha_i\left(\frac{\lambda_i}{\lambda_1}\right)^kv_i \rightarrow\alpha_iv_1$$ So the conclusion holds as long as $\alpha_1\ne 0$.