0
$\begingroup$

Assume that $A \in \mathbb R^{n×n}$ has $n$ linearly independent eigenvectors $u_1, u_2, . . ., u_n ∈ \mathbb C^n$ with associated eigenvalues $λ_1, λ_2, . . ., λ_n$ with $λ_1 = λ, λ_2 = −λ$, for some $λ \in \mathbb R$, and $λ_j \in \mathbb C$ with $|λ_j| < λ$ for $j = > 3, 4, . . . , n$. Assume $x \in \mathbb R^n$ has a non-zero component in the two-dimensional subspace spanned by the eigenvectors associated with the eigenvalues $λ$ and $−λ$. Prove that $$\lim_{k\rightarrow > \infty} \frac{\|A^{k+2}x\|}{\|A^{k}x\|}=\lambda^2.$$

My attempt: I suppose this is related to the power iteration method. $x = c_1u_1+c_2u_2+...+c_nu_n,$ then $A^k x = c_1\lambda_1^ku_1+c_2\lambda_2^ku_2+...+c_n\lambda_n^ku_n=\lambda^k(c_1u_1+c_2u_2)+...+c_n\lambda_n^ku_n.$ How to proceed?

$\endgroup$
1
  • 1
    $\begingroup$ Just divide the numerator and the denumerator by $\lambda^k$, then you are almost done. $\endgroup$ Apr 29, 2019 at 13:22

1 Answer 1

1
$\begingroup$

First start showing it is true for the case in which A is a diagonal.

After that, since A has n Eigen values, A is diagnosable. Make a change of basis (that does not change the norm) and then you are done

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .