Convergence of the power of a matrix with one eigenvalue equal to $1$ and the others strictly less in magnitude

Question

I have an $ n \times n$ matrix $A$ that has eigenvalues $\lambda_1 =1 > | \lambda_i|, \forall 1<i\leq n$ Then let $$A^{\infty }= \lim_{k\rightarrow \infty } A^k$$ What can we say about the eigenvalues of $A^{\infty }$? Note that $A$ not necessary diagonalizable?

Answer 1

In general if $A$ has an eigenvalue $\lambda$ then there exists some eigenvector $v$ such that

$$ Av = \lambda v $$

so that $$ A Av = \lambda Av = \lambda^2 v$$

hence the $k$th powers of the eigenvalues are the eigenvalues of $A^k$ you can conclude from there that the spectrum of $A^\infty$ will be all zero apart from $\lambda_1=1$.

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