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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?

Rodrigo de Azevedo
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MrX
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  • @ MrD , if you are a young man in a hurry, then do running and not mathematics. On the other hand, the problem that you were asked was certainly not the one you offer; the right question would be: show that $(A^k)_k$ converges to a projector. In fact, it is not obvious that the sequence converges; in particular, if $A$ has two eigenvalues equal to $1$ (and not only one), then the result is false. –  Sep 18 '16 at 22:57
  • @loupblanc , thanks for the advice I will consider that, however, as you might have noticed we have one eigenvalue that is equal to 1 and the rest are strictly less than one in magnitude. Yet the question that you've raised and answered is interesting ... – MrX Sep 20 '16 at 05:34

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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 $\lambda_1=1$.

Related question which may be of interest: Eigenvalues and power of a matrix

Community
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tibL
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    Thanks for the edit and the answer, I was in hurry, I should have thought about it before posting the question... – MrX Sep 17 '16 at 08:27