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Does $\lim_{n\to\infty} A^n = 0$, when the maximum of the norms of the eigenvalues of $A$ is less than $1$? Background: I am not familiar with matrix theory yet. Thanks in advance!

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    Over what field are you considering matrices? Complex numbers? Real numbers? Note that for example in the real case, you have to take nonreal eigenvalues into account. Because you say you are not familiar with matrix theory, I suppose you have not seen Jordan form? If you just want the answer, it is "yes."2012-05-01
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    @JonasMeyer: Thank you! I would like to know why? The matrix can be real or complex, and its eignevalues can be real or complex. I have seen Jordan form in linear algebra, but not quite remember the details.2012-05-01
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    If you are familiar with the concept of operator norm, this follows directly...2012-05-01
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    @sos440: Thanks! Can you elaborate? I think I can understand.2012-05-01
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    If the matrix is diagonalizable, this question is trivial.2012-06-14

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If you know that the spectral radius of $A$ is the limit, when $n$ goes to infinity, of $||A ^n ||^{1 / n}$, then this is immediate.

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    You only need to know that the spectral radius is at most this limit, which is easier to prove.2012-06-14
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    Can't we just say that since the eigenvalues of $A^n$ are $\lambda^n$ so they all approach $0$. the only matrix with $n$ eigenvalues $0$ is the zero matrix and so we are done ?2012-08-13