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Suppose A is a square matrix.

Does $A^n$(matrix multiplication) converge when n is an infinite big number?

Is it always true or under certain circumstances?

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2 Answers 2

up vote 4 down vote accepted

If the Jordan normal form of the matrix $A$ is $J$, then you have $A=PJP^{-1}$ and this yields $A^n=PJ^nP^{-1}$. So we only have to ask when powers of the Jordan blocks of the given matrix converge. The structure of powers of Jordan blocks is relatively simple.

It is relatively easy to see that the power $A^n$ converges to zero matrix if $|\lambda|<1$ for all eigenvalues of $A$. (See e.g. the result at the end of this text.)

If the only eigenvalue with absolute value 1 is 1 and the corresponding Jordan blocks have size 1, then it converges. If there is a Jordan block corresponding to 1 of size at least 2, then the power does not converge. (This was pointed out by Ted in the comments, thanks for the correction.)

If there are complex eigenvalues different from 1 with $|\lambda|=1$ then the power does not converge.

If it has an eigenvalue with $|\lambda|>1$, it does not converge.

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In which case they converge to the zero matrix –  jspecter Sep 25 '11 at 7:39
@jspecter Thanks for your commment, I've added it to the answer. –  Martin Sleziak Sep 25 '11 at 7:42
If there is an eigenvalue 1 in a Jordan block $J$ of size > 1, then the powers of the Jordan block won't converge: $J=I+N$ where $N$ is a nonzero nilpotent matrix with entries in $\{0,1\}$, so $(I+N)^k$ equals $I + kN + C(k,2)N^2 + \ldots + N^k$. The off-diagonal entries go to $\infty$ as $k \to \infty$. –  Ted Sep 25 '11 at 7:50
@Ted I have overlooked this possibility, thanks for noticing and notifying me. I've corrected this part of my answer. –  Martin Sleziak Sep 25 '11 at 7:58

The convergence properties of $A^n$ are fully understood by studying the eigenspectrum of $A$. The work http://arxiv.org/pdf/1407.0671.pdf by H.H. Bauschke thoroughly covers the subject, and beyond.

For example, a sufficient condition for convergence is that the spectral radius of $A$ be strictly less than $1$. In fact, in this case, $\underset{n \rightarrow \infty}{lim}A^n = 0$.

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