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Hi all,

I heard a claim that if I have a matrix $A\in\mathbb R^{n\times n}$ such that $A^n \to 0 \ (\text{for }n\to\infty )$ (that is, every entry of $A^n$ converges to $0$ where $n\to \infty$) then $I-A$ is invertible.

anyone knows if there is a name for such a matrix or how (for general knowledge) to prove this ?

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  • $\begingroup$ Indeed, for some $n$ the norm of $A^n$ is less than 1, so $I-A^n$ is invertible, and a fortiori so is $I-A$, since it is a factor of it. $\endgroup$
    – Pietro Majer
    Commented Apr 11, 2011 at 18:42

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The matrices you are looking for are exactly those that have spectral radius (the max. of the absolute value of the eigenvalues) strictly less than one. I do not know whether there is a more specific name. (A matrix such that a finite power would be exactly the zero-matrix would be called nilpotent; but this is a different property.)

Regarding the invertibility of $I-A$. Note that (first only formally) $(I-A) (I + A + A^2 + \dots )=I$

To make this rigorous it suffices to show that $(I + A + A^2 + \dots )$ converges.

This can be done by noting that the spectral radius is 'almost' a matrix norm; more precisely, for $\varepsilon>0$ and all sufficiently large $k$ one has $||A^k|| \le (r + \varepsilon)^k$ where $r$ is the spectral radius. Now, you just have to sum a geometric series. For some more details and or background see e.g. http://en.wikipedia.org/wiki/Spectral_radius and http://en.wikipedia.org/wiki/Matrix_norm

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  • $\begingroup$ As for the name, I've heard them called ''convergent'' matrices (e.g., in the Berman-Plemmons book) or ''discrete-time stable'' matrices (common in the dynamical systems and control theory literature). The latter sounds more descriptive. $\endgroup$
    – Federico Poloni
    Commented Apr 11, 2011 at 18:57
  • $\begingroup$ Hi, thanks for the answer. I agree with everything you wrote, but the only thing I don't understand is why (sufficiently large k) ||A^k||<=(r+e)^k can you,please, explain why this is true ? $\endgroup$
    – user13743
    Commented Apr 11, 2011 at 19:00
  • $\begingroup$ @unknown: You are welcome. If you follow the link Spectral radius I gave and scroll down to the Theorem of Gelfand you will find (in its proof) a detailed argument for this. @Federico Poloni: thank you for suplementing this information. $\endgroup$
    – user9072
    Commented Apr 11, 2011 at 19:09
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    $\begingroup$ Also, by the spectral radius formula, $A^n\to 0$ immediately implies that $r(A)=\inf_{n\ge 0} \|A^ n \| ^ {1/n} < 1 $ because we have $\|A^n\|<1$ for some $n$. $\endgroup$
    – Pietro Majer
    Commented Apr 11, 2011 at 20:08
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No need for an infinite series. If $I - A$ was not invertible, there would be a nonzero vector $v$ with $A v = v$, and then $A^n v = v$ for all $n$, implying $A^n$ can't go to 0 as $n \to \infty$.

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It is quite easy:

Consider the sum $\sum_{n=0}^\infty A^n$. Your condition makes sure that this converges. At the same time, pretend that this is a usual, geometric series. Then the sum is given by $1/(1-A)$ or, if you wish, multiplicative inverse of $I-A.$

So in short, $I-A$ has an inverse, and it is given by the converging sum $\sum_{n=0}^\infty A^n$.

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