invertability of a matrix 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 ?
 A: 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 $n \to \infty$.   
A: 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 
A: 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 $\sum_{n=0}^\infty A^n$.
