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 ?



invertability of a matrix
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> 


