invertability of a matrix - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-23T17:10:22Z http://mathoverflow.net/feeds/question/61316 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61316/invertability-of-a-matrix invertability of a matrix unknown (google) 2011-04-11T18:23:00Z 2011-04-11T19:53:15Z <p>Hi all,</p> <p>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.</p> <p>anyone knows if there is a name for such a matrix or how (for general knowledge) to prove this ?</p> http://mathoverflow.net/questions/61316/invertability-of-a-matrix/61322#61322 Answer by quid for invertability of a matrix quid 2011-04-11T18:44:49Z 2011-04-11T18:44:49Z <p>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.)</p> <p>Regarding the invertibility of $I-A$. Note that (first only formally) <code>$(I-A) (I + A + A^2 + \dots )=I$</code></p> <p>To make this rigorous it suffices to show that <code>$(I + A + A^2 + \dots )$</code> converges. </p> <p>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 <code>$||A^k|| \le (r + \varepsilon)^k$</code> 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. <a href="http://en.wikipedia.org/wiki/Spectral_radius" rel="nofollow">http://en.wikipedia.org/wiki/Spectral_radius</a> and <a href="http://en.wikipedia.org/wiki/Matrix_norm" rel="nofollow">http://en.wikipedia.org/wiki/Matrix_norm</a> </p> http://mathoverflow.net/questions/61316/invertability-of-a-matrix/61323#61323 Answer by Per Alexandersson for invertability of a matrix Per Alexandersson 2011-04-11T18:45:38Z 2011-04-11T18:45:38Z <p>It is quite easy:</p> <p>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 $I-A.$</p> <p>So in short, $I-A$ has an inverse, and it is given by the converging sum $\sum_{n=0}^\infty A^n$.</p> http://mathoverflow.net/questions/61316/invertability-of-a-matrix/61330#61330 Answer by Robert Israel for invertability of a matrix Robert Israel 2011-04-11T19:53:15Z 2011-04-11T19:53:15Z <p>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 $n \to \infty$. </p>