invertability of a matrix - MathOverflow [closed]

invertability of a matrix

2011-04-11T18:23:00Z

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 ?

Answer by quid for invertability of a matrix

2011-04-11T18:44:49Z

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

Answer by Per Alexandersson for invertability of a matrix

2011-04-11T18:45:38Z

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 $\sum_{n=0}^\infty A^n$.

Answer by Robert Israel for invertability of a matrix

2011-04-11T19:53:15Z

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 $n \to \infty$.