Re: Simple matrix proof

From: joccis (not.adomain_at_foo.com)
Date: 09/27/04 

Date: Mon, 27 Sep 2004 20:22:02 +0000 (UTC)

Thank you all for your imput. I figured out how to do it somewhat properly through MX=0 <=> X=0 by considering that you can always choose k so that there is at least one A^(k-1) which is not zero. However, the series approach was a complete and nice surprise after a day of worh with the problem :)

joccis

On 26 Sep 2004, Alain Verghote wrote:
>On 26 Sep 2004, joccis wrote:
>>Let A be a (n x n) matrix, such that A^k=0 for some k>1.
>>Show that I-A is non-singular.
>>
>>Proof
>>Let M=I-A, then
>> A^(k-1)M = A^(k-1)I - A^(k-1)A^k
>> A^(k-1)M = A^(k-1)-A^k
>> A^(k-1)M = A^(k-1)
>><=> M=I, hence M is non-singular.
>>
>>Am I missing something trivial here? If M=I, then A=0, and A^k=0 on every k and the statement turns out quite uninteresting.
>>
>>regards
>>joccis
>
>Dear joccis,
>
>We choose k = n and A^n=0 .
> Proof I/(I-A) exists .Because A^n=0;I/(I-A)=I+A+A^2+..A^(n-2)+A^(n-1)
> Look you don't need bigger power than n-1.
> If (I-A)^(-1) is defined then I-A is non-singular.
> Hope It helps.
>
>Alain.