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 September 20th, 2012, 06:49 AM #1 Member   Joined: Aug 2012 Posts: 30 Thanks: 0 invertible matrix I have some difficulty proving the bolded part of the exercise: If A is an nxn matrix establish the identity Isub(n)-A^(k+1)=(Isub(n)-A)(Isub(n)+A+A^2+...+A^k). Deduce that if some power of A is the zero matrix then Isub(n)-A is invertible. Suppose now that A=2 2 -1 -1 -1 0 0 0 -1 -1 1 0 0 1 -1 1 Compute the powers (Isub(n)-A)^i for i=1,2,3,4 and, by considering A=Isub(4)-(Isub(4)-A), prove that A is invertible and determine A-1. I would be grateful for any help you are able to provide... September 21st, 2012, 08:47 AM   #2
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Re: invertible matrix

Quote:
 Originally Posted by shine123 I have some difficulty proving the bolded part of the exercise: If A is an nxn matrix establish the identity Isub(n)-A^(k+1)=(Isub(n)-A)(Isub(n)+A+A^2+...+A^k).
That's pretty much direct multiplication: and $-A(I_n+ A+ A^2+ \cdot\cdot\cdot+ A^k)= -A- A^2- A^3-\cdot\cdot\cdot- A^{k-1}[/tex]. Now add those. Quote:  Deduce that if some power of A is the zero matrix then Isub(n)-A is invertible. If , the taking k= j- 1 in the formula above [latex]I_n- A^j= I_n= (I_n-A)(I_n+ A+ A^2+ \cdot\cdot\cdot+ A^{j-1}$.

Quote:
 Suppose now that A=2 2 -1 -1 -1 0 0 0 -1 -1 1 0 0 1 -1 1 Compute the powers (Isub(n)-A)^i for i=1,2,3,4 and, by considering A=Isub(4)-(Isub(4)-A), prove that A is invertible and determine A-1. I would be grateful for any help you are able to provide...
With that A, and you can calculate the powers directly. What do you get? Tags invertible, matrix Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Magnesium Linear Algebra 2 December 11th, 2013 02:09 AM frankpupu Linear Algebra 3 March 6th, 2012 11:45 AM problem Linear Algebra 3 August 31st, 2011 05:30 AM 450081592 Linear Algebra 1 June 24th, 2010 07:13 AM paulyc2010 Algebra 4 April 18th, 2010 02:50 PM

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