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Nilpotent What is its index?

  • Thread starter Bob Ho
  • Start date
18
0
1. Homework Statement
A square(nxn) matrix is called nilpotent of index k if A[tex]\neq[/tex]0, A^2[tex]\neq[/tex]0,....A^(k-1)[tex]\neq[/tex]0, But A^k=0 for some positive integer K

Verify that A={{{021,002,000}}} is nilpotent. What is its index? Show that for this matrix (I-A)-1= I + A + A^2




3. The Attempt at a Solution

I am unsure how different values of k affects the matrix... but For the equation (I-A)-1= I + A + A^2, I found the inverse of (I-A) which was {{{125,012,001}}}, which then gave me A^2 as {{{004,000,001}}}.
 
Last edited:

Answers and Replies

Dick
Science Advisor
Homework Helper
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You got the inverse of (I-A) ok. I'm assuming by A2 you mean A^2 (i.e. A squared, A times A). That didn't come out right. To find the index of A just keep computing successive powers A until you get zero. What is the index? To show (I-A)^(-1)=(I+A+A^2) you just want to show (I-A)*(I+A+A^2)=I. To do that you don't even need the numerical value of A once you've found it's index.
 
18
0
Thanks alot! I figured out A^3=0 so im assuming the nilpotent matrix is of index 3.
I'm not sure how you managed to get the equation "(I-A)*(I+A+A^2)=I." I just did I=(I-A)^-1 -A - A^2. Would that also be suitable?
 
Dick
Science Advisor
Homework Helper
26,258
618
You've got the index too. So A^3=0. If C^(-1)=D then that just means C*D=I. So to check if (I-A)^(-1)=(I+A+A^2) just check whether (I-A)*(I+A+A^2)=I. That's where I got it. To check that just multiply that expression out. Remembering A^3=0.
 

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