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If A is nilpotent , prove that the matrix ( I+ A ) is invertible

  1. Sep 19, 2012 #1
    if A is nilpotent " A^k = 0 , for some K > 0 " , prove that the matrix ( I+ A ) is invertible ..

    ****
    I found more than a topic in the website talk about this theorem biu every one of them didn't produce a complete proof !!

    I found the question in artin book and I tried to solve this problem and I solved it when K is even , I could calculate the inverse but I hope to find the hole proof here ..
     
  2. jcsd
  3. Sep 19, 2012 #2

    micromass

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    Please show an attempt at the solution. For example, how did you solve the even case?
     
  4. Sep 19, 2012 #3

    My attempt is wrong !!

    I've found that it's wrong , any way here is it

    ( 1 + A ) ( 1 - A ) = 1 - A^2 : 1 denote I matrix

    ( 1 - A^2 ) ( 1 + A^2 ) = 1 - A^4

    continue in that way , if k is even , so the power of A is even to and the power of A " let us denote it by n " will take every possible even number

    when you multiply time ater time n will be equal to K then A^k = 0 so
    ( 1 + A ) ( 1 - A ) ( 1 + A^2 ) .... ( 1 + A^ c ) = 1
    where 1 + 1 + 2 + 4 + 4 + 8 + ..+ c = K

    so " ( 1 - A ) ( 1 + A^2 ) .... ( 1 + A^ c ) " is the inverse

    but that is not true because K may be 6 for example but you can't get 6 with this way

    so this way is useful when K = 2^ m

    that's my attempt
     
  5. Sep 19, 2012 #4

    Hurkyl

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    What way can you get 6?

    Alternatively, if you can't make 6, can you still do with 4 or 8 or something else?
     
  6. Sep 19, 2012 #5

    micromass

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    So, if I understand you right, you want to find an inverse of the form

    [tex]a_0 I+ a_1 A + a_2 A^2 + .... + a_n A^n[/tex]

    where [itex]A^{n+1}=0[/itex].

    Now, why don't you multiply

    [tex](I+A)(a_0 I+ a_1 A + a_2 A^2 + .... + a_n A^n)[/tex]

    this to see what you can take as [itex]a_i[/itex]?
     
  7. Sep 19, 2012 #6
    I read your reply before you edit it :))

    you are right I can make A^ 8 = A^6 * A^2 = 0

    so it's useful for even numbers :))

    thanks alot , I forgot that if A^k = 0 then A^(k+n) = 0 : n >0

    Thanks , But do you know a complete proof ?
     
  8. Sep 19, 2012 #7

    micromass

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    Yes, we know the complete proof. But the idea is that you should figure out yourself instead of us telling you what it is.
     
  9. Sep 19, 2012 #8

    I understand , I tried to solve the problem when K is odd , but my mistake was that I use the inverse of A and I forgot that A is not invertible !!
     
  10. Sep 19, 2012 #9

    micromass

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    Did you read Hurkyl's suggestion and my suggestion? Certainly that will give you something to do?
     
  11. Sep 19, 2012 #10
    I found that : if we make

    a(n) = -a(n+1)

    then ,
    a0 I + a1 A + a2 A^2 + .. + an A^n

    the sum will be ( a0 * I )

    so , let a0 = 1 then a1 = -1 then a2 = 1 and so on I will get the I matrix

    so

    the matrix

    ( 1 - A + A^2 - A^3 + ... (-+)A^n )
    is the inversr of ( I + A)

    Is that right ?!
     
  12. Sep 19, 2012 #11
    yes , I did .. I replied both of you
     
  13. Sep 19, 2012 #12

    micromass

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    Yes, that is right. That is indeed the inverse of I+A.
     
  14. Sep 19, 2012 #13

    ok , Thank you very much :))

    and my greetings to Hurkyl too ..

    I want to say something ,

    the way that you follow is great " i mean that you made me get the inverse myself and didn't gave it to me " , I tried so much for 2 days to get this inverse but I couldn't , now i'm so happy :))


    thank tou again ..

    my last question !

    Is there a proof which prove that ( 1 + A ) is invertible without getting the inverse ?

    I mean , is there a way prove that it's invertible without knowing what is the inverse ?!!
     
  15. Sep 20, 2012 #14
    Hi everyone .

    Maths Boy I advice you to use the Latex codes its very cute .

    something like that :

    $$I_{3 \times 3} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
     
  16. Sep 20, 2012 #15
    I don't know how to use latex codes !
     
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