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Nilpotent Matrix Proof

  1. Mar 29, 2008 #1
    1. The problem statement, all variables and given/known data
    If B is any nilpotent matrix, prove that I-B is invertible and find a formula for (I-B)^-1 in terms of powers of B.




    3. The attempt at a solution
    If I make a matrix <<ab,cd>> then if 1/(ad-bc)[tex]\neq[/tex]0 then the matrix has an inverse. Since I think all nilpotent matrices have a 0,0,0 leading diagonal with the other diagonal being not fully "0"s. Wouldn't it be impossible for nilpotent matrices to not have an inverse? I think I may have my wording jumbled.
     
  2. jcsd
  3. Mar 29, 2008 #2
    A nilpotent matrix cannot have an inverse. Say B^n = 0 where n is the smallest positive integer for which this is true. Now suppose it were invertible and let C be it's inverse. Then CB = I. But then 0 = CB^n = B^(n-1), a contradiction.


    As to your original problem, you know B^n = 0 for some n.

    Start small, suppose B^2 = 0 then notice (I - B)(I + B) = I. Now suppose B^3 = 0 what's an inverse for I - B in this case? Generalize this.
     
    Last edited: Mar 29, 2008
  4. Mar 29, 2008 #3
    Hmm. Why would you start out by assuming B^2 = I? Shouldn't it be 0 rather than I?
     
    Last edited: Mar 29, 2008
  5. Mar 29, 2008 #4
    that was a typo thanks, fixed, should be B^2 = 0
     
  6. Mar 30, 2008 #5
    I do not understand how this leads to me getting a formula for the inverse of (I-B).
    The only equation I can think of that relates to this is (I-B)^-1(I-B)=I.... and I wouldn't have a clue how to change that if the powers of B were changing... any other ideas to get my brain into gear?
     
  7. Mar 30, 2008 #6

    HallsofIvy

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    (I-B)(I+B)= I tells you that I+ B is the inverse of I-B; its product with I- B is I- that's all an inverse does.

    If you don't get the idea yet, look at B3= 0. Then B3- I= -I or I-B3= I. But I- B3= (I- B)*what? Since their product is I, whatever that factor is is the inverse of I- B.

    Now, can you find the general formula?
     
  8. Mar 30, 2008 #7
    From that I'm assuming the general formula I need is I-B^k=(I-B)(I+B)?
    I hope thats right, thanks alot you are very helpful
     
  9. Mar 30, 2008 #8

    HallsofIvy

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    ?? Surely that's not what you meant to say! (I- B)(I+ B)= I- B2, not I- Bk for any k! Do you know the general factorization formula for xn- 1?
     
  10. Apr 9, 2008 #9
    Similar to an old calculus problem...

    You've already written "(I-B)^-1". What sorts of things have you learned in previous classes about scalar expressions like (1-B)^-1? (... or, perhaps more familiarly, 1/(1-x)?) You have seen a way to write this that depends only on powers of B, and because of the extra thing you know about B, you can ignore almost all of those powers.

    In general, directly using results from scalar equations is treacherous. Nevertheless, taking an equation or expression involving less familiar objects and rewriting it in terms of more familiar objects (like using "1" in place of "I" above) may provide a starting direction for a proof.
     
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