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Proving that for nilpotent A, A+I is invertible

  1. Jul 15, 2010 #1
    Suppose A is nilpotent, i.e. [tex]A^k=0[/tex], [tex]k>0[/tex]. Show that [tex]A+I[/tex] is invertible.

    Here's my go at it... although I'm unsure about step marked with asterisks... is it ok? I'm not sure how to show that it's true for sure.

    Contrapositive: Suppose [tex]A+I[/tex] isn't invertible. Then [tex](A+I)X =0 [/tex] has a non-zero solution, say [tex]X = X_0[/tex].
    Then

    [tex](A+I)X_0=0[/tex]

    [tex]AX_0+X_0=0[/tex]

    [tex]A^kX_0 +A^{k-1}X_0=0[/tex]

    If [tex]A^k=0[/tex] then [tex]A^{k-1}X_0=0[/tex]... but [tex]X_0[/tex] is not a solution of [tex]A^{k-1}X=0[/tex].************

    Hence, [tex]A[/tex] must not nilpotent, QED

    Thanks
     
  2. jcsd
  3. Jul 15, 2010 #2

    Office_Shredder

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    Why?
     
  4. Jul 15, 2010 #3
    Ok... I was bluffing. But I don't know how else to proceed!

    I just assumed that since [tex]X_0[/tex] was the solution of one equation it couldn't solve another different equation.
     
  5. Jul 15, 2010 #4

    Office_Shredder

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    It helps to gather some intuition from the one dimensional case here.

    What is the inverse of (1+x) when x is small (x is a real number)?
     
  6. Jul 15, 2010 #5
    If you mean multiplicative inverse isn't it

    [tex]\frac{1}{1+x} = 1-x+x^2-...[/tex]

    Can this be used for matrices too...?

    EDIT: Heh, I guess it can, thanks office_shredder
     
    Last edited: Jul 15, 2010
  7. Jul 15, 2010 #6

    Office_Shredder

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    Of course you have to be worried about convergence, but in the nilpotent case that's not a big deal
     
  8. Jul 27, 2010 #7
    I am a little slow, so could someone please humor me: just using 1/(1 + x) = 1 - x + x^2 - ...
    and seeing x as A and 1 as I, isn't it apparant that this series converges since it is geometric and can be easily calculated? At what point does the nilpotency of A matter, and how can you use it in proof using the mentioned series?
     
  9. Jul 27, 2010 #8

    Mark44

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    For some integer power k, Ak = 0 (the zero matrix).

    1/(1 + x) in the reals corresponds to (I + A)-1.
     
  10. Jul 27, 2010 #9

    Office_Shredder

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    If A is just a real number and A=2, the series doesn't converge. The mere fact it is geometric isn't enough; you need your matrix to somehow be "small enough". What exactly small enough is is a lot less obvious for matrices than for real numbers, but if a matrix is nilpotent it qualifies
     
  11. Aug 4, 2010 #10
    [tex](A+I)v=0[/tex]

    [tex]Av=-v[/tex]

    [tex]A^2v=v[/tex]

    [tex]\dots[/tex]

    [tex]A^kv=(-)^{k+1}v=0[/tex]

    [tex]v=0[/tex]

    [tex]v=0[/tex] is the only vector in [tex]\textrm{Ker}(A+I)[/tex]

    [tex]A+I[/tex] is invertible.
     
  12. Aug 8, 2010 #11
    Its not true. If X^k=0

    [tex]e^X=Id+A+A^2/2+\dots A^{k-1}/(k-1)![/tex]
     
  13. Aug 8, 2010 #12
    First, X^k = 0 is equivalent to being nilpotent. Second, you may want to replace all your As by Xs, but you are right. Sometimes you do get Exp(X)=X+id but not always. I forgot you could have something like a Heisenberg matrix where exponentiation puts ones on the diagonal but does alter one entry.
     
  14. Aug 8, 2010 #13
    Suppose (I+A) is not invertible. Then there exists a vector v such that (I+A)v = 0.

    Then v=-Av. Then -1 is an eigenvalue which is a contradiction since A is nilpotent i.e. all eigenvalues are zero.
     
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