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Linear Algebra: Proof involving basic properties

  1. Sep 25, 2008 #1
    1. The problem statement, all variables and given/known data

    A square matrix A is called nilpotent if A^k =0 for some k>0. Prove that if A is nilpotent then I+A is invertible.


    3. The attempt at a solution

    My guess would be to do a proof by induction (on the size of the matrix)

    So for the trivial cases:

    Let A be a 2x2 Nilpotent matrix... thus it is of the form

    [0 x] [0 0]
    [0 0] Or [x 0]


    Clearly when we add I to A in this case, we get get a matrix, whose det =/= 0


    Im having trouble doing the more general cases, seeing as that i cannot mentally see what nilpotent matrices of a larges size look like.

    Is this even the best way to approach this problem?

    Any help is appreciated.
     
  2. jcsd
  3. Sep 25, 2008 #2
    Would using the fact that the trace of a nilpotent matrix is equal to 0 help in any way? perhaps coupled with the trace of I= n?
     
  4. Sep 26, 2008 #3

    Defennder

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    I can think of one way to do this. If I+A is invertible, then det(I-A) is non-zero, and that implies that 1 is not an eigenvalue of A. So we need to prove that last statement. Do it by contradiction:

    Suppose 1 is an eigenvalue of A, that implies there is a nonzero vector v such that Av=v. Given that A is nilpotent, can such a non-zero vector v exist?

    Unfortunately this method isn't constructive; while it does show that I+A is invertible, it cannot come up with the inverse of I+A itself. So does anyone else have a better idea?
     
  5. Sep 26, 2008 #4

    HallsofIvy

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    Oh, I think it can be done directly.

    Remember that [tex]1- x^k= (1+ x)(1- x+ x^2+ \cdot\cdot\cdot+ (-1)^k x^{k-1})[/tex]? If that still works for matrices, then taking x= A and k such that Ak= 0, you are done.
     
  6. Sep 29, 2008 #5
    How does this statement prove that it is invertible?
     
  7. Sep 29, 2008 #6

    gabbagabbahey

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    If [itex]I+A[/itex] is invertible then [itex]I=(I+A)(I+A)^{-1}[/itex] ...does that look anything like what Hallsofly posted?:wink:
     
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