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Homework Help: Formula for the inverse problem

  1. Sep 6, 2005 #1
    If [tex]A^2 = 0[/tex], show that [tex]I - A[/tex] is invertible.

    So we know that [tex]\det(A^2) = (\det A)^2 = \det 0 \Leftrightarrow \det A = 0[/tex]

    We should now show that [tex]\det(I-A) \not= 0[/tex].

    But I'm not sure how to do that. Could someone kick me in the right directon?
     
  2. jcsd
  3. Sep 6, 2005 #2
    There's no need for determinants here. There is a very simply formula for the inverse of I - A (when A^2 = 0). Any further hints will practically give away the entire solution, but (I - A)^(-1) "looks" a lot like I - A...
     
  4. Sep 6, 2005 #3

    matt grime

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    forget matrices for one second, think taylor series

    what is (1-x)^{-1} as a series expansion when |x|<1?

    This can be suitably altered to tell us what the inverse of 1-A is for any A up to some convergence questions. Since A^2=0 these convergence questions vanish entirely as they do if instead of A^2=0 we had A^r=0

    obviously since you have been given r=2 i could be using something, talyor series, you've never heard of, and instead you are supposed to use something that, up to making a sign change, is what Muzza said (difference of two squares, anyone?)
     
  5. Sep 6, 2005 #4

    mathwonk

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    and you don't have to be clever. try showing I-A has no kernel. this suffices at least in finite dimensions.
     
  6. Sep 7, 2005 #5
    http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/series/taylor.html

    [tex](I-A)^{-1}=(I+A)\text{, since }A^n=A^2\cdot A^{n-2} = 0\cdot A^{n-2}=0[/tex]

    [tex](I - A)(I + A) = II + IA - AI - AA = I + A - A - 0 = I[/tex]
    [tex](I + A)(I - A) = II - IA + AI - AA = I - A + A - 0 = I[/tex]

    Now when I've solved this problem I would like to see your solutions (if it's different). It's allways good to learn more ways to solve a problem :smile:

    mathwonk: exactly what do you mean by 'kernel'?
     
    Last edited by a moderator: Apr 21, 2017
  7. Sep 7, 2005 #6

    HallsofIvy

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    The "kernel" of a linear transformation (or any function, for that matter) is the subset of the domain that get taken to 0: {x| f(x)= 0}. If a function is invertible, it must be one-to-one and (since for any f a linear transformation, f(0)=0) so the kernel of an invertible linear transformation can only be {0}.
     
  8. Sep 7, 2005 #7

    matt grime

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    Mathwonk's solution goes a little like this:

    suppose that (1-A)x=0, that is x is in the kernel, then that is the same as x=Ax, but now apply A to both sides, what happens? Now, if U and V are zero what is U+V?
     
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