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Invertible matrices

  1. Mar 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Let A be an nxn matrix such that A^k=0 for some natural integer k (0 is the nxn zero matrix). Show that I + A is invertible, where I is the nxn identity matrix.

    2. Relevant equations

    Invertible implies det(I+A) not equal zero.

    3. The attempt at a solution

    I really don't know where to start with this one. I can see that A itself must be non-invertible, but I can't seem to get any more conditions on A based on that fact that A^k=0. Could anyone give me a hint please?

    Thanks.
     
  2. jcsd
  3. Mar 4, 2009 #2
    A hint :
    [tex]I-A^k=(I+A)(I-A+A^2-A^3+...+(-A)^{k-1})[/tex]
     
  4. Mar 4, 2009 #3

    tiny-tim

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    Homework Helper

    Hi kidsmoker! :smile:

    Forget determinants … use algebra, and construct an inverse!

    Hint: I = I - Ak :wink:

    Edit: ooh, boaz … that's too near a complete solution! :wink:
     
  5. Mar 4, 2009 #4

    statdad

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    Homework Helper

    To get you thinking, suppose [tex] A^3 = 0 [/tex]. What happens when you fully multiply and collect terms for the product

    [tex]
    (I + A) (I - A + A^2)
    [/tex]

    Use the distributive rule for multiplication, remember that [tex] A^3 = 0 [/tex]. If you make this work, you will have the idea for general case. (The pattern should remind you of the geometric series for numbers:

    [tex]
    \frac 1 {1 + x} = 1 -x + x^2 - x^3 + \dots
    [/tex]

    with the big difference that the matrix problem doesn't involve an infinite series.)
     
  6. Mar 4, 2009 #5

    statdad

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    Man, some of these other people are FAST on the keyboard.
     
  7. Mar 4, 2009 #6
    Okay thanks you guys, got it :p
     
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