Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Lenear algebra question

  1. Jan 30, 2009 #1
    i am given a square matrix called A(they dont specify how it looks)
    i am given that
    A^2=0

    I-unit matrix

    find (A-I)^-1
    ??


    -1 is the inverse command

    i can pick many matrices which A^2=0
    and do this easy subtraction and inverse function

    but i need here a general solution
    ??
     
  2. jcsd
  3. Jan 30, 2009 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Maybe this question can be of help to you :smile:

    Or, if you want another approach, notice the real number identity
    [tex]\frac{1}{x - 1} = \frac{1}{x - 1} \times \frac{x + 1}{x + 1}[/tex]
     
  4. Jan 30, 2009 #3

    Mark44

    Staff: Mentor

    I believe that where they're heading with this problem is expressing (A - I)^(-1) as an infinite series. Another poster had a similar problem yesterday.

    A problem that is somewhat analogous is the series representation for 1/(1 -a) = 1 + a + 1/2! * a^2 + 1/3! * a^3 + ... + 1/n! * a^n + ...

    You're trying to find (A - I)^(-1), which is (-(I - A))^(-1). This latter expression can be shown (I think) to be equal to -(I - A)^(-1), which should be equal to
    -[I + A + 1/2! * A^2 + 1/3! * A^3 + ... + 1/n! * A^n + ...]

    As a sanity check, I picked a 2 X 2 matrix A = [0 1; 0 0] (shown row by row), and things worked out, which gave me some evidence that I was on the right track.


    BTW, I is usually called an identity matrix.
     
  5. Jan 30, 2009 #4

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Yes, only it won't become infinite. And I posted a link to that question in my earlier post.
     
  6. Jan 30, 2009 #5

    Mark44

    Staff: Mentor

    Didn't want to give everything away:smile:
     
  7. Jan 30, 2009 #6
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook