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Linear Algebra - polynomial functions of matrices

  1. Sep 16, 2011 #1
    1. The problem statement, all variables and given/known data
    Given p(x) = x4+2x2+1 and
    A =
    [[1 1 -2 0]​
    [0 1 0 2]​
    [1 1 -1 1]​
    [0 0 -2 -1]]​
    p(A) = 0
    Find a polynomial q(x) so that q(A) = A-1
    a) What is q(x)?
    b) Compute q(A) = A-1

    2. Relevant equations
    I found the Cayley-Hamilton theorem, which states: p(x) = det(A-xIn).


    3. The attempt at a solution
    I found the inverse of A, which is:
    [[-1 -1 2 0]​
    [4 1 -4 -2]​
    [1 0 -1 -1]​
    [-2 0 2 1]]​
    From here I thought I could set q(x) = A-1.
    [[-1-x -1 2 0]​
    [4 1-x -4 -2]​
    [1 0 -1-x -1]​
    [-2 0 2 1-x]]​
    Then I should just solve out to find the det(). I used the 2nd column as a reference point.
    Doing this I end up with the equation, x4-6x2+1.
    so q(x) = x4-6x2+1.

    This seems to be the wrong answer. I'm not at all sure if I did this right, or if I made an error in my calculations. Would really appreciate if someone could look over and point out my error.

    Thanks
     
  2. jcsd
  3. Sep 16, 2011 #2

    Mark44

    Staff: Mentor

    You went at it the hard way.

    Did you check that p(A) = 0? I did, and it turns out that this is so.

    p(A) = 0 ==> A4 + 2A2 + I = 0
    ==> -A4 - 2A2 = I
    If you can factor an A out of the left side, you'll be almost where you need to go.
     
  4. Sep 16, 2011 #3
    I have a similar problem, and was wondering what part b was asking for.
     
  5. Sep 16, 2011 #4

    Mark44

    Staff: Mentor

    It's asking for A-1.
     
  6. Sep 16, 2011 #5
    That's what I thought, but it wants a polynomial. Thank you.
     
  7. Sep 16, 2011 #6

    Mark44

    Staff: Mentor

    And A-1 will be that polynomial in A; i.e., q(A).
     
  8. Sep 17, 2011 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    As Mark44 said before, [itex]-A^4 - 2A^2 = A(-A^3- 2A)= I[/itex]
    Now what is [itex]A^{-1}[/itex]?
     
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