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Related to inverses

  1. Feb 28, 2005 #1
    I have not come across this kind of problem, hence, not sure:

    A is nxn, such that A^2 + 3A + I = 0
    show that A is invertible and A^(-1) = -A-3I

    So far I have this:

    -A^2 - 3A = I

    but now I am not sure if I can factor out A:

    (-1)A [A + 3] = I

    because I do not think that A + 3 is defined for matrices, since 3 is a scalar and not actually a vector.

    Any hints are very much appreciated.
     
  2. jcsd
  3. Feb 28, 2005 #2

    Hurkyl

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    3 A = 3 A I
     
  4. Feb 28, 2005 #3
    Thanks I will try to work it out.
    But just to confirm, it's wrong to factor in this case?
    So far I have this solution, even though I factored afterall :redface:

    (-1) A [A + 3] = I

    (-1)A A^(-1) [A + 3] = I A^(-1)

    (-1) [A + 3] = I A^(-1)

    I^(-1) (-1)[A + 3] = I^(-1) I A^(-1)

    -AI - 3I = A^(-1)

    :confused:
     
    Last edited: Feb 28, 2005
  5. Feb 28, 2005 #4

    Hurkyl

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    Factoring is fine for matrices, they satisfy the distributive law:

    P(Q + R) = PQ + PR
    and
    (P + Q)R = PR + QR
     
  6. Feb 28, 2005 #5
    Oh, so I can represent 3A as 3 A I and then factor out A and end up with 3 I which is theoretically a matrix?
     
  7. Feb 28, 2005 #6

    Hurkyl

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    It's not just theoretically a matrix. :smile:
     
  8. Feb 28, 2005 #7

    Hurkyl

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    Anyways, an easier way to do the problem is to simply multiply A and the hypothesized A^-1, and check if you get I. (Don't forget to apply the polynomial identity A satisfies)

    However, the work you did would be necessary if you weren't given a guess for A^-1.
     
  9. Mar 1, 2005 #8
    Shish! I see it now :bugeye: .
    Thanks!
     
  10. Mar 1, 2005 #9

    HallsofIvy

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    However, you appear to be assuming that A has an inverse in your work which is what you want to prove.

    What you can do is simply start with A^2 + 3A + I = 0, write that
    as -A^2- 3A= I or (-A-3I)A= A(-A-3I)= I. Now what is the DEFINITION of inverse?
     
  11. Mar 2, 2005 #10
    so, then from this by DEFINITION of inverse:
    AB=I=BA, where B is inverse, and given A(-A-3I)=I i can conclude that B in this case is (-A-3I).
    I hope I am not making assumptions again...
    Thank you all.
     
  12. Mar 3, 2005 #11

    HallsofIvy

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    Yes, of course. The inverse of A is just -A-3I.
     
  13. Mar 3, 2005 #12

    mathwonk

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    so the general principle is that if A satisfies a polynomial with non zero constant coiefficient then A is invertible. (over a field).

    think about a diagonal matrix, with diagonal entries c1,....cn.

    Then A satisfies (A-c1)(A-c2)....(A-cn), which has non zero constant term if and only if all the diagonal entries ci were non zero.
     
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