1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Adjoint of an adjoint?

  1. Apr 8, 2007 #1

    daniel_i_l

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    I have a general question about adjoints, does:
    adj(adj(A)) = A?


    2. Relevant equations



    3. The attempt at a solution
    Intuitivly it looks right and the examples that i tried worked out but is it always true?
    Thanks.
     
  2. jcsd
  3. Apr 8, 2007 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What kind of object is A, and which definition of adjoint are you studying?
     
  4. Apr 8, 2007 #3

    daniel_i_l

    User Avatar
    Gold Member

    A is a nxn matrix and the definition that I'm using is the transpose of the cofactor matrix.
     
  5. Apr 8, 2007 #4

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    That isn't the adjoint, that is the adjugate.

    Adj(A) is the matrix that satisfies

    A*Adj(A)=det(A)I

    From this we see that (det(Adj(A))/det(A))*A is Adj(Adj(A)), when det(A) is non-zero. I leave it to you to determine what the det of Adj(A) is.
     
  6. Apr 8, 2007 #5

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    As I recall, one of the principal properties of the adjoint is
    A * adj(A) = det(A) * I​
    which should allow you to easily settle your question, at least for invertible matrices.

    (P.S. I too have heard this called the adjoint before)
     
  7. Apr 8, 2007 #6

    daniel_i_l

    User Avatar
    Gold Member

    matt grime: The det of adj(A) is det(A)^n-1 if A is invertible. So that means that adj(adj(A)) = det(A)^n-2 * A =/= A. But what happens is A is singular?

    Hurkyl: As i said above, this would mean that it isn't true for invertible matrices but what about the singular ones?
    Thanks!
     
  8. Apr 8, 2007 #7

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If you just want a specific example, there's the zero matrix.
     
  9. Apr 8, 2007 #8

    daniel_i_l

    User Avatar
    Gold Member

    That would be a case where it is true, is there a case where it isn't true?
    Thanks.
     
  10. Apr 8, 2007 #9

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You haven't proven they're unequal. In fact, you said yourself you know of examples where they are equal.
     
  11. Apr 8, 2007 #10

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Well, have you really tried many examples? The formula you derived suggests that most matrices do not satisfy the given equation... so the fact that you did not find a counterexample suggests that you really haven't tried much!
     
  12. Apr 8, 2007 #11

    daniel_i_l

    User Avatar
    Gold Member

    Well, they're only equal when det(A) = 1 and when A is invertible, not in general, right?
    Thanks.
     
  13. Apr 8, 2007 #12

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You have shown that, if you assume A is invertible, then they're equal iff det(A)^(n-2) = 1.

    In particular, it says nothing about the noninvertible case, and you missed some cases in the invertible case.

    Wait, is your arithmetic right? Shouldn't that be det(A)^(2n-2)?
     
    Last edited: Apr 8, 2007
  14. Apr 8, 2007 #13

    daniel_i_l

    User Avatar
    Gold Member

    What invertible cases did I miss? And how do I approach the non-invertible cases?
    And I think that the arithmetic is right:
    |adjA|/|A| = |A|^(n-1) / |A| = |A|^(n-1-1) = |A|^(n-2)
    Thanks!
     
    Last edited: Apr 8, 2007
  15. Apr 8, 2007 #14

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    In the non-invetible cases, there are differences depending on what the rank of A. It is easy to find cases where Adj(A) is the zero matrix. As Hurkyl has said, the number of cases where youir conjecture is true is vanishingly small, so you really can't have tried that many cases.
     
  16. Apr 8, 2007 #15

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    How did you deduce det(A) = 1? Is that something you're always allowed to do?
     
  17. Apr 9, 2007 #16

    daniel_i_l

    User Avatar
    Gold Member

    Your're right:) My problem was that I only tried 2x2 matrices, but for this invertable 3x3 matrix:
    [1 0 0]
    [0 1 0]
    [0 0 2]
    The equation isn't true. and for the non-invertable matrix where all the numbers are 1 it also isn't true.
     
  18. Apr 9, 2007 #17

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    It isn't true for many 2x2 matrices either.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Adjoint of an adjoint?
  1. Self adjointness (Replies: 7)

  2. Adjoint operators (Replies: 16)

  3. Adjoint of an operator (Replies: 1)

  4. Adjoint question (Replies: 4)

Loading...