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Determinant proof

  1. Aug 12, 2005 #1
    Is this proof correct?

    Show that det(A + B^T) = det(A^T + B)

    det(A + B^T) = det(A^T + B)
    det(A) + det(B^T) = det(A^T) + det(B)
    det(A) + det(B) = det(A) + det(B)

    Thanks.
     
  2. jcsd
  3. Aug 12, 2005 #2

    HallsofIvy

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    det(A+B) is not, in general, equal to det(A)+ det(B).

    What is true is that (A+B)T= AT+ BT so that, in particular (AT+ B)T= A+ BT.
     
    Last edited: Aug 12, 2005
  4. Aug 12, 2005 #3

    AKG

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    Well your "proof" is backwards for one, and second it's invalid. It seems you're suggesting that det(X + Y) = det(X) + det(Y). What if X = I2 (the 2x2 identity matrix) and Y = -I2?

    The determinant is not a linear function, however transposition is linear. Use that fact plus the trivial fact that matrix addition is commutative, plus the fact that the det(AT) = det(A), to get the desired result.
     
  5. Aug 12, 2005 #4
    Hmm, so is this the correct method of doing it?

    det(A + B^T) = det(A^T + B)
    det(A + B^T)^T = det(A^T + B)^T
    det(A^T + B) = det(A + B^T)

    I can't do this, correct? Thanks for the hints though.
     
  6. Aug 12, 2005 #5

    TD

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    Well, I think it will be enough if you do it on one side :wink:

    det(A + B^T) = det(A^T + B)
    det(A + B^T)^T = det(A^T + B)
    det(A^T + B) = det(A^T + B)
     
  7. Aug 12, 2005 #6

    AKG

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    det(AT + B) = det((AT + B)T) = det(ATT + BT) = det(A + BT)
     
  8. Aug 12, 2005 #7

    HallsofIvy

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    This proof, as TD wrote:
    det(A + B^T) = det(A^T + B)
    det(A + B^T)^T = det(A^T + B)
    det(A^T + B) = det(A^T + B)
    which starts with what you want to show and arrives at a true statement is valid if you make sure each step is reversible! (That's often called "synthetic proof".)

    But AKG's suggestion:
    det(AT + B) = det((AT + B)T) = det(ATT + BT) = det(A + BT)
    is much nicer!
     
  9. Aug 12, 2005 #8
    Thanks everyone!
     
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