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Prove the derivative of the determinant equals the sum of the derivatives of ea/row

  1. Sep 7, 2011 #1

    syj

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    1. The problem statement, all variables and given/known data

    PROVE:
    If A(t) is nxn with elements which are differentiable functions of t
    Then:
    [itex]\frac{d}{dt}[/itex](det(A))=[itex]\sum[/itex]det(Ai(t))
    where Ai(t) is found by differentiating the ith row only.

    2. Relevant equations
    I know I should prove this by induction on n



    3. The attempt at a solution
    Consider the matrix A1 being a 1x1 matrix
    So n=1
    the derivative of the determinant is the same as the derivative of that one row, therefore the theorem holds for n=1

    assume the proof will hold true for n=k call this matrix Ak

    now prove the theorem holds true for n=k+1
    [itex]\frac{d}{dt}[/itex](det(Ak+1)) =[itex]\frac{d}{dt}[/itex](det(A1)+[itex]\frac{d}{dt}[/itex](det(Ak))
    AND
    [itex]\sum[/itex](det(Ak+1) = [itex]\sum[/itex]det(A1)+[itex]\sum[/itex]det(Ak)

    Is this it?
    have I proved it?
     
  2. jcsd
  3. Sep 7, 2011 #2
    Re: Prove the derivative of the determinant equals the sum of the derivatives of ea/r

    I had posted the exact same question a while ago and was too lazy to solve it.

    I can't understand how you justify your first equation.
    I can assure you the proof is much longer.
     
  4. Sep 7, 2011 #3

    syj

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    Re: Prove the derivative of the determinant equals the sum of the derivatives of ea/r

    The determinant of a 1x1 matrix is equal to that one entry in the matrix isnt it?
    so the derivative of the determinant is equal to the derivative of the one function in A.
     
  5. Sep 7, 2011 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Re: Prove the derivative of the determinant equals the sum of the derivatives of ea/r

    Sounds like an excellent plan!
    Good.

    Now, you have lost me. A "k+1 by k+ 1" determinant is NOT the sum of a 1 by 1 matrix and a k by k determinant which what you appear to be saying since you just use the "sum rule" for the derivative.

    What is true is that a determinant can be "expanded" on one row. That is, we can, for example, expand by the first row. The k+1 by k+1 determinant is the sum of each entry in the first row time (plus or minus) the k by k matrix got by removing the first row and appropriate column. Use that to do a proof by induction.
     
  6. Sep 7, 2011 #5

    syj

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    Re: Prove the derivative of the determinant equals the sum of the derivatives of ea/r

    :blushing:
    I knew there was something fishy about my proof.
    It just seemed too simple
    I shall try to follow your advice ;)
    I guarantee I shall be posting questions soon :)
    thanks again
     
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