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

  • Thread starter syj
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  • #1
syj
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Homework Statement



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.

Homework Equations


I know I should prove this by induction on n



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?
 

Answers and Replies

  • #2
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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.
 
  • #3
syj
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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.
 
  • #4
HallsofIvy
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Homework Statement



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.

Homework Equations


I know I should prove this by induction on n
Sounds like an excellent plan!

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
Good.

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))
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.

And [itex]\sum[/itex](det(Ak+1) = [itex]\sum[/itex]det(A1)+[itex]\sum[/itex]det(Ak)

Is this it?
have I proved it?
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.
 
  • #5
syj
55
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: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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