Suppose f : R 3 → R 3 such that
f ( x , y , z ) = ⎣ ⎡ x + y + z x y + y z + z x x y z ⎦ ⎤
Determine the determinant of the total derivative of f at each of the points ( 0 , 0 , 1 ) , ( 0 , 1 , 0 ) , ( 0 , 1 , 1 ) , ( 1 , 0 , 0 ) , ( 1 , 0 , 1 ) , ( 1 , 1 , 0 ) , and ( 1 , 1 , 1 ) .
What is the value of the sum of the determinants?
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How can we tell that the Jacobi determinant is 0 when two coordinates are equal, without working through all of the calculations?
Hint: Why isn't the map invertible?
How can we tell that the Jacobi determinant is 0 when two coordinates are equal, without working through all of the calculations?
Hint: Why isn't the map invertible?
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Well the map isn't invertible as it is not one-to-one as f ( 1 , 0 , 0 ) = f ( 0 , 1 , 0 ) = f ( 0 , 0 , 1 ) and f ( 0 , 1 , 1 ) = f ( 1 , 0 , 1 ) = f ( 1 , 1 , 0 ) due to the order of x , y , and z not mattering by how f is defined.
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Not quite. Invertible is a local property, whereas you are are comparing global (far away) points.
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@Calvin Lin – Could you provide a bit more insight into this? Invertibility is not exactly my strong point at this time. I realize that I used global invertibility in my last comment rather than local invertibility, since I couldn't exactly think of how to apply an argument for local invertibility.
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@Jequil Hartz – Because the functions are symmetric, when you are at the point f ( 1 , 1 , 0 ) = ( 2 , 1 , 0 ) , and you get a slight perturbation which sends you to the point which evaluates to ( 2 + ϵ , 1 + ϵ , 0 ) , you do not know whether you are at the point ( 1 + ϵ , 1 , 0 ) or ( 1 , 1 + ϵ , 0 ) .
Hence, this map isn't invertible when x = y , and thus we can conclude that the Jacobian is 0.
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@Calvin Lin – Ah. I see. Thanks for the explanation.
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The total derivative or Jacobian, J , of a function is defined as the matrix with the partial derivatives of the component functions of f .
f ( x ) = ⎣ ⎢ ⎡ f 1 f 2 f 3 ⎦ ⎥ ⎤ ⟹ J = ⎣ ⎢ ⎢ ⎡ ∂ x ∂ f 1 ∂ x ∂ f 2 ∂ x ∂ f 3 ∂ y ∂ f 1 ∂ y ∂ f 2 ∂ y ∂ f 3 ∂ z ∂ f 1 ∂ z ∂ f 2 ∂ z ∂ f 3 ⎦ ⎥ ⎥ ⎤
⟹ J = ⎣ ⎢ ⎡ 1 y + z y z 1 x + z x z 1 x + y x y ⎦ ⎥ ⎤
⟹ d e t ( J ) = x y ( x − y ) + y z ( y − z ) + z x ( z − x )
= − ( x − y ) ( y − z ) ( z − x ) .
Thus, the determinant is zero whenever any two of the components of the point are equal which is true for every point given.
Therefore the answer is 0 .