WWJD (What would Jacobi do?)

Calculus Level 5

Suppose f : R 3 R 3 f: {\mathbb{R}}^{3} \rightarrow {\mathbb{R}}^{3} such that

f ( x , y , z ) = [ x + y + z x y + y z + z x x y z ] f(x,y,z) = \begin{bmatrix} x + y + z \\ xy + yz + zx \\ xyz\end{bmatrix}

Determine the determinant of the total derivative of f f at each of the points ( 0 , 0 , 1 ) , ( 0 , 1 , 0 ) , ( 0 , 1 , 1 ) , ( 1 , 0 , 0 ) , ( 1 , 0 , 1 ) , ( 1 , 1 , 0 ) , (0,0,1),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0), and ( 1 , 1 , 1 ) (1,1,1) .

What is the value of the sum of the determinants?


The answer is 0.

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1 solution

Jequil Hartz
Aug 21, 2015

The total derivative or Jacobian, J J , of a function is defined as the matrix with the partial derivatives of the component functions of f f .

f ( x ) = [ f 1 f 2 f 3 ] J = [ f 1 x f 1 y f 1 z f 2 x f 2 y f 2 z f 3 x f 3 y f 3 z ] \large f(x) = \begin{bmatrix} {f}_{1} \\ {f}_{2} \\ {f}_{3} \end{bmatrix} \Longrightarrow J = \begin{bmatrix} \frac{\partial {f}_{1}}{\partial x} & \frac{\partial {f}_{1}}{\partial y} & \frac{\partial {f}_{1}}{\partial z} \\ \frac{\partial {f}_{2}}{\partial x} & \frac{\partial {f}_{2}}{\partial y} & \frac{\partial {f}_{2}}{\partial z} \\ \frac{\partial {f}_{3}}{\partial x} & \frac{\partial {f}_{3}}{\partial y} & \frac{\partial {f}_{3}}{\partial z} \end{bmatrix}

J = [ 1 1 1 y + z x + z x + y y z x z x y ] \large \Longrightarrow J = \begin{bmatrix} 1 & 1 & 1 \\ y + z & x + z & x + y \\ yz & xz & xy \end{bmatrix}

d e t ( J ) = x y ( x y ) + y z ( y z ) + z x ( z x ) \large \Longrightarrow det(J) = xy(x - y) + yz(y - z) + zx(z - x)

= ( x y ) ( y z ) ( z x ) \large = -(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 \boxed{0} .

Moderator note:

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?

Calvin Lin Staff - 5 years, 9 months ago

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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 ) 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 ) f(0,1,1) = f(1,0,1) = f(1,1,0) due to the order of x , y , x, y, and z z not mattering by how f f is defined.

Jequil Hartz - 5 years, 9 months ago

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Not quite. Invertible is a local property, whereas you are are comparing global (far away) points.

Calvin Lin Staff - 5 years, 9 months ago

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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.

Jequil Hartz - 5 years, 9 months ago

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@Jequil Hartz Because the functions are symmetric, when you are at the point f ( 1 , 1 , 0 ) = ( 2 , 1 , 0 ) 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 ) (2 + \epsilon, 1+ \epsilon, 0 ) , you do not know whether you are at the point ( 1 + ϵ , 1 , 0 ) (1 + \epsilon, 1, 0 ) or ( 1 , 1 + ϵ , 0 ) ( 1 , 1 + \epsilon, 0 ) .

Hence, this map isn't invertible when x = y x = y , and thus we can conclude that the Jacobian is 0.

Calvin Lin Staff - 5 years, 9 months ago

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@Calvin Lin Ah. I see. Thanks for the explanation.

Jequil Hartz - 5 years, 9 months ago

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