Jacobian matrix determinant formula

The jacobian matrix has some really interesting properties when applied to surface and volume geometry. What is the formula that would give out the surface/volume element of any dimensional object? As example let's consider calculating the jacobian of a 26 dimensional sphere. Calculating this by hand would take years and without the help of a general formula for spherical determinant, it would be quite hard. Another problem comes out, what is the proof for this kind of formula? Lets start by writing an n by n square matrix.

\left[ {\begin{array}{*{20}{c}} {\frac{{\partial {x_1}}}{{\partial {\theta _1}}}}&{\frac{{\partial {x_1}}}{{\partial {\theta _2}}}}&{\frac{{\partial {x_1}}}{{\partial {\theta _3}}}}&{\frac{{\partial {x_1}}}{{\partial {\theta _4}}}}& \ldots &{\frac{{\partial {x_1}}}{{\partial {\theta _n}}}}\\ {\frac{{\partial {x_2}}}{{\partial {\theta _1}}}}&{\frac{{\partial {x_2}}}{{\partial {\theta _2}}}}&{\frac{{\partial {x_2}}}{{\partial {\theta _3}}}}&{\frac{{\partial {x_2}}}{{\partial {\theta _4}}}}& \ldots &{\frac{{\partial {x_2}}}{{\partial {\theta _n}}}}\\ {\frac{{\partial {x_3}}}{{\partial {\theta _1}}}}&{\frac{{\partial {x_3}}}{{\partial {\theta _2}}}}&{\frac{{\partial {x_3}}}{{\partial {\theta _3}}}}&{\frac{{\partial {x_3}}}{{\partial {\theta _4}}}}& \ldots &{\frac{{\partial {x_3}}}{{\partial {\theta _n}}}}\\ {\frac{{\partial {x_4}}}{{\partial {\theta _1}}}}&{\frac{{\partial {x_4}}}{{\partial {\theta _2}}}}&{\frac{{\partial {x_4}}}{{\partial {\theta _3}}}}&{\frac{{\partial {x_4}}}{{\partial {\theta _4}}}}& \ldots &{\frac{{\partial {x_4}}}{{\partial {\theta _n}}}}\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ {\frac{{\partial {x_n}}}{{\partial {\theta _1}}}}&{\frac{{\partial {x_n}}}{{\partial {\theta _2}}}}&{\frac{{\partial {x_n}}}{{\partial {\theta _3}}}}&{\frac{{\partial {x_n}}}{{\partial {\theta _4}}}}& \cdots &{\frac{{\partial {x_n}}}{{\partial {\theta _n}}}} \end{array}} \right]

The determinant of a matrix can be obtain from the sum of all the product of the right-oriented diagonals then subtract the sum of all the product of the left-oriented diagonals. The first line of diagonal is made completely of matching numbers, X 1, Theta 1, X 2 Theta 2, so it's easy to find the rule for this one. When it comes to the other diagonals, numbers are following another order each time, the formula I found to calculate it is this:

∣DetJ∣=∣∏k=1n(∂xk∂θk−∂xk∂θn+1−k)+∑m=1n−1(∏k=1n−m∂xk∂θk+m∏k=1m∂xn−m+k∂θk−∏k=1n−m∂xk∂θn−k−m+1∏k=1m∂xn−m+k∂θn−k+1)∣\left| {De{t_J}} \right| = \left| {\prod\limits_{k = 1}^n {\left( {\frac{{\partial {x_k}}}{{\partial {\theta _k}}} - \frac{{\partial {x_k}}}{{\partial {\theta _{n + 1 - k}}}}} \right)} + \sum\limits_{m = 1}^{n - 1} {\left( {\prod\limits_{k = 1}^{n - m} {\frac{{\partial {x_k}}}{{\partial {\theta _{k + m}}}}\prod\limits_{k = 1}^m {\frac{{\partial {x_{n - m + k}}}}{{\partial {\theta _k}}} - \prod\limits_{k = 1}^{n - m} {\frac{{\partial {x_k}}}{{\partial {\theta _{n - k - m + 1}}}}\prod\limits_{k = 1}^m {\frac{{\partial {x_{n - m + k}}}}{{\partial {\theta _{n - k + 1}}}}} } } } } \right)} } \right|

This is where my interrogation comes from. How do I prove this formula? It seems very complex and I have no reference or any idea how to prove it. I tested in a 3 dimensional space with coordinates change from Cartesian to spherical to compute the volume of a sphere and it worked without any problem. I'm wondering if anyone have an idea how to prove it or if it's wrong, I would like to know where the equation breaks.

#HelpMe! #Advice #Math

Note by Samuel Hatin
7 years, 9 months ago

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Comments

That method of taking the sum of all the products of the right-oriented diagonals minus the sum of all the products of the left-oriented diagonals only works for 1×11 \times 1, 2×22 \times 2, and 3×33 \times 3 matrices.

You can verify that ∣2100120000210012∣=9\left| \begin{matrix} 2 & 1 & 0 & 0 \\ 1 & 2 & 0 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 1 & 2 \end{matrix}\right| = 9, but that method yields 1515 as the determinant.

Jimmy Kariznov - 7 years, 9 months ago

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I didn't know that it only worked for 2 and 3 dimension matrix. The other method is like splitting 1 matrix up with smaller ones with the sign matrix. Thanks for making me realize this. I will try to compute a general formula using the sum of the product of the determinant of smaller matrix.

Samuel Hatin - 7 years, 9 months ago

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The determinat of an n×nn\times n matrix, whose (i,j)(i,j)th component is ai,ja_{i,j}, is ∑π∈Snsgn(π)a1,π(1)×a2,π(2)×a3,π(3)×⋯×an,π(n) \sum_{\pi \in S_n} \mathrm{sgn}(\pi) a_{1,\pi(1)}\times a_{2,\pi(2)}\times a_{3,\pi(3)} \times \cdots \times a_{n,\pi(n)} where the sum is over all permutations π\pi of nn symbols, and sgn(π)\mathrm{sgn}(\pi) is the sign of the permutation. Every permutation can be written as a product of transpositions (permutations that just swap two elements), and the parity (evenness/oddness) of the number of transpositions required is fixed for each permutation. We say that sgn(π)=1\mathrm{sgn}(\pi)=1 if an even number of transpositions is required, and −1-1 if an odd number is needed.

Each term a1,π(1)×a2,π(2)×⋯×an,π(n) a_{1,\pi(1)}\times a_{2,\pi(2)} \times \cdots \times a_{n,\pi(n)} is one of the n!n! ways of simultaneously taking one element from each row and each column, and multiplying them together. The sgn(π)\mathrm{sgn}(\pi) term tells whether you should add or subtract it in.

I am afraid you are not going to find any much simpler general formula than this.

Mark Hennings - 7 years, 9 months ago
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