Why determinant gives area enlargement?

• Jul 8th 2009, 07:59 AM
Twig
Why determinant gives area enlargement?
Hi!

Iīm trying to fully understand why the determinant gives the area enlargement when switching between variables.

This might belong in calculus, but the book says linear algebra "shows this" so here it is.

Like when we do polar coordinates: $x = r\cdot cos(\theta) \mbox{ and } y = r\cdot sin(\theta)$

Then we multiply with the jacobian-determinant, and get an additional $r$ .

Thx
• Jul 8th 2009, 08:56 AM
Random Variable
The simplest explanantion is that the new area element after the transformation from rectangular coordinates is approximately a paralleogram. (In three dimensions, the new volume element is approximately a parallelepiped.)

Do you know the the relationship between determinants and the area of parallograms (and the volume of parallelepipeds)?
• Jul 8th 2009, 09:46 AM
Twig
Hi!

For two dimensions, isnt the determinant area of parallellogram?

About the parallellepipeds, I belive itīs a triple scalar product or something like that.
• Jul 8th 2009, 10:52 AM
Random Variable
Here's the best link I could find:

http://syssci.atu.edu/math/faculty/f...934/cal167.pdf
• Jul 8th 2009, 11:44 AM
Laurent
The answer is twofold. A piece of differential calculus, and a piece of linear algebra.

Let me first set notations. Consider a diffeomorphism $\phi:U\to V$ where $U$ and $V$ are open subsets of $\mathbb{R}^n$, and $f:V\to\mathbb{R}$ be a measurable function (piecewise continuous, for instance) that is integrable on $V$. The change of variable formula is

$\int_U f(\phi(x)) |J\phi(x)| dx=\int_V f(y) dy$

where $J\phi(x)$ is the Jacobian of $\phi$ at $x$.

In particular, if $V$ has finite area, then

${\rm area}(V)=\int_U |J\phi(x)| dx$.

Thus, indeed, $|J\phi(x)|$ measures the area enlargement at point $x$. Let me try to explain how this can be understood. (Note that I write ${\rm area}$ even if this would rather be called "volume" in dimension higher than or equal to 3)

First, remember (?) that $J\phi(x)=\det \left(\frac{\partial\phi_i}{\partial x_j}\right)_{1\leq i,j\leq n}$ can also be written $J\phi(x)=\det(d_x\phi)$, where $d_x\phi$ is the differential of $\phi$ at point $x\in U$. The differential of $\phi$ is the linear function that approximates $\phi$ at the best near $x$. In dimension 1, this amounts to approximating $\phi$ by its tangents. To state it rigorously, we have (by definition) $\phi(x+h)=\phi(x)+(d_x\phi)(h)+o_{h\to 0}(\|h\|)$ hence $\phi(x+h)\simeq \phi(x)+(d_x\phi)(h)$.

Imagine $U$ is a very small neighbourhood of $x_0$. Then $\phi(x_0+h)$ can be approximated by $\phi(x_0)+(d_{x_0} \phi)(h)$ for any $h$ (necessarily small) such that $x_0+h\in U$. Hence $V=\phi(U)\simeq \phi(x_0)+(d_{x_0}\phi)(U)$, and thus ${\rm area}(V)\simeq {\rm area}((d_{x_0}\phi)(U))$.

We are thus reduced to finding how the linear map $A=d_{x_0}\phi$ transforms the area of $U$. We need to find ${\rm area}(A(U))$ where $A$ is a linear (non-singular) map.

a) If $A(x)=\lambda x$ then it is well-know that ${\rm area}(A(U))=|\lambda|^n {\rm area}(U)$, hence ${\rm area}(A(U))=|\det A| {\rm area}(U)$ (remember I work in dimension $n$). Good point.

b) If $A$ is a rotation, then of course ${\rm area}(A(U))={\rm area}(U)$, hence ${\rm area}(A(U))=|\det A|{\rm area}(U)$ since the determinant of a rotation is 1. The same works for any orthogonal map $A$ (rotation + symmetry, determinant $\pm 1$).

c) If $A$ is a diagonal matrix $A=\begin{pmatrix} a_1 & & \\ & \ddots & \\ & & a_n\end{pmatrix}$, then $A$ maps the cube $[0,1]^n$ to the "rectangle" $[0,a_1]\times [0,a_2]\times\cdots\times [0,a_n]$ which has volume $|a_1|\times \cdots\times |a_n|$, and similarly ${\rm area}(A(C))=|a_1\cdots a_n|{\rm area}(C)=|\det A|{\rm area}(C)$ for any other cube (change its sidelength using paragraph a) and rotate it using b)). Then if we decompose $U$ into small cubes (up to an error tending to 0), we see that we have as well ${\rm area}(A(U))=|\det A|{\rm area}(U)$, even if $U$ is not a cube.

From a),b),c), we can deduce that ${\rm area}(A(U))=|\det A|{\rm area}(U)$ for any linear map $A$. Indeed, any linear map can be decomposed into a product of orthogonal and diagonal maps. This results from the "polar decomposition" (for matrices) or "OS-decomposition". Any linear map is the product of an orthogonal map and a symmetric map, and a symmetric map is orthogonally diagonalizable (on $\mathbb{R}$). This was the linear algebra part. Maybe someone can provide an intuitive explanation of the polar decomposition for matrices if you ask for one. I don't have one in mind right now...

To conclude the connection between both parts, we finally have ${\rm area}(\phi(U))={\rm area}(V)\simeq {\rm area}( (d_{x_0}\phi)(U))= |\det d_{x_0}\phi|{\rm area}(U)$ $=|J\phi(x_0)|{\rm area}(U)$ for a small set $U$ containing $x_0$. Heuristically, if " ${\rm area}(U)=dx$", " ${\rm area}(\phi(U))=\phi(dx)$" and we have thus proved " $\phi(dx)=|J\phi(x)|dx$". Summing over small pieces where $\phi$ is almost linear, we deduce that the integral formula ${\rm area}(V)=\int_U |J\phi(x)|dx$ is true for "large" sets $U$.

Differential calculus always plays with interactions between calculus and linear algebra: near a point, any (smooth) map is almost linear, hence we can take its determinant, invert it,... Looking more closely, we can use a second-order approximate of $\phi$, which is a quadratic map instead of a linear map. Then we can apply bilinear algebra to study local extrema... Plenty of nice things. (Nod)
• Jul 8th 2009, 12:06 PM
Twig
I really liked the link you gave me RandomVariable.

I have one question about its contents though.

In the picture below, I dont quite follow from the first double sum to the last. The first says area of $R_{ij}$ , which feel reasonable.

But in the last I interpret it as the whole area R. That canīt be right.

Thx!
• Jul 8th 2009, 12:34 PM
Laurent
Quote:

Originally Posted by Twig
I really liked the link you gave me RandomVariable.

I have one question about its contents though.

In the picture below, I dont quite follow from the first double sum to the last. The first says area of $R_[ij}$ , which feel reasonable.

But in the last I interpret it as the whole area R. That canīt be right.

Thx!

You're correct. The factor $\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\D elta u\Delta v$ corresponds to the parallelogram $R_{ij}$ hence it should depend on $i,j$. You should add indices $i,j$. Or: the parallelograms are assumed to have the same size, perhaps, hence $\Delta u\Delta v$ wouldn't depend on the indcies. The ratio however is evaluated at point $(u_{ij},v_{ij})$, which depends on $i,j$ anyway, it should have been specified.

The fact that we split the area $R$ into small parallelograms allows to apply the first part which was about a "small $R$" (the smallness is because we need to identify the mapping with its differential: this is local).
• Jul 8th 2009, 02:47 PM
Twig
Thanks!