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I have a function $f: \Re^2 \to \Re^2$ and would like to understand why

$$|Jf(x)|=\max_\theta|D_\theta f(x)|\cdot\min_\theta|D_\theta f(x)|$$

that is, why the determinant of the Jacobian of $f$ at $x$ is equal to the product of the minimum and the maximum of the directional derivatives at $x$.

I think it must be a more general matrix fact, but I'm stuck at the moment.

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    $\begingroup$ Very briefly (I hope this makes sense): in polar coordinates, $\theta\mapsto D_\theta f(x)$ is an ellipse, and determinant of the Jacobian is (proportional to) the area of this ellipse $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jan 31, 2018 at 5:57
  • $\begingroup$ Where does this result come from? Do you know if there is a version for $n$ dimensions? $\endgroup$
    – Jose Brox
    Commented Jan 31, 2018 at 6:26
  • $\begingroup$ @JoseBrox I believe follows easily from$$\begin{align*}D_\theta f(x)&=\left.\frac d{dt}f(x_1+t\cos(\theta),x_2+t\sin(\theta))\right|_{t\leftarrow0}\\&=(\partial_{x_1}f_1(x)\cos(\theta)+\partial_{x_2}f_1(x)\sin(\theta),\partial_{x_1}f_2(x)\cos(\theta)+\partial_{x_2}f_2(x)\sin(\theta))\end{align*}$$ $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Feb 1, 2018 at 9:51
  • $\begingroup$ In $n$ dimensions similarly you get the unit sphere dilated in the $\theta$ direction by $D_\theta f(x)$, so you must get volume of an ellipsoid in that case. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Feb 1, 2018 at 9:59

3 Answers 3

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Any matrix $A$ can be written as $B.U$ where $B=\sqrt{AA^\top}$ is posititive semidefinite and $U$ is orthogonal (polar decomposition). Thus $|\det(A)| =|\det(B)|.|\det(U)|$ is the product of the eigenvalues of $B$ which are $\ge 0$ and which are also called the singular values or s-numbers of $A$. Your max.min formula multiplies the two singular values of the Jacobian.

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  • $\begingroup$ I have the impression that, when properly rephrased, this should be equivalent to the fact that the Gaussian curvature of a surface in 3-space is the product of the principal curvatures, but I can't quite put my finger on it. ? $\endgroup$
    – Mikhail Katz
    Commented Apr 10 at 15:54
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This is an addendum to the answer of Peter Michor simply to point out that the situation is even simpler since the Jacobean matrix is symmetric. It is thus just the basic fact in linear algebra---the determinant of a symmetric matrix is the product of its eigenvalues (so in the $2\times 2$ case the product of the larger and the smaller one). There is a direct extension to the general case using the minimax characterisation of the eigenvalues of a symmetric matrix (for which see the classical text of Courant and Hilbert).

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    $\begingroup$ Not every Jacobian matrix is symmetric. Consider $f(x) = (x_2, 0)$. $\endgroup$
    – gerw
    Commented Jan 31, 2018 at 12:22
  • $\begingroup$ If (and, locally, only if) $f$ is the gradient of a function $g$, then the Jacobian of $f$ equals the Hessian of $g$ and is thus symmetric. $\endgroup$
    – Peter Michor
    Commented Feb 1, 2018 at 9:44
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As you point out, there are some general matrix facts in here. Consider the eigendecomposition of a matrix: $A = S \Lambda S^{-1}$, where $S$ and $S^{-1}$ are orthonormal (implying their determinant is 1), so we have $|A| = |\Lambda|$. Since $\Lambda$ is a diagonal matrix (with entries equal to eigenvalues), the determinant is the product of the eigenvalues. In the two dimensional case: $|A| = \lambda_1 \lambda_2$.

How does this connect with minimum and maximum directional derivatives? Any matrix is a linear transformation, which if we stick to the direction of the eigenvectors $v_i$, it just means applying some stretching; think of the definition of eigenvalues: $Av_i = \lambda_i v_i, \; \lVert Av_i\rVert = |\lambda_i|$ (where we are considering unitary vectors $\lVert v_i \rVert = 1$). Any other direction can be expressed as an average of eigenvectors (again we are working with unitary vectors here), and it will also be strectched by an average of the eigenvalues: $$x = av_1 + (1-a)v_2 \; \rightarrow \; Ax = A(av_1 + (1-a)v_2) = a\lambda_1 v_1 + (1-a)\lambda_2 v_2 \; \rightarrow \\ \rightarrow \lVert Ax \rVert = |A| \lVert av_1 + (1-a)v_2 \rVert = a|\lambda_1| + (1-a)|\lambda_2|, \; a \in (0,1). $$

(We can pull the vectors $v_i$ out of the norm $\lVert \cdot \rVert$ preserving equality because they are orthogonal). This means the amount of stretching will be something in between $\lambda_1$ and $\lambda_2$, or in other words, that $\lambda_1$ and $\lambda_2$ represent the extreme (maximal and minimal) amounts of stretching.

So in summary, the determinant equals the product of the eigenvalues, which represent (in 2D) the maximal and minimal amounts of stretching of a linear transformation and in the case of the Jacobian these are maximal and minimal directional derivatives.

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