I know determinants tell you the oriented volume of the parallelepiped after the linear transformation, but if you define the derivative as I do below then it seems equivalent, at least in terms of the “slope” of the linear transformation. Or at least I hope someone can answer what exactly it is I’m defining below. It gives the derivative in single variable but doesn’t give best linear approx. in multivariate case. Is it a kind of directional derivative?

The typical definition of a derivative is $$f'(x) \approx \frac{f(x+dx) - f(x)}{dx}$$ where $dx$ is an infinitesimal (appealing to nonstandard analysis here for simplicity).

This is usually motivated as the slope of the tangent line to a point at $f(x)$, but the more general formulation of the derivative is the ratio of oriented intervals:

Let $I_{ab}$ be an oriented interval in the domain of a function $f$, in most cases we will want this to be an infinitesimal interval so an interval of $[a, a+dx]$. More generally an oriented interval is an oriented n-dimensional hypercube/hyper-rectangle/parallelotope, if $f: \mathbb{R} \rightarrow \mathbb{R}$ then $I_{ab}$ is the interval $[a,b]$ where $a,b \in \mathbb{R}$, so it is a 1-dimensional oriented hypercube, such that $ab = -ba$ (suppresed $I$ notation, treating the interval algebraically like a simplicial complex).

The magnitude of the coefficient to the interval is the size of the interval (i.e. the volume of the interval hypercube) and the sign of the coefficient is the orientation, so we start with a positive unit interval $I_{ab}$ and see how the function changes the interval size and orientation. Then the derivative of a function $f: \mathbb{R}\rightarrow \mathbb{R}$ is:

$$ f'(x) \approx \frac{f(I_{ab})}{I_{ab}} $$

For example, if $f(x) = -2x$ then $f(I_{ab}) = 2I_{ba} = -2I_{ab}$ and hence $f'(x) = \frac{-2I_{ab}}{I_{ab}} = -2$. So more generally for a function $f: \mathbb{R}^M \rightarrow \mathbb{R}^N, f' = \frac{f(I^N)}{I^M}$ where $I^N$ refers to an N-dimensional oriented interval.

For a 2x2 matrix, representing a linear map from $g: \mathbb{R}^2 \rightarrow \mathbb{R}^2$, the interval is an oriented 2-dimensional hypercube (a square), with corners denoted $abcd$, read clockwise for a positive orientation, such that $abcd = -adcb$.

Consider the linear map $F: \begin{bmatrix}-2&0\\0&2 \end{bmatrix}$, which simply inverts one orthonormal basis and scales by 2. We know the determinant is -4. Label an arbitrary 2-hypercube with corner points $abcd$ read in clockwise orientation.

This linear map will map this interval such that $F(I_{abcd}) = 4I_{adcb} = -4I_{abcd}$ (suppresing $I$ for easier reading, $F(abcd) = 4adcb = -4abcd$, since the map flips the originally clockwise orientation of the corners to counter-clockwise, hence negative sign. Now we calculate $F' = \frac{-4I_{abcd}}{I_{abcd}} = -4$, the same as the determinant. In the general case, $f(I^N)$ may map an oriented unit hyper-cube to an arbitrary N-parallelotope, but it is still a kind of interval with orientation which gives the sign of the determinant.

I already knew determinants tell you the volume of the parallelepiped after the mapping but in the way I've defined a derivative here, in what sense is the determinant any different than the slope of the derivative of the linear map represented by the matrix?

The determinant seems analogous to the derivative to me. The derivative tells you how much an (infinitesimal) interval is scaled (magnitude of derivative) and if its orientation changes (sign of derivative), which is exactly what the determinant is telling you about a linear map.


Short answer, no. The derivative of the linear map is the map itself.

What you are computing is how the linear map transforms volumes, which of course is the determinant.

  • $\begingroup$ This says it all (+1), but one could easily write a two-page long answer unpacking everything that is in here... $\endgroup$ – Vincent Nov 29 '18 at 19:08
  • 2
    $\begingroup$ @Vincent I know, but I have to catch a plane and writing an essay would be too long! :D $\endgroup$ – Federico Nov 29 '18 at 19:09
  • $\begingroup$ I hope someone can explain more. I think the way I’ve setup the derivative as a ratio between oriented n-dimensional hypercube intervals (which after a linear map may become an n-dimensional parallelotope) is equivalent to the ordinary derivative definition and also works for linear maps from arbitrary dimensional spaces and using this definition the derivative of a linear map is exactly the same as the determinant of the matrix. $\endgroup$ – Brandon Brown Nov 29 '18 at 21:14
  • $\begingroup$ Reading a bit more, I can see now that I'm using the low-level calc definition of derivative, whereas higher-derivative is seen as the best linear approximation to a point, which of course makes sense that f'(x) = f(x) if f(x)=ax. But then the determinant of this 1x1 matrix [a] = a which is the slope of the line, so there still seems to be a very short conceptual distance from derivative to determinant. $\endgroup$ – Brandon Brown Nov 30 '18 at 1:09
  • $\begingroup$ "I think the way I’ve setup the derivative as a ratio between oriented n-dimensional hypercube intervals (which after a linear map may become an n-dimensional parallelotope) is equivalent to the ordinary derivative definition". I don't think so. In your definition, the Lebesgue measure is playing a crucial role, being the measure which you use to compute the volumes, whereas the standard definition of derivative is purely metric and not measure theoretic. $\endgroup$ – Federico Dec 3 '18 at 14:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.