Geometric Meaning of the Jacobian of a Linear Transformation

Question

Consider the multivariable function $f(x, y) = \begin{bmatrix}-y \\ x \end{bmatrix}$, whose geometry is shown here. For any point $(x,\ y)$, this function's Jacobian matrix is always $\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}$. There is no need to even plug in the point $(x,\ y)$; all of the appropriate partial derivatives are constants.

According to this video from the same series, "the Jacobian matrix is fundamentally supposed to represent what a transformation looks like when you zoom in near a specific point."

However, while the Jacobian matrix of the transformation we're considering does coincide with the global view of the transformation as a whole (a counterclockwise rotation), I wouldn't expect it to describe the appearance when zoomed into an arbitrary point, as the Jacobian matrix suggests. Things might work out correctly if we zoom in near the origin, but this transformation should look completely different when we zoom into the points $(0,\ 999)$ versus $(999,\ 0)$. The former would look like almost straight leftward movement, and the latter nearly straight upward movement.

What is the Jacobian matrix saying geometrically in this case?

Answer 1

The Jacobian of a differentiable map $f$ at a point $p$ is the matrix of the linear map $L$ which behaves as close as possible to $f$ near $p$. So, if $f$ is linear (which is the case here), that linear map is $f$ itself. And I hope that you will agree that what $f$ does to the point $(999,0)$ is precisely what the linear map whose matrix (with respect to the canonical basis) is $\left[\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right]$ does to it.

Answer 2

Really it is the mapping $$ x \mapsto f(x_0) + f'(x_0)(x - x_0) $$ which approximates $f$ well when you zoom in near the point $x_0$. (Here $x_0$ is a point in $\mathbb R^n$, and the $n \times n$ matrix $f'(x_0)$ is the derivative (i.e., Jacobian) of $f$ at $x_0$.) This approximation $$ f(x) \approx f(x_0) + f'(x_0)(x - x_0) $$ is accurate near $x_0 = (999,0)$ and also near $x_0 = (0,999)$.

Answer 3

(Edit: In the original version the letters $X$ and $Y$ were used for two different purposes.)

You are talking about the "Jacobian of a linear transformation", and are mentioning various true facts in this respect. What has remained unclear is mainly due to some small abuse of notation that you'll get accustomed to. Assume that we have a linear map $A:\>V\to W$ between finite dimensional vector spaces $V$ and $W$. Such an $A$ maps the full space $V$ to the space $W$, and thereby warps in some way figures you have drawn in $V$. Note that both $V$ and and $W$ have a declared origin $0$ and a vector space structure you are familiar with.

Now it may be the case that you are particularly interested how the map $A$ behaves in the neighborhood of a particular point $p\in V$. This point $p$ has an image point $q:=A(p)\in W$. It is then a familiar move to view $p$, resp. $q$, as origins of new coordinate systems in $V$, resp. $W$. A general point $x\in V$ would then be written as $x=p+X$ with an increment variable $X$, and similarly a general $y\in W$ would be written as $y=q+Y$. The map $A$ assigns to each (maybe small) increment $X$, measured from $p$, an increment $Y$, measured from $q$. Call the effect of $A$ on the increments the Jacobian of $A$ at $p$, and write $dA(p):\>X\mapsto Y$. The global linearity of $A$ allows for a simple computation of $dA(p)$: From $A(p+X)=q+Y$ we get $$Y=A(p+X)-q=A(p)+A(X)-q=A(X)\ .$$ This shows that $dA(p)=A$. But note that the domain of $dA(p)$ is the so-called tangent space at $p$, and not the original $V$; and similarly for the codomain. These two shifts of origin are glossed over in the notation.