Geometric Interpretation of the Jacobian Matrix and Its Eigenvectors

Question

I understand that for scalar-valued functions $g: \mathbb{R}^n \to \mathbb{R}$, the gradient represents the direction of maximum ascent. Similarly, for vector-valued functions $f: \mathbb{R}^n \to \mathbb{R}^n$, each row of the Jacobian matrix $J_f(x)$ provides the direction of maximum ascent for the corresponding component function.

I have the following questions:

1. Geometric interpretation of eigenvectors of the Jacobian matrix: What is the geometric interpretation of the eigenvectors of the Jacobian matrix at a point? How do they affect the behavior of the function near that point?

2. Same eigenvectors at every point: Suppose the Jacobian matrix of $f$ has the same set of eigenvectors at every point in $\mathbb{R}^n$. Does this imply that the function is linear or Jacobian diagonal, or if not is there some special structure to the function $f$? What are the nontrivial examples of vector fields with same eigen vectors at every point in $\mathbb{R}^n$

Any insights or references would be greatly appreciated!

Answer 1

Although this does not answer the question in general, we can consider the transformation from Cartesian Coordinates to Polar Coordinates as a toy example. The transformation is:

$$x=r \cos \theta, \quad y=r \sin \theta$$

[ This video is my favorite introduction to the Jacobian matrix in this context: https://www.youtube.com/watch?v=hhFzJvaY__U ]

If you could encode the transformation uniquely as a matrix (you can't, since this is a nonlinear transformation) this matrix would represent how points are transformed geometrically when moving from one coordinate system (Cartesian) to another (polar) in a global sense.

The Jacobian matrix is used to represent the local rate of change in the transformation, capturing how small changes in polar coordinates $(r, \theta)$ affect Cartesian coordinates $(x, y)$ or vice versa. For this, you can definitely write a matrix since you are making a local transformation (at any given point) from $dx$ and $dy$ into $dr$ and $d\theta$ .

Jacobian Matrix for the transformation from polar to Cartesian coordinates:

$$ J=\left[\begin{array}{ll} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{array}\right]=\left[\begin{array}{cc} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{array}\right] $$

From which we can calculate:

$$\lambda_{1,2}=\cos \theta \pm \sqrt{\cos ^2 \theta-r}$$

The Jacobian tells you how small changes in $r$ and $\theta$ translate into small changes in $x$ and $y$, effectively giving the local stretching and rotation behavior of the transformation.

Now recall that eigenvectors tell you the directions that are only scaled after the transformation.

Thus, in this specific transformation from polar to Cartesian coordinates, the eigenvalues and eigenvectors of the Jacobian matrix provide a local understanding of how the transformation behaves in terms of scaling and rotation around a point. The eigenvectors indicate directions that are only scaled (not rotated) by the transformation, and the eigenvalues give the magnitude of that scaling.

The Jacobian matrix captures the local behavior of the nonlinear transformation, and the eigenvalues and eigenvectors describe how the transformation affects points in space, particularly identifying directions that experience pure scaling without distortion.

Answer 2

Let $F : \mathbb R^n \to \mathbb R^n$ be a $C^1$ function. It is convenient to think of $F$ as a vector field, that is, $F(x)$ is a vector based at $x.$ Let's say that $F(x)$ has units of speed. If a particle starts at a point $x_0$ at $t = 0$ and its velocity is given by the field $F,$ then the particle follows a trajectory $x(t)$ according to the differential equation $$\frac d{dt} x = F(x).$$ This flow equation always has a unique solution according to the theory, which is defined on some interval $(a, b)$ around $0.$

Fix a critical point $x_c$ where $F(x_c) = 0.$ Then the differential $dF$ (or the Jacobian) tells us about the behaviour of $x$ near $x_c.$ For example, if all the eigenvalues have positive real part, then $x$ is repelled from $x_c.$ If they have negative real part, then $x$ is attracted to $x_c.$

Take a look at these pictures for more examples according to whether the matrix is diagonalizable.

If $F$ is linear, give the matrix the same name $F(x) = Fx.$ Then we may solve the equation (loosely speaking) like $x(t) = e^{tF}x_0$ for some scalar constant $C.$ Even if you are concerned about the exponential of the matrix, try differentiating both sides. Using the motto "exponential sums into products", calculate $\det(e^{tF}) = e^{t\operatorname{tr}(F)}.$ Thus the trace of $F,$ which is the sum of its eigenvalues, represents the speed at which the size of a region gets enlarged or shrinked following the flow.

If $F$ is not linear, then the differential $dF$ at each point represents the best linear approximation to $F,$ so the interpretation of $\operatorname{tr}(dF)$ as the "divergence" of the flow still makes sense. Note that $dF_x$ has units of $1/t.$ Also, given an eigenvector $dF_{x_0}(v) = \lambda v,$ it's not hard to show that $e^{dF_{x_0}}v = e^\lambda v.$ Thus the distinction according to the sign of $\Re(\lambda)$ becomes a distinction about the module of $e^\lambda,$ which is more natural.

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Now, for the second question, one way to phrase it is:

What can we tell about $F$ if $dF_{x}$ is diagonalizable with the same basis of eigenvectors for each $x$?

Clearly linear or affine transformations satisfy the above (diagonalizable ones for simplicity).

For simplicity assume that the canonical basis is a basis of eigenvectors at each point. Then the Jacobian is a diagonal matrix at each point. Another example is $F(x, y) = (x^3+x, y).$ The Jacobian is diagonal, which implies the above (but with a change of coordinates you can obtain an example where the Jacobian is not diagonal).

If the Jacobian in the canonical basis is diagonal at every point, then we can say that $F(x, y) = (f(x), g(y))$ for appropriate functions $f, g$, because for example $\frac{\partial}{\partial y} f(x, y) = 0$ means that $f$ does not depend on $y.$

For the non-diagonalizable case, I don't know the generalisation that you have in mind.