[Math] Understanding the Jacobian – meaning of columns – best linear approximation

calculuslinear algebra

I am trying to get my head around the Jacobian matrix I suppose.

If we have a map $f:\Bbb R^n\to \Bbb R^m$ in variables $x_1,\dots x_n$ we can form the Jacobian matrix:

$$J=(\partial f_i/\partial x_j)_{ij}$$
1) Why do we hope that this has rank $n$? I understand that each column is a vector encoding how that position of the function behaves under the first derivative(with respect to each variable).

I.e. column one is:
$$\begin{pmatrix}\frac{\partial f_1}{\partial x_1}\\\vdots\\\frac{\partial f_n}{\partial x_1}\end{pmatrix}$$
and thus is there is dependence, the tangent space should be lacking information in some way.

2) If we evaluate this at some point that is continuously differentiable, say $a$ i.e. evaluate $J(a)$, we are meant to be able to use this as the best linear approximation near $a$. How would we use this? Compute $J(a)$ and obtain a matrix $J(a):\Bbb R^n \to \Bbb R^m$ and multiply this by what?


Example: If I map $\Bbb R^1\to \Bbb S^1$ via $f(x)=(\cos(x),\sin(x))$ I can take $f:(0,\frac\pi2)\mapsto $ the first quarter arc of the circle (counterclockwise) (open arc) – with Jacobian:
$$\begin{pmatrix}-\sin(x)\\\cos(x)\end{pmatrix}$$ and I have a vector $(-\sin(x),\cos(x))$, what good does that do me? I can write this vector as a tangent to my circle by translating it from the origin to my point?

Best Answer

This is mostly clarification, need not address all your concerns.

1) Why do we hope that this has rank n?

... column 1 is $ (\frac{\partial f_1}{\partial x_1} \frac{\partial f_2}{\partial x_1} \ldots \frac{\partial f_n}{\partial x_1})^T $ and thus is there is dependence

Our concern here is linear dependence not just dependence. For instance, $ x^2 $ and $ x^4 $ are obviously dependent but not linearly.

2) How would we use this (Jacobian matrix)?

First, Jacobian matrix is not approximation for any function as you guessed, but a matrix gives us a linear transformation, we are talking about using this linear transformation (usually called differential) as a linear approximation. Also differential is not the (local) linear approximation for the original function, but

It’s the best linear approximation to the change in the (original) function’s value near the point

as pointed out by @amd in his answer to the question How is the Jacobian the derivative if they have different dimensions. There, he explains it well with examples. The differential of $ \Delta(x) $ is $ J(a)\Delta(x) $. So, we are basically multiplying the Jacobian with displacement $ \Delta(x) $.

I hope this helps. Thank you.

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