How to write Jacobian matrix in more vector-like format

Question

Let $\mathbf{f}: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ be a vector field. Suppose I wish to compute $\frac{\partial \mathbf{f(x)}}{\partial \mathbf{x}}$. Then we have the Jacobian.

$$\frac{\partial \mathbf{f(x)}}{\partial \mathbf{x}} = \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \cdots & \frac{\partial f_1}{\partial x_n} \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \cdots & \frac{\partial f_2}{\partial x_n} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial f_m}{\partial x_1} & \frac{\partial f_m}{\partial x_2} & \cdots & \frac{\partial f_m}{\partial x_n} \end{bmatrix}$$ Now, the directional derivative of a scalar field $f$ can be define as

$$f'(\mathbf{x}; \mathbf{y}) = \lim_{h \to 0} \frac{f(\mathbf{x} + h\mathbf{y}) - f(\mathbf{x})}{h} $$

Here's my issue. In principle, I should be able to make $\mathbf{f}$ a scalar field by simply setting $m=1$. This would then become $$\frac{\partial \mathbf{f(x)}}{\partial \mathbf{x}} = \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \cdots & \frac{\partial f_1}{\partial x_n} \\ \end{bmatrix}$$

My confusion is why this would be a vector here, but then not for operation for the directional derivative above (at least it doesn't look like it).

Can someone clarify this for me?

Answer 1

I think the confusion stems from the fact that the directional derivative for scalar fields is scalar itself. To see that, you just need to write down the directional derivative in coordinates: $$ D_vf(x) = \sum_{i=1}^{n}v^{i}\frac{\partial f}{\partial x^i} = v\,\cdot \nabla f(x) $$ In this case, there is no vector notation to apply since it's just a number.

Basically, using the coordinate-based presentation of the directional derivative, you can write it down in the vector notation in the following way: $$ [\nabla_v\mathbf{f}(\mathbf{x})]_{j} = \sum_{i=1}^{n}v^{i}\frac{\partial f_j}{\partial x^i}. $$

It is obvious from here that you can express a directional derivative using matrix multiplication: $$ \nabla_{\mathbf{v}}{\mathbf{f}}(\mathbf{x})=\mathbf{J}v $$ which in case $m=1$ is indeed a scalar.