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According to the definition of the Jacobian on Wikipedia:

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I've seen it written elsewhere that the members of a Jacobian matrix are all real numbers. But in case $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ for some $m > 1$, then aren't each of the $\partial f_i / \partial x_j \in \mathbb{R}^m$?

Question:

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  • $\begingroup$ Each $f_i$ is a function from $\mathbb{R}^n\rightarrow\mathbb{R}$, so the derivative with respect to $x_j$ is another function from $\mathbb{R}^n\rightarrow\mathbb{R}$. $\endgroup$ – Michael Burr May 17 '19 at 21:34
  • $\begingroup$ $\partial f_i/\partial x_j$ is the partial derivative of $f_i$ with respect to $x_j$. It is a scalar-valued function. $\endgroup$ – littleO May 17 '19 at 21:35
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No, because each $f_i$ is a real function, and therefore each $\frac{\partial f_i}{\partial x_j}$ is a real number.

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No, for example let $f(x,y) = (x^2,y^2,xy)$ then $$ \mathcal{J} = \begin{pmatrix} \frac{\partial \left[x^2\right]}{\partial x} & \frac{\partial \left[x^2\right]}{\partial y} \\ \frac{\partial \left[y^2\right]}{\partial x} & \frac{\partial \left[y^2\right]}{\partial y} \\ \frac{\partial [xy ]}{\partial x} & \frac{\partial [xy ]}{\partial y} \end{pmatrix} = \begin{pmatrix} 2x & 0 \\ 0 & 2y \\ y & x \end{pmatrix} $$

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