# Are the members of a Jacobian matrix necessarily real numbers?

According to the definition of the Jacobian on Wikipedia: 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:

• 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 $\mathbb{R}^n\rightarrow\mathbb{R}$. – Michael Burr May 17 '19 at 21:34
• $\partial f_i/\partial x_j$ is the partial derivative of $f_i$ with respect to $x_j$. It is a scalar-valued function. – littleO May 17 '19 at 21:35

No, because each $$f_i$$ is a real function, and therefore each $$\frac{\partial f_i}{\partial x_j}$$ is a real number.
No, for example let $$f(x,y) = (x^2,y^2,xy)$$ then $$\mathcal{J} = \begin{pmatrix} \frac{\partial \left[x^2\right]}{\partial x} & \frac{\partial \left[x^2\right]}{\partial y} \\ \frac{\partial \left[y^2\right]}{\partial x} & \frac{\partial \left[y^2\right]}{\partial y} \\ \frac{\partial [xy ]}{\partial x} & \frac{\partial [xy ]}{\partial y} \end{pmatrix} = \begin{pmatrix} 2x & 0 \\ 0 & 2y \\ y & x \end{pmatrix}$$