Basic vector calculus: Show that $\nabla \vec{r} = \vec{1}$ [closed]

Question

Show that $\nabla \vec{r} = \vec{1}$

My instructor in my E & M class put the $r$ and $1$ in bold. I am not sure what a bold one means. From my work I get $1ii + 1jj + 1zz$.

Answer 1

Taking the gradient of a vector field gives a tensor of rank 2: $$\nabla(\mathbf r) = \nabla(x \mathbf{\hat x} + y \mathbf{\hat y} + z \mathbf{\hat z}) = \begin{bmatrix} \frac{\partial (x)}{\partial x} & \frac{\partial (x)}{\partial y} & \frac{\partial (x)}{\partial z} \\ \frac{\partial (y)}{\partial x} & \frac{\partial (y)}{\partial y} & \frac{\partial (y)}{\partial z} \\ \frac{\partial (z)}{\partial x} & \frac{\partial (z)}{\partial y} & \frac{\partial (z)}{\partial z} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ which is the unit tensor, $\overleftrightarrow 1$, in this case. In general, $$\nabla(\mathbf F) = \nabla(F_x \mathbf{\hat x} + F_y \mathbf{\hat y} + F_z \mathbf{\hat z}) = \begin{bmatrix} \frac{\partial F_x}{\partial x} & \frac{\partial F_x}{\partial y} & \frac{\partial F_x}{\partial z} \\ \frac{\partial F_y}{\partial x} & \frac{\partial F_y}{\partial y} & \frac{\partial F_y}{\partial z} \\ \frac{\partial F_z}{\partial x} & \frac{\partial F_z}{\partial y} & \frac{\partial F_z}{\partial z} \end{bmatrix} \text{or} \begin{bmatrix} \frac{\partial F_x}{\partial x} & \frac{\partial F_y}{\partial x} & \frac{\partial F_z}{\partial x} \\ \frac{\partial F_x}{\partial y} & \frac{\partial F_y}{\partial y} & \frac{\partial F_z}{\partial y} \\ \frac{\partial F_x}{\partial z} & \frac{\partial F_y}{\partial z} & \frac{\partial F_z}{\partial z} \end{bmatrix}$$ depending on the context. This, however, does not seem to be the result that you're trying to show, so I suspect that there's a typo.