# Gradient of a vector field

We did in lectures gradient of a scalar field and I am wondering how is the grad of a vector field. I tried the following Let

$$V = f(x,y,z)\hat{\imath} + g(x,y,z)\hat{\jmath} + h(x,y,z)\hat{k} \, .$$

Then by definition

$$\mathrm{grad}(V)= \hat{\imath} V_x + \hat{\jmath}V_y + \hat{k}V_z \, ,$$

where $$V_x$$, $$V_y$$, $$V_z$$ denote the partial derivatives of $$V$$ with respect to to $$(x,y,z)$$. But $$i V_x = f(x,y,z)x$$ since $$\hat{\imath}\cdot\hat{\imath}=1$$, $$\hat{\imath}\cdot\hat{\jmath}=0$$, and $$\hat{\imath}\cdot\hat{k}=0$$.

Similarly, $$\hat{\jmath}V_y = g(x,y,z)y$$ and $$\hat{k}V_z=h(x,y,z)$$. So we end up with

$$\mathrm{grad}(V) = f(x,y,z)x + g(x,y,z)y + h(x,y,z)z = \mathrm{div}(V) \, .$$

I guess that is wrong but what is wrong with the method.

The gradient of a vector field in ordinary $R^3$ is a tensor field. You will need to use the tensor product to represent it $\otimes$.
$$\nabla V = \left( \hat{e}_i \frac{\partial}{\partial x_i} \right ) (V_m \hat{e}_m) \\ = \frac{\partial V_m}{\partial x_i} \hat{e}_i \otimes \hat{e}_m$$