Divergence Operator on Vector Space is Dot Product of Del Operator

Theorem
Let $R$ be a region of Cartesian $3$-space $\R^3$.

Let $\map {\mathbf V} {x, y, z}$ be a vector field acting over $R$.

Let $\tuple {i, j, k}$ be the standard ordered basis on $\R^3$.

Then
 * $\operatorname {div} \mathbf V = \nabla \cdot \mathbf V$

where:
 * $\operatorname {div} \mathbf V $ denotes the divergence of $\mathbf V$
 * $\nabla$ denotes the del operator.

Proof
We have by definition of divergence of $\mathbf V$:


 * $\ds \operatorname {div} \mathbf V = \dfrac {\partial V_x} {\partial x} + \dfrac {\partial V_y} {\partial y} + \dfrac {\partial V_z} {\partial z}$

Now: