Vector operator and scalar operator I'm confused about what it means by a scalar operator and a vector operator. We call $\nabla$ , $\nabla\cdot$ ,$\nabla\times$ are all called vector operators. But, the Laplacian $\nabla^2$ is a scalar operator. But, why is $\nabla\cdot$ a vector operator although it generates a scalar field by operating on a vector field? And why is $\nabla^2$ is a scalar operator although it generates scalar and vector field when it operates on scalar and vector field, respectively. 
What does decide which one is a scalar operator and which one is a vector operator?
 A: The first thing I learned while using the del operator is that del operator should always be written like: $$\overrightarrow \triangledown=\frac{\partial}{\partial x}\widehat{i}+\frac{\partial}{\partial y}\widehat{j}+\frac{\partial}{\partial z}\widehat{k}$$
Del operator is a vector differential operator.Now let us say there is a function $f(x,y,z)$, and we have to operate the del on it the function can be a vector valued function or a scalar valued function.


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*Del operator on a scalar valued function is like :
$$\overrightarrow \triangledown(f(x,y,z))=\frac{\partial f}{\partial x}\widehat{i}+\frac{\partial f}{\partial y}\widehat{j}+\frac{\partial f}{\partial z}\widehat{k}=grad(f(x,y,z))$$
Notice this is analogous to a vector simply multiplied to a constant scalar.

*Del operator on a vector valued function resulting in a scalar is :
$$\overrightarrow \triangledown .\overrightarrow {f(x,y,z)}=(\frac{\partial}{\partial x}\widehat{i}+\frac{\partial}{\partial y}\widehat{j}+\frac{\partial}{\partial z}\widehat{k}).(f_x \widehat i + f_y \widehat j + f_z \widehat k)$$
$$\overrightarrow \triangledown .\overrightarrow {f(x,y,z)}=\frac{\partial f_x}{\partial x}+\frac{\partial f_y}{\partial y}+\frac{\partial f_z}{\partial z}=div(\overrightarrow f(x,y,z))$$
Notice this is analogous to the dot product of two vectors.


*Del operator on a vector valued function resulting in a vector is like:
$$\overrightarrow \triangledown X\overrightarrow {f(x,y,z)}=\begin{vmatrix} \widehat i & \widehat j&\widehat k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\f_x&f_y&f_z \end{vmatrix}=curl(\overrightarrow f(x,y,z))$$
This is analogous to the cross product of two vectors.


Now Laplacian operator is second order differential operator:
$$\triangledown^2f(x,y,z)=div(grad(f(x,y,z))=\triangledown . (\triangledown \overrightarrow {f(x,y,z)})$$
$$\triangledown ^2f(x,y,z)=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}$$
It becomes a scalar operator because the gradient gives a vector and divergence is just the dot product on gradient giving a scalar.
Hope this helps...
