How to think of gradient of a vector?

Question

I have this introductory problem:

Calculate the gradient:

$$ \nabla_x \boldsymbol{a}^T \boldsymbol{x} = \boldsymbol{a} $$

$\boldsymbol{a}$ and $\boldsymbol{x}$ are vectors.

There are several things that were confusing to me in this exercise. First, I had trouble understanding the notation $\nabla_x$ i.e. what is gradient with superscript $x$ supposed to mean but according to this answer it is a vector of variables with respect to which you calculate the partial derivatives. Second thing is the presence of the right hand side of the equation - at first I thought it's like solving an unknown in e.g. $5+4x=16$. But now I'm convinced the RHS is just the answer I'm supposed to get by evaluating LHS.

I still don't how I should think about the question asking me to calculate gradient of what evaluates to be a vector - according to the Gradient wikipedia page you calculate gradient of a function, not a vector like here. Does it have some special meaning?

Edit: I just realized the matrix resulting from $\boldsymbol{a}^T \boldsymbol{x}$ is a 1x1 matrix, so if I treat it as just an expression it makes more sense.

Answer 1

Let $\boldsymbol a=(a,b,c)$, so that $\boldsymbol a^T\boldsymbol x=ax+by+cz$.

Then

$$\nabla_{\boldsymbol x}(\boldsymbol a^T\boldsymbol x)=\left(\frac{\partial}{\partial x}(ax+by+cz),\frac{\partial}{\partial y}(ax+by+cz),\frac{\partial}{\partial z}(ax+by+cz)\right)=(a,b,c).$$

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Geometric interpretation:

The locus of points such that $ax+by+cz=d$ where $d$ is some constant (i.e. an iso-surface), is a plane. The gradient is a vector normal to this plane. More generally, the gradient of a scalar function is a vector orthogonal to the iso-surfaces.