I have a function
$$ f(\mathbf{x}) = \mathbf{x}^\intercal \mathbf{a} $$
where $\mathbf{x}, \mathbf{a} \in \mathbb{R}^d$. Now, suppose that I don't know the values of $\mathbf{a}$. I can fully determine them by taking a set of non-linearly dependent vectors $\mathbf{e}_1, \mathbf{e}_2, \dots, \mathbf{e}_d$ where $\mathbf{e}_i \in \mathbb{R}^d$. If I construct a (square) matrix
$$ E = \left[\begin{matrix} \mathbf{e}^\intercal_1\\ \mathbf{e}^\intercal_2 \\ \dots\\ \mathbf{e}^\intercal_d \end{matrix}\right] $$
I can claim that
$$ \mathbf{a} = E^{-1}\mathbf{y} $$
where
$$ \mathbf{y} = \left[\begin{matrix} f(\mathbf{e}_1)\\ f(\mathbf{e}_2)\\ \dots \\ f(\mathbf{e}_d) \end{matrix}\right]. $$
The first question is: how is this Theorem called?
The second question is: how it is called the Theorem that ensures me that for any $\mathbf{x}$ there is a vector $\mathbf{h} \in \mathbb{R}^d$ such that $\mathbf{x} = E^\intercal \mathbf{h}$.
The third question is: suppose that you have a non-linear mapping $\phi:\mathcal{X}\to\mathbb{R}^d$, therefore the codomain $C$ of $\phi$ is $C \subseteq \mathbb{R}^d$. I still suppose that there is a function $f$
$$ f(\mathbf{x}) = \phi^\intercal(\mathbf{x})\mathbf{a} $$
I want still to say that it exists a set of vectors ${\mathbf{c}_i \in C}_{i=1}^d$ such that every pair of vectors is non-linear, and find a solution, as I did before, by inverting a matrix where each row contains $\mathbf{c}_i^\intercal$.
What is the condition that I need to impose on $\phi$ and on $\mathcal{X}$ to obtain these results? I imagine that for some choice of $\phi$, the codomain does not contain $d$ non-linear vectors, and therefore I cannot build my matrix $C$.
As you see, I miss some nomenclature, and I need it to communicate the assumptions that I want to make in a clean and readable way.
