Is there a non-constant continous function $f:\mathbb{R}\rightarrow \mathbb{R}$ and matrices $A=\begin{pmatrix}
a\_1 & 0\\
0 & a\_2\\
\end{pmatrix}$ and 
$B=\begin{pmatrix}
b\_1 & 0\\
0 & b\_2\\
\end{pmatrix}$ for which there does not exist any matrices $C\in \mathrm{Mat}\_{d\times 2},D \in \mathrm{Mat}\_{2\times d}$ and vectors $c \in \mathbb{R}^d,e\in \mathbb{R}^2$ such that:


* $
D f\_d\left(C
\begin{pmatrix}
x\\
x
\end{pmatrix}
 +c
\right) + e = Af\_2
\left(B
\begin{pmatrix}
x\\
x
\end{pmatrix}
\right) \qquad (\forall x \in \mathbb{R}) 
$
* $D,e,C,$ and $c$ only have non-zero entries,


where $f\_i(x)= (f(x\_1),\dots,f(x\_n))$.



Non-existent matrices with "essential zeros"
**Yes**, we can find such a triple $(f, A, B)$. 


Let us first observe that OP's question can be rephrased as follows.



> 
> **Question.** Find a continuous function $f: \mathbb{R} \rightarrow \mathbb{R}$
>  such that $f$ does not lie in the $\mathbb{R}$-linear span $L(f)$ of $\{ x \mapsto f(cx + c') \}\_{(c, c') \in \mathbb{R} \times \mathbb{R} \setminus \{0\}}$.
> 
> 
> 


For instance, the continuous extension $f$ of $x \mapsto \text{exp}(- 1 / x^2)$ is such that $f \notin L(f)$. Indeed, any function in $L(f) $ is analytic in a neighbourhood of 0 whereas $f$ isn't. I believe that many analytic functions, including $f(x) = \text{exp}(x^2)$, can be shown to satisfy $f \notin L(f)$, but a simple proof of this fact still eludes me. 


By contrast, if $f$ is a real-valued polynomial function over $\mathbb{R}$ then $L(f)$ is the $\mathbb{R}$-vector space of the polynomial functions of degree at most $\deg(f)$, so that $f \in L(f)$. To see this, one may use the Taylor series of $f(x + c)$ together with a well-known result on Vandermonde matrices.


It is also immediate to check that periodic functions and the exponential function $f(x) = \text{exp}(x)$ satisfy $f \in L(f)$. They can be used to build algebras of functions satisfying this property.


