$F:{\mathbb{R}}_{3}[x]\text{}\mathbb{]}\to {\mathbb{R}}_{3}[x]$

$F(v)=\frac{{d}^{2}v}{d{v}^{2}}$

Basis: $1,x,{x}^{2},{x}^{3}$ and ${\mathbb{R}}_{3}[x]$ - the set of all third degree polynomials of variable $x$ over $\mathbb{R}$ Assume that all coefficients of the polynomials are $1$

The first thing that springs to my mind is to calculate this derivative by hand, and so we got

$\frac{{d}^{2}y}{d{y}^{2}}=2+6x$

Now, we need to put these values - $2$ and $6$ in such a matrix that - when multiplied by the basis vector -will give us $2+6x$ But there are many ways I can think of, for example

$\left[\begin{array}{cccc}0& 0& 2& 0\\ 0& 0& 0& 6\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]$

Or maybe

$\left[\begin{array}{cccc}2& 0& 0& 0\\ 0& 6& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]$

Because both of them, when multiplied by$\left[\begin{array}{c}1\\ x\\ {x}^{2}\\ {x}^{3}\end{array}\right]$

Will give the correct answer. Thus, what is the correct way to solve this?