I know that the basis for first derivative would be { 1,x,$\left(\frac{x^{2}}{2!}\right)$,$\left(\frac{x^{3}}{3!}\right)$.......$\left(\frac{x^{n}}{n!}\right)$} for the matrix to look like $$ \begin{matrix} 0 & 1 & 0 \cdots 0 \\ 0 & 0 & 1 \cdots 0\\ 0 & 0 & 0 \ddots 1\\ 0 & 0 & 0 \ddots 0\\ \end{matrix} $$ (I know the matrix is not looking perf but I dunno how to space it better. Sorry.)
Does the same basis apply for the fourth derivative to achieve the same form of having 1 on the next diagonal? I thought so but a friend of my told me it cannot be.
Could someone help me with it? Does it apply? If not what would apply? I know that to put it in the block diagonal style with Jordan forms for the fourth derivative endomorphism would be just swapping some basis vectors. But I can't find the basis.
Thank you for taking the time to answer if you do. I would appreciate is a lot.