As mentioned in a previous post (I think), I've been trying to learn some linear algebra, and so I've begun to post little questions whose answers I'm sure are obvious to most here; this is just a way for me to clarify my thinking and reassure myself that I am on the right track/get myself back, as it were, on the right track.
Let $D = \partial /\partial t$ be the differentiation map on the space $V$ of smooth functions $\mathbb{R} \to \mathbb{R}$. Let $U$ be the subspace generated by the set $\{1, t, e^{t}, e^{2t}\}$. I am trying to find the matrix associated to $D \big |_U$ (restricting $D$ to $U$) relative to the ordered basis $\{1, t, e^{t}, e^{2t}\}$. I am pretty sure that this matrix should be
\begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{pmatrix}
considering the derivatives of each basis vector. However, I was wondering if the matrix
\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 1/t & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{pmatrix}
might also be an alternative candidate.
Your alternative doesn't work because the matrix should not depend on $t$.
At least, as long as you view the function space as a real vector space the matrix elements have to be members of $\mathbb R$. Your $1/t$ is (at best) a function rather than a real number, and therefore does not belong as a matrix element.
If you want to be all fancy about it, you could view your vector space as one over a larger scalar field than $\mathbb R$, for example the field of rational functions in $t$. If you do that, you'll be allowed to write $1/t$ as a matrix element -- but on the other hand your basis is then not a basis anymore, because $1$ and $t$ are linearly dependent when $t$ is just a scalar. And differentiation is then not a linear map anymore, so suddenly no matrix will work to represent it.