35. $T\big(f(t)\big) = \big(f(0),f'(0),f''(0),f'''(0),\dotsc\big)$ from $P$ to $V$, where $P$ denotes the space of all polynomials
Hi,
I am not sure how to show this is not an isomorphism. Is it because the $\dim(V)$ does not equal the $\dim(P)$?
Thanks!
I am assuming here that $V$ is the infinite product $\prod_{i = 0}^{\infty} F$ for some field $F$; if $V$ is instead the infinite direct sum $\bigoplus_{i=0}^{\infty} F$, then what I'm about to write is false. Note that $T$ clearly isn't surjective, since every polynomial has finitely many nonzero coefficients, which means taking sufficiently (finitely) many derivatives will always annihilate your polynomial. Therefore, the image of any element always has only finitely many nonzero coordinates. A vector like $(1, 1, 1, \ldots)$ is not in the image of $T$ (neither is any vector with infinitely many nonzero entries).
It is true that $P$ and $V$ do not have the same dimension, but this is a rather sophisticated issue. Clearly, $P$ has countable dimension over its field of coefficients (explicitly, $\{1, x, x^{2}, \ldots,\}$ is a countable basis). On the other hand, the dimension of $V$ is $|F|^{\aleph_{0}}$, which is quite a bit larger. You can find a discussion of this here and here.