The question is as follows:
Give an example of a non-nilpotent linear transformation for some vector space with the following property, for every $v \in V$ there is an $n$ for which $A^n(v)=0$.
I really don't know how such a matrix would exist, since you can find a basis for this vector space, let's say $v_1,v_2,…$. Then we know for every $v_i$ there is an $n_i$ for which $A^{n_i}(v_i)=0$.
Then write $v$ as $\sum_{i}^{k} \alpha_i v_i$ and consider $A^{\sum_{i}^{k} n_i} (v)$, which gives $\sum_{i}^{k} \alpha_i \cdot A^{\sum_{i}^{k} n_i} (v_i)$ which should give zero for every $v_i$.
Maybe they meant an infinite vector space, however I doubt it since we never touched the infinite vector spaces in our algebra course.
Then you can consider the basis $B=v_1,v_2, …$ an infinite basis over some field and let $A(v_1)=0$ and $A(v_i)=v_{i-1}$ then since there exist online finite amount of non-zero coordinates then there must be an $n$ for which $A^n(v)=0$, however the transformation is not nilpotent.
So my question is whether it's really true that such a transformation doesn't exist for finite-dimensional vector spaces.
Best Answer
Let $V$ be the space of all polynomial functions from $\Bbb R$ into $\Bbb R$ and let $A\bigl(p(x)\bigr)=p'(x)$. If $\deg p(x)=n$, then $A^{n+1}\bigl(p(x)\bigr)=0$, but $A$ is not nilpotent: for any $n\in\Bbb N$,$$A\left(x^{n+1}\right)=(n+1)!\ne0.$$