Rank-Nullity confusion

Question

Consider a linear map $T\colon V\to V$ and the composition $T^2$. Would the rank-nullity theorem applied to $T^2$ be $$\mathop{\mathrm{rank}}(T^2)+\mathop{\mathrm{nullity}}(T^2)=\dim V?$$
or can we say, let $U$ be the map restricted to $\mathrm{im}(T)$, then $$\mathop{\mathrm{rank}}(U)+\mathop{\mathrm{nullity}}(U)=\mathop{\mathrm{rank}}(T)?$$ or do both hold?

Answer 1

Both are correct, but you are applying the dimension theorem first to $T^2\colon V\to V$, and then to $T\restriction \mathrm{im}(T)=U\colon\mathrm{im}(T)\to\mathrm{im}(T)$, which are different maps.

To answer your question, the dimension theorem "applied to $T^2$" would be the first equality $$\mathop{\mathrm{rank}}(T^2)+\mathop{\mathrm{nullity}}(T^2)=\dim V.$$