= Solution
Put $n=\dim V>0$ and $r=\operatorname{rank}T$. The <direct sum> in part (i) gives
$$
\boxed{\chi_T(x)=x^{n-r}(x-1)^r.}
$$
The <eigenvalues> are the members of $\{0,1\}$ actually present: zero occurs exactly when $r<n$, and one exactly when $r>0$. Since $T$ is <diagonalizable>, its <minimal polynomial> contains each present <eigenvalue> once:
$$
\boxed{m_T(x)=\begin{cases}
x,&T=0,\\
x-1,&T=I,\\
x(x-1),&T\ne0,I.
\end{cases}}
$$
In particular, a nonzero <vector space> does not force both <eigenvalues> to occur. The zero and <identity maps> are important special cases.
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