= Solution
After choosing a basis of $V$, a linear <endomorphism> is specified by an $n$ by $n$ matrix, with $n^2$ freely chosen coefficients. Therefore
$$
\boxed{\dim\operatorname{End}(V)=n^2}.
$$
The $n^2+1$ <endomorphisms> $I,\alpha,\ldots,\alpha^{n^2}$ are linearly dependent. Some nonzero coefficient list gives a nonzero <polynomial> $p$ satisfying $p(\alpha)=0$.
For nonzero $V$, the <minimal polynomial> $m_\alpha$ is the monic <polynomial> of least degree annihilating $\alpha$. It is unique: subtracting two monic least-degree choices would give a lower-degree annihilator. More generally division of any annihilator $p$ by $m_\alpha$ gives $p=qm_\alpha+r$, with $\deg r<\deg m_\alpha$, and evaluation implies $r(\alpha)=0$. Minimality forces $r=0$, so every annihilator is divisible by $m_\alpha$. If $V=\{0\}$, the convention $m_\alpha=1$ handles the degenerate case.
The arguments work over the underlying scalar field. In part (b), roots in that field correspond to <eigenvalues> there; to discuss other roots one extends scalars, for example by complexifying a real <vector space>.
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