 Solution

The minimal polynomial of $T$ is the monic nonzero polynomial of smallest degree annihilating $T$. Such a polynomial exists, since sufficiently many powers of $T$ are linearly dependent in the finite-dimensional space $\mathrm{End}(V)$. If two monic annihilators had the same least degree, their difference would be an annihilator of smaller degree and hence zero. Thus it is unique. Division with remainder also shows that every annihilating polynomial is a multiple of it.

The primary decomposition theorem states that if $m={\prod }_{j=1}^r{f}_j^{e_j}$ with distinct monic irreducible polynomials, thenThese subspaces are $T$-invariant, and the restriction to $V_j$ has minimal polynomial $f_j^{e_j}$.

Let $r=\gcd (q,m)$, chosen monic. The Bézout's identity for polynomials gives $r=aq+bm$. If $q(T)v=0$, then $r(T)v=a(T)q(T)v+b(T)m(T)v=0$. Conversely, $q=rh$, so $r(T)v=0$ implies $q(T)v=0$. ThereforeThe following Roman scopes correspond to the three labelled conditions, with their implications proved from the indicated condition.