= Minimal polynomial of a vector under a linear operator
{title2=$I_v=(m_v),\quad I_v=\{f:f(T)v=0\}$}
For a <linear operator> $T$ on an $n$-dimensional <vector space> over a <field>, the polynomials annihilating a fixed vector $v$ form an ideal of $K[X]$. It is nonzero because $v,Tv,\ldots,T^nv$ are linearly dependent. Since $K[X]$ is a <principal ideal domain>, its unique monic generator $m_v$ exists and has degree at most $n$. It divides $f$ exactly when $f(T)v=0$. For $v=0$ it equals one. Its degree is the dimension of the cyclic subspace generated by $v$: the vectors up to $T^{d-1}v$ are independent by minimality and span an invariant space by the annihilating recurrence.
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