Minimal polynomial of a vector under a linear operator ($I_v=(m_v),I_v=\{f:f(T)v=0\}$)

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 $v,Tv,\dots ,T^{n}v$ 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.