= Solution
The <minimal polynomial> $m_\beta(t)$ of the <linear map> $\beta$ is the unique <monic polynomial> of smallest degree satisfying $m_\beta(\beta)=0$. An annihilating <polynomial> exists because the powers $I,\beta,\beta^2,\ldots$ lie in the finite-dimensional <vector space> $\operatorname{End}(V)$ and are therefore <linearly dependent>. If two monic annihilating <polynomials> had the same least degree, their difference would be an annihilating <polynomial> of smaller degree, so they coincide.
Write $m_\beta(t)=a_0+tq(t)$. If $a_0\ne0$, evaluation at the <linear map> gives $\beta q(\beta)=-a_0I$. Since $\beta$ commutes with every <polynomial> in itself, this is a two-sided inverse:
$$
\boxed{\beta^{-1}=-\frac{q(\beta)}{a_0}}.
$$
Conversely, if $\beta$ is an <invertible linear map> and $a_0=0$, then $m_\beta(t)=tq(t)$ and multiplication of $\beta q(\beta)=0$ by $\beta^{-1}$ gives $q(\beta)=0$, contradicting the defining least degree. Thus \b[invertibility is equivalent to a nonzero constant term of the minimal polynomial]. In the zero-dimensional case the <minimal polynomial> is $1$ and the same conclusion holds.
Back to article page
