Section 9.3 Minimal polynomial of a linear map
Let \(F\) be a field and let \(F[x]\) the polynomial ring in one variable over \(F\text{.}\) Suppose that \(V\) is a finite-dimensional vector space over \(F\text{,}\) i.e., \(V\) is a left \(F\)-module. Consider an \(F\)-linear map \(T\in\End_F(V)\text{.}\) We make \(V\) as an \(F[x]\)-module using \(T\) as follows. The scalar multiplication \(F[x]\times V\to V\) is defined by
\begin{equation*}
\left(a_0+a_1x+\cdots+a_nx^n\right)\cdot v= a_0v+a_1T(v)+\cdots+a_nT^n(v).
\end{equation*}
Thus for \(f(x)=a_0+a_1x+\cdots+a_nx^n\in F[x]\) if we consider \(f(T)=a_0\unit_V+a_1T+\cdots+a_nT^n\) then we have, for any \(v\in V\text{,}\)
\begin{equation*}
f(x)\cdot v=f(T)(v).
\end{equation*}
It is left to the reader to verify that with the scalar multiplication defined above \(V\) becomes an \(F[x]\)-module. Given \(f\in F[x]\) we define a group endomorphism of an abelian group \((V,+)\) as follows:
\begin{equation*}
f(T)\colon V\to V\quad\text{given by}\quad v\mapsto f(x)\cdot v=f(T)(v).
\end{equation*}
As in Example 9.2.6 we consider the following ring homomorphism:
\begin{equation*}
\varphi\colon F[x]\to\End_{\rm Grps}(V)\quad\text{given by}\quad f(x)\mapsto f(T).
\end{equation*}
Note that \(f(T)\) is not only a group endomorphism but it is an \(F\)-linear map. Thus, \(f(T)\in\End_F(V)\subset\End_{\rm Grps}(V)\text{.}\) Since \(\varphi\) is a ring homomorphism its kernel, \(\ker(\varphi)\) is an ideal in \(F[x]\text{.}\) By the fundamental isomorphism theorem we get a ring and a vector space isomorphism
\begin{equation*}
F[x]/\ker(\varphi)\simeq\left\{f(T):f\in F[x]\right\}.
\end{equation*}
Since \(\left\{f(T):f\in F[x]\right\}\subseteq\End_F(V)\) and \(\dim_F(\End_F(V))\lt\infty\text{,}\) while \(\dim_F(F[x])=\infty\text{,}\) the \(\ker(\varphi)\) is non-trivial.
As \(F[x]\) is a PID there exists a monic polynomial \(m_T(x)\in\ker(\varphi)\) such that \(\ker(\varphi)=\langle m_T(x)\rangle\text{.}\) Thus \(m_T(x)\) is characterized by each of the following properties:
- Let \(f(x)\in F[x]\) and \(m_T(x)\) is monic. Then \(f(T)\equiv 0\) if and only \(m_T(x)\mid f(x)\text{.}\)
- The polynomial \(m_T(x)\) is a monic polynomial of the least degree such that \(m_T(T)=0\text{.}\)
The polynomial \(m_T(x)\) is called the minimal polynomial of \(T\).
We reprove the following result from ‘Linear Algebra’ course.
Proof.
\begin{equation*}
0=m_T(T)(v)=(T-a\unit_V)\left(g(T)(v)\right).
\end{equation*}
In other words, \(a\in F\) is an eigenvalue of \(T\) with an eigenvector \(g(T)(v)\in V\text{.}\) Now we show the converse. Suppose that \(a\in F\) is an eigenvalue of \(T\text{.}\) Thus there exists \(0\neq v\in V\) such that \(T(v)=av\text{.}\) By division algorithm there exists a constant \(c\in F\) and \(q(x)\in F[x]\) such that
\begin{equation*}
m_T(x)=q(x)(x-a)+c.
\end{equation*}
Hence,
\begin{equation*}
0=m_T(T)(v)=q(T)\left((T-a\unit_V)(v)\right)+c\cdot v=c\cdot v.
\end{equation*}
Since \(v\neq 0\) we get that \(c=0\in F\) and hence \(x-a\mid m_T(x)\text{,}\) i.e., \(a\in F\) is a root of \(m_T(x)\text{.}\)
