= Solution
Suppose some exponent $q_j\geq2$. Define
$$
h(x)=\frac{m_T(x)}{f_j(x)^2},\qquad N=f_j(T).
$$
Part (iii) makes $N$ a <normal operator>. For every $v\in V$, put $w=h(T)v$. The <minimal polynomial> identity gives
$$
N^2w=f_j(T)^2h(T)v=m_T(T)v=0.
$$
Part (ii), applied to $N$, forces $Nw=0$. As this holds for every $v$,
$$
0=Nh(T)=\left(\frac{m_T}{f_j}\right)(T).
$$
This contradicts minimality: $m_T/f_j$ is a <monic polynomial> of smaller degree annihilating $T$. Therefore the <minimal polynomial of a normal operator> is a <square-free polynomial> over either of the given <fields>:
$$
\boxed{q_1=\cdots=q_r=1,\qquad m_T=f_1\cdots f_r}.
$$
This argument does not require the <irreducible polynomials> to be linear, so it also covers the real case.
Back to article page
