 Solution

Suppose some exponent $q_j\ge 2$. DefinePart (iii) makes $N$ a normal operator. For every $v\in V$, put $w=h(T)v$. The minimal polynomial identity givesPart (ii), applied to $N$, forces $Nw=0$. As this holds for every $v$,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:This argument does not require the irreducible polynomials to be linear, so it also covers the real case.