= Solution
Suppose $m$ has no repeated irreducible factor. For every <polynomial> $p$ and <integer> $n\ge1$, its monic greatest common divisors satisfy $\gcd(p^n,m)=\gcd(p,m)$, since raising $p$ to a power changes no membership of an irreducible factor of the square-free $m$. This is the <square-free minimal polynomial criterion for polynomial kernel stabilization>. The kernel identity proved above gives
$$
\ker(p(T)^n)=\ker\gcd(p^n,m)(T)
=\ker\gcd(p,m)(T)=\ker p(T).
$$
Thus \b[(ii) implies (iii)] over every <field>.
Over $\mathbb C$, the distinct irreducible factors of $m$ are linear. The <primary decomposition theorem> therefore reads $V=\bigoplus_j\ker(T-\lambda_jI)$, a direct sum of <eigenspaces>. Taking a <basis> in each summand gives an <eigenbasis>, proving \b[(ii) implies (i) over the complex numbers].
Over $\mathbb R$, take $T=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$. Its <minimal polynomial> is $X^2+1$, which is an <irreducible polynomial> over $\mathbb R$ and has no repeated factor. But it has no real <eigenvalues>, so no real <eigenbasis>. This is the required counterexample. Over a general <field>, the phrase “no repeated factors” here refers to irreducible factors in $K[X]$, rather than a prior assumption that those factors split.
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