Suppose there is a linear transform T from V to V where V is the finite dimensional vector space, what will be the conditions that the minimal polynomial splits? I know if it splits then T is diagonalizable and all the T-invariant subspaces will have an eigenvector of T, will these implies the minimal polynomial splits?
Thanks in advance
First, the minimal polynomial$~\mu_T$ is split if and only if the characteristic polynomial$~\chi_T$ is split, as both have the same irreducible factors (since $\mu_T$ divides $\chi_T$, which divides a power of$~\mu_T$). If none of these irreducible factors have degree${}\geq2$ then $\mu_T$ and $\chi_T$ are split, otherwise they are not.
Clearly this condition depends on the field one is working over. Over an algebraically closed field like $\Bbb C$ all irreducible polynomials are of degree$~1$, so $\mu_T$ and $\chi_T$ are always split. Working over other fields$~K$ they are not always split.
If $\mu_T$ has an irreducible factor$~P\in K[X]$ with $\deg P>1$ (so in particular $P$ has no roots in$~K$) then $\ker(P[T]$) is a nonzero $T$-stable subspace without any eigenvectors of$~T$. So indeed, if there are no such subspaces, that implies that $\mu_T$ and $\chi_T$ were split. Conversely, if $\mu_T$ splits, then $V$ is a direct sum of generalised eigenspaces of$~T$, and the projections on the factors of this decomposition are given by certain polynomials in$~T$; this implies that any $T$-stable subspace contains its own projections on the generalised eigenspaces, and is the direct sum of those projections. Moreover any nonzero $T$-stable subspace of a generalised eigenspace for$~\lambda$ contains an eigenvector for$~\lambda$, so by the preceding any nonzero $T$-stable subspace of$~V$ contains an eigenvector.
So indeed this property characterises $T$ with split $\mu_T$.