 Solution

If $p=m_{{\alpha }_W}$ and $q=m_{{\alpha }_{V/W}}$, then $m_{\alpha }$ is divisible by both. Also $q(\alpha )V\subseteq W$, so $p(\alpha )q(\alpha )=0$; hence $\mathrm{lcm}(p,q)\mid m_{\alpha }\mid pq$. A single size-two nilpotent Jordan block with its invariant eigenline makes the lcm too small; a diagonal operator with the same eigenvalue on $W$ and $V/W$ makes the product too large. Square-free $m_{\alpha }$ implies both induced minimal polynomials square-free, so diagonalizability descends. The nilpotent-block example disproves the converse.