Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 2 / 8E (Linear Algebra)

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.

Each eigenspace of ${\alpha }_1$ is stable under every commuting ${\alpha }_i$. Restrict the remaining diagonalizable maps to each eigenspace and induct; concatenating their simultaneous eigenbases proves the result.

Yes. In finite dimension, construct a common-eigenspace decomposition by repeatedly refining with members of the family. Every proper refinement increases the finite number of summands, so finitely many maps already produce a decomposition stable under all remaining maps; diagonalize their restrictions as above.