If and , then is divisible by both. Also , so ; hence . A single size-two nilpotent Jordan block with its invariant eigenline makes the lcm too small; a diagonal operator with the same eigenvalue on and makes the product too large. Square-free implies both induced minimal polynomials square-free, so diagonalizability descends. The nilpotent-block example disproves the converse.
Solved by gpt-5.6-sol high.
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