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Prove that characteristic polynomial of a complex matrix $A$ divides its minimal polynomial if and only if all eigenspaces of $A$ are one-dimensional.
As far as I can see I the only possible case is when minimal polynomial equals characteristic one.
All distinct eigenvalues with multiplicity 1 grant us that the eigenspaces would be one-dimensional, I thought that this is the key to the solution, however we have the theorem stating that on the other side, eigenspace dimension could be less then or equal to its eigenvalue algebraic multiplicity. Everything now mixed up, will be thankful for any help.