**Problem 1**. Let be a vector space over some field and suppose that Let be a linear transformation. Prove that the minimal and the characteristic polynomials of are equal if and only if there exists some such that are linearly independent over

**Solution**. Let be the minimal polynomial of .

*Proof of *. We cannot have because then, since we’d have which contradicts our assumption that are linearly independent. Thus

*First proof of *. For every let See that is an ideal of and thus, since for some Clearly for all and so But the number of divisors of is finite, which means that the number of distinct elements of the set is finite. Let be those distinct elements and put

See that is a subspace of for all and Therefore for some Thus for all i.e. That implies by the minimality of and so because we already know that and that is a monic polynomial. Finally, suppose that

for some , and let . Then , which gives us

that is Therefore , because and hence for all

*Second proof of* . Give a structure of an module by defining the multiplication by Then see that becomes a finitely generated torsion module and thus, by the fundamental theorem for finitely generated modules over PIDs, there exists some such that It’s easy now to show that are linearly independent over

An important result of Problem 1 is the following:

**Problem 2.** Let be a vector space over and Let be a linear transformation. If are linearly independent over then there exists some such that are linearly independent over

**Solution.** Since the operators are linearly independent, the degree of the minimal polynomial of is and thus the characteristic and the minimal polynomials of are equal. Now the result follows by Problem 1.