Michael H. answered 05/28/19
Instructor in linear algebra courses with much tutoring experience
This will not be a complete answer, and probably any complete answer should include examples.
The fact that a matrix is a root of its own characteristic polynomial (the Cayley–Hamilton theorem) means there is at least one polynomial of which the matrix is a root.
Degrees of polynomials are in the set { 0, 1, 2, 3, . . . }, and that is a well-ordered set, meaning every non-empty subset of it has a smallest member. The set of degrees of polynomials having the matrix as a root is non-empty because the degree of the characteristic polynomial belongs to it, by the Cayley–Hamilton theorem.
Therefore there is a smallest such degree. Call that number m. There is some polynomial p(λ) of degree m that has our matrix M as a root, i.e. we have p(M) = 0. Dividing both sides of that equation by its leading coefficient yields a monic polynomial, i.e. a polynomial with leading coefficient 1, of degree m, with M as a root. Now one would like to prove there cannot be more than one such monic polynomial. Suppose p(λ) and q(λ) are two monic polynomials of degree m with M as a root. Then p(λ) – q(λ) has M as a root. But the leading terms of p(λ) and q(λ) are the same, both being λm, so they cancel out when one subtracts. This gives us a polynomial whose degree is less than m. But no polynomial of degree less than m can have M as a root because, as seen above, m is the smallest. Therefore there can be only one minimal polynomial.