Jordan Form and Minimal PolynomialsBy necessity the minimal polynomial has a zero at each eigenvalue.
Lemma: If A is an nxn matrix and p(x) is any polynomial, then the eigenvalues of p(A) are p(lambda), where lambda is an eigenvalue of A. The proof is not hard, but we will try to justify the lemma with an example instead.
Lemma: If A is an nxn matrix and ...
Lemma: If lambda is an eigenvalue of A, then (x-lambda) is a factor of m(x).
Lemma: If lambda is an eigenvalue of A, ...
These two lemmas show that the minimal polynomial has a zero at each eigenvalue.
Note that by the Cayley-Hamilton Theorem [if c(x) is the characteristic polynomial of A, c(A)=0], the minimal polynomial divides the characteristic polynomial.
These two facts will enable us to search for the minimal polynomial. We will also discover how the minimal polynomial relates to the Jordan form.
Example 1
Example 2
Example 3
Up to Jordan Canonical Form
Example 4