# How to prove that the rank of a nonderogatory matrix is at least its order $-1$

If $$A \in M_{n}$$ is nonderogatory, why is rank $$A \geq n-1 ?$$

• @coffeemath According to Wikipedia, a matrix with minimal polynomial equal to the characteristic polynomial. Jul 16 '20 at 12:59
• @Arthur Thanks. Just saw that... deleted my c omment. Jul 16 '20 at 13:00
• If nullity is at least 2, that means in the JNF there are at least two blocks with eigenvalue 0, hence min poly and char poly are not the same. Jul 16 '20 at 13:14
• For more on non-derogatory matrices, see also this post. Jul 16 '20 at 13:29

Hint: If $$A$$ is non-derogatory, then all eigenvalues of $$A$$ must have geometric multiplicity $$1$$.

• I know that. Could you give a detailed proof?
– Ben
Jul 16 '20 at 22:26

Consider the Jordan canonical form of $$A$$, we have $$A = SJ_AS^{-1}$$.

Assume that $$J_A=J_{n_{1}}\left(\lambda_{1}\right) \oplus \cdots \oplus J_{n_{k}}\left(\lambda_{k}\right)$$.

Since the geometric multiplicity of a given eigenvalue of a Jordan matrix is equal to the number of Jordan blocks corresponding to that eigenvalue, a matrix is nonderogatory if and only if each of its distinct eigenvalues corresponds to exactly one block in its Jordan canonical form. Thus, $$\lambda_1, \cdots , \lambda_k$$ are mutually distinct.

Case 1: $$A$$ has $$0$$ eigenvalue. Then by the rank-nullity theorem, we have $$n - rank(A-0I) = 1$$, we have $$rank A = n-1\ge n-1$$

Case 2: $$A$$ doesn't have $$0$$ eigenvalue, i.e., $$\lambda_i \ne 0, \forall i = 1,\dots, k$$. Thus, the rank of A is the number of nonzero rows of $$J_A$$, i.e., n

To summary, the rank of a nonderogatory matrix is at least its order $$-1$$.