# 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. – Arthur Jul 16 '20 at 12:59
• @Arthur Thanks. Just saw that... deleted my c omment. – coffeemath 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. – user10354138 Jul 16 '20 at 13:14
• For more on non-derogatory matrices, see also this post. – Ben Grossmann 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 $$-1$$.