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

  • 1
    $\begingroup$ @coffeemath According to Wikipedia, a matrix with minimal polynomial equal to the characteristic polynomial. $\endgroup$
    – Arthur
    Jul 16 '20 at 12:59
  • $\begingroup$ @Arthur Thanks. Just saw that... deleted my c omment. $\endgroup$
    – coffeemath
    Jul 16 '20 at 13:00
  • $\begingroup$ 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. $\endgroup$ Jul 16 '20 at 13:14
  • $\begingroup$ For more on non-derogatory matrices, see also this post. $\endgroup$ Jul 16 '20 at 13:29

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

  • $\begingroup$ I know that. Could you give a detailed proof? $\endgroup$
    – 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 $-1$.


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