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Let $A$ be a square matrix. Prove the equivalence of the following two statements:

(1). No two Jordan block of the Jordan normal form of $A$ corresponding to the same eigenvalue of $A$;

(2). A matrix $B$ commutes with $A$ if and only if $B=p(A)$ for some polynomial $p$.

My try: From cyclic vector theorem, (2) is equivalent to $A$ has a cyclic vector $v$, i.e., $\{v,Av,...,A^{n-1}v\}$ is a basis of vector space. By Rational canonical form, $A$ is then similar to $C(\mu)$, where $\mu$ is the minimal polynomial of $A$ and $C$ is the companion matrix. This means $A$ has only one invariant factor $\mu$ and thus equivalent to (1).

Can we prove the equivalence of (1) and (2) directly?

Q-Y
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  • This is a very interesting mathematical fact! – Greg Martin Dec 19 '19 at 20:47
  • The negation of (1) is equivalent to "$A$ has an eigenvalue $\lambda$ with two corresponding eigenvectors that are linearly independent", while the negation of (2) of course is "There exists a matrix $B$, not of the form $p(A)$, that commutes with $A$". I feel it should be easy to use the first negation to imply the second negation. In the extreme case where all vectors are eigenvectors for $\lambda$ (and $A$ is already in Jordan form), we get such a matrix $B$ by deleting some but not all of the entries $\lambda$ in $A$; something similar should work in general. – Greg Martin Dec 19 '19 at 20:49

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