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Let $A$ be a complex matrix and $A_c$ the companion matrix of its characteristic polynomial. From what I have read, I believe the following two statements to be true: (1) not every $A$ is similar to $A_c$, and (2) $A$ is similar to $A_c$ if $A$ has a simple spectrum. My question would be then, is $A$ similar to $A_c$ if and only if the spectrum of $A$ is simple? or, Are there other conditions under which we can have $A \sim A_c$? I am particularly interested in complex Hermitian matrices.

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Are you familiar with the theory of Jordan normal form? The problem reduces to computing the Jordan normal form of a companion matrix. – Qiaochu Yuan Feb 6 '13 at 17:21
This explains it clearly – Andreas Caranti Feb 6 '13 at 17:27
@QiaochuYuan Do you mean by "Theory of Jordan normal form" that every complex matrix can be decomposed as $A = PJP^{-1}$ where $J$ is the Jordan canonical form of $A$ and $P$ is nonsigular? I know that. If $A$ and $A_c$ have the same Jordan form, then they must be similar. But, can we say something about $A \sim A_c$ without making reference to the Jordan form of $A_c$? Perhaps the answer is simply "no" but I am not sure at this point. – Goku Feb 6 '13 at 17:28
@AndreasCaranti Ok, I think I get it now. The statements $A \sim A_c$ and "the minimal polynomial of $A$ has degree $n$" are equivalent. If I am not wrong, the second statement implies that the spectrum of $A$ is simple so that "$A \sim A_c$ if and only if $A$ has a simple spectrum." – Goku Feb 6 '13 at 17:58
@Goku: $A \sim A_c$ is a statement about the Jordan normal form of $A_c$. It is not true that this condition implies that $A$ has simple spectrum; you can find examples with $A$ nilpotent. – Qiaochu Yuan Feb 6 '13 at 19:42
up vote 1 down vote accepted

(For sake of having an answer, let's turn the comments to this question into one.)

The minimal polynomial and characteristic polynomial of a companion matrix are equal. So, if $A\sim A_c$, the minimal and characteristic polynomials of $A$ are equal, too. In other words, in the Jordan form of $A$, every eigenvalue is associated with only one Jordan block. Alternatively speaking, the geometric multiplicity of each eigenvalue is equal to $1$.

When $A$ is Hermitian, geometric multiplicities and algebraic multiplicities coincide. Hence the above necessary and sufficient condition reduces to that all eigenvalues of $A$ are simple, i.e. $A$ has distinct eigenvalues.

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