TITLE: Do characteristic polynomials exhaust all monic polynomials?
QUESTION [24 upvotes]: Let $A$ be an $n\times n$ matrix, then $\mathrm{char}_A(x):=\det(xI-A)$ is a monic polynomial of degree $n$. It is called the characteristic polynomial of $A$. My question is the converse:

Let $p(x)$ be a monic polynomial of degree $n$. Can we always find an $n\times n$ matrix such that $p(x)=\mathrm{char}_A(x)$?

REPLY [13 votes]: Not only does the companion matrix exist, but it is quite easy to motivate its construction. Fix a field $k$ and consider the action of $x$ on $k[x]/p(x)$, which (by monicity of $p$) is a $k$-vector space of dimension $n$ with basis $\{ 1, x, x^2, ... x^{n-1} \}$. The action of $x$ in this basis is the companion matrix. Moreover, by construction $p(x) = 0$ is the minimal polynomial of $x$, and it has degree $n$ so it is also the characteristic polynomial.