Does a matrix exist whose eigenpolynomial equals to any polynomial $f(x)$?

Question

For any polynomial $f(x) = \sum _{i = 0} ^{n} a_i x_i $ $(n \ge 1, a_n = 1)$, does a matrix $A$ exist so that $\left \vert \lambda I - A \right \vert = f(\lambda)$ ?

I verified some cases:

1. For $n = 1,$ $f(x) = x + a_0 \Rightarrow f(\lambda) = \left \vert \lambda I - \begin{pmatrix} -a_0 \end{pmatrix} \right \vert$

2. For $n = 2,$ $f(x) = x^2 + a_1 x + a_0 = x(x + a_1) + a_0 \Rightarrow f(\lambda) = \left \vert \begin{pmatrix} \lambda & a_0 \\ -1 & \lambda + a_1 \end{pmatrix} \right \vert = \left \vert \lambda I - \begin{pmatrix} 0 & -a_0 \\ 1 & -a_1 \end{pmatrix} \right \vert$

3. If it is true for any irreducible polynomial, then it is true for any polynomial, since assuming $f(x) = \prod _{i = 1} ^{m} p_i(x), $ where $p_i(x)$ is irreducible, then $ f(\lambda) = \left \vert \lambda I - \begin{pmatrix} A_1(\lambda) & & \\ & \ddots & \\ & & A_m(\lambda) \end{pmatrix} \right \vert$ where $p_i(x) = \left \vert \lambda I - A_i(\lambda) \right \vert$

Answer 1

Yes, for example the companion matrix of the polynomial $f(x)$ has characteristic polynomial $f(x)$.

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Hope this helps.