Sufficient condition for a polynomial to be a characteristic polynomial

Question

Let $A\in \operatorname{Mat}_{n\times n}(F),~F$ being a field, satisfies $p(x)\in F[x]$ where $\deg p(x)=n$ and $p(x)$ is a monic polynomial. Can we say $p(x)=\chi_A(x)?$

Answer 1

Just to provide a simple example, the $2 \times 2$ zero matrix is a root of all polynomials $$ x^2 + a x, $$ for all $a \in F$.

Its characteristic polynomial is $x^2$, of course.

Answer 2

No.

While it is true that $\chi_A$ is monic of degree $n$, in general there are many such polynomials, and only a single one of them is $\chi_A$.

Answer 3

If I understand correctly, I think you mean for any monic polynomial $p(x)$ of degree $n$ over $F$, there is always a $n$-dimensional matrix corresponding to. This is true as you can always take the corresponding companion matrix.

Answer 4

No. If $\chi_A(x)$ is the characteristic polynomial of $A$, then $\chi_A(x)+1$ is a degree-$n$ polynomial that is necessarily not equal to $\chi_A(x)$. Hence there always exists a degree-$n$ monic polynomial that is not the characteristic polynomial of $A$.

Alternatively, if every arbitrary degree-$n$ monic polynomial $p(x)$ is a characteristic polynomial of $A$, then characteristic polynomials are not unique. Yet, by definition, $\chi_A(x)=\det(xI-A)$, meaning that the characteristic polynomial is uniquely defined.