Polynomials $\&$ Matrices Assume $A$ is a matrix of order $n$.  We know that the characteristic polynomial of matrix $A$ is obtained as follows
$$
P(x)=\det (A-x\,I)\, .
$$
Where $I$ is an identity matrix of order $n$. What about inverse?  For a given polynomial 
like $P(x)$, Is there an efficient method to find a 
matrix like $A$ where the  characteristic polynomial of matrix $A$ be the polynomial $P(x)$.
I know that the number of matrices that have the same characteristic polynomials  are uncountable. Because, If 
we assume that all entries of matrix $A$ be indeterminates then the number of variables in 
the equation 
$$
\det (A-x\,I)=P(x)
$$
 are more than equations. Thanks for any suggestion.
Edit:
My motivation of this question is that if $A$ be a non-derogatory matrix (in other words, its minimal and characteristic polynomials coincide) then Frobenius normal form of matrix $A$ is a companion matrix. Now, If we have a companion matrix like $C$, how to find a matrix like $A$, such that the companion matrix $C$ be the Frobenius normal form of matrix $A$. 
The user @Jack answered the obviously solution of my original question and because of this I edit my question.

Yes. What you are looking for is the companion matrix.
