Determining a matrix from its characteristic polynomial Let $A\in\mathcal{M}_{n}(K)$, where $K$ is a field. Then, we can obtain the characteristic polynomial of $A$ by simply taking $p(\lambda)=\det(A-\lambda I_n)$, which give us something like
$$p(\lambda) = (-1)^n\lambda^n + (-1)^{n-1}(\text{tr } A)\lambda^{n-1} +  \cdots + \det A$$
Now, how can we obtain the matrix $A$ knowing the characteristic polynomial?
 A: Any two similar matrices $B = P A P^{-1}$ will have the same characteristic polynomial.  Also $A$ and $A^T$ have the same characteristic polynomial.
While the matrix $A$ which has a given characteristic polynomial is not unique, it is often convenient to choose an upper Hessenberg matrix called the (Frobenius) companion matrix or its (lower Hessenberg) transpose.
That is, the $n\times n$ matrix:
$$ \begin{bmatrix}
0 & 0 & \ldots & 0 & -c_0 \\
1 & 0 & \ldots & 0 & -c_1 \\
0 & 1 & \ldots & 0 & -c_2 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \ldots & 1 & -c_{n-1}
 \end{bmatrix} $$
has characteristic polynomial $p(x) = x^n + c_{n-1}x^{n-1} + \ldots + c_1 x + c_0$.
One approach to finding the roots of a polynomial is to apply eigenvalue solvers to the companion matrix for the polynomial.
A: There is no way you can determine the matrix from its characteristic polynomial because the characteristic polynomial is determined by $n$ numbers and the matrix by $n^2$ numbers.
A: If $A$ and $S$ are $n\times n$ matrices, with $S$ invertible, then $A$ and $SAS^{-1}$ have the same characteristic polynomial. But even non similar matrices can have the same characteristic polynomial: consider
$$
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix},\qquad
\begin{bmatrix}
1 & 1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix},\qquad
\begin{bmatrix}
1 & 1 & 0 \\
0 & 1 & 1 \\
0 & 0 & 1
\end{bmatrix}
$$
So you cannot find the matrix having a given characteristic polynomial.
