Problem on characteristic polynomial of a matrix Without computing $\det(xI-A)$, how to find the characteristic polynomial of $A$ where $A$ is a $4 \times 4$ matrix given by:
$$A = \begin{bmatrix} 0 & 0 & 0 & -4 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 1 & 0 \end{bmatrix}$$

To compute the characteristic polynomial one generally use the Faddeev-LeVerrier algorithm

In your peculiar case however, your matrix has the form of a companion matrix:
$$
A=\begin{bmatrix}0&0&\dots &0&-c_{0}\\1&0&\dots &0&-c_{1}\\0&1&\dots &0&-c_{2}\\\vdots &\vdots &\ddots &\vdots &\vdots \\0&0&\dots &1&-c_{{n-1}}\end{bmatrix}
$$
and by identification the characteristic polynomial is
$$
P(\lambda)=\sum_{k=0}^m c_k \lambda^k=4-5\lambda^2+\lambda^4
$$

In the general case if you want to play the game of not using $\det(\lambda I -A)$ you can use the Jacobi formula which can be obtained by "unfoding" the recurrence relation of the Faddeev-LeVerrier algorithm:
For a $A$ a $n\times n$ matrix, you have
$$
c_{n-m}={\frac {(-1)^{m}}{m!}}{\begin{vmatrix}\text{tr} A&m-1&0&\cdots \\\ \text{tr} A^{2}&\text{tr} A&m-2&\cdots \\\ \vdots &\vdots &&&\vdots \\\ \text {tr} A^{m-1}&\operatorname {tr} A^{m-2}&\cdots &\cdots &1\\\ \text {tr} A^{m}&\operatorname {tr} A^{m-1}&\cdots &\cdots &\operatorname {tr} A\end{vmatrix}}
$$
However there are a lot of computations and it is easier to directly compute $\det(\lambda I -A)$. 
Anyway, here is the details for your example:
$$
A=\left(
\begin{array}{cccc}
 0 & 0 & 0 & -4 \\
 1 & 0 & 0 & 0 \\
 0 & 1 & 0 & 5 \\
 0 & 0 & 1 & 0 \\
\end{array}
\right)
$$


*

*$m=0$ you have $c_4=1$ (the domimant coefficient of the polynomial is always $1$)

*$m=1$, $\text{tr}(A)=0$, hence $c_{4-1}=c_3=0$

*$m=2$, $\text{tr}(A^2)=10$
$$
c_{4-2}=\frac{(-1)^2}{2!}\left|
\begin{array}{cc}
0 & 2-1 \\ 10 & 0
\end{array}
\right|=\frac{1}{2}(-10)=-5
$$

*$m=3$, $\text{tr}(A^3)=0$
$$
c_{4-3}=\frac{(-1)^3}{3!}\left|
\begin{array}{ccc}
0 & 3-1 & 0 \\ 10 & 0 & 3-2 \\
0 & 10 & 0
\end{array}
\right|=\frac{-1}{6}(0)=0
$$

*$m=4$, $\text{tr}(A^4)=34$
$$
c_{4-4}=\frac{(-1)^4}{4!}\left|
\begin{array}{cccc}
0  & 4-1 & 0 & 0 \\ 
10 & 0   & 4-2 & 0 \\
0  & 10  & 0 & 4-3 \\
34 & 0 & 10 & 0 \\
\end{array}
\right|=\frac{1}{24}(96)=4
$$
Your characteristic polynomial is
$$
P(\lambda)=4-5\lambda^2+\lambda^4
$$
which is hopefully the same answer as the "companion matrix" identification trick.
