Dtyjlui

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
$$
P(lambda)=sum_{k=0}^m c_k lambda^k=4-5lambda^2+lambda^4
$$

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 $ntimes n$ matrix, you have
$$
c_{n-m}={frac {(-1)^{m}}{m!}}{begin{vmatrix}text{tr} A&m-1&0&cdots \ text{tr} A^{2}&text{tr} A&m-2&cdots \ vdots &vdots &&&vdots \ text {tr} A^{m-1}&operatorname {tr} A^{m-2}&cdots &cdots &1\ text {tr} A^{m}&operatorname {tr} A^{m-1}&cdots &cdots &operatorname {tr} Aend{vmatrix}}
$$

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
$$
P(lambda)=4-5lambda^2+lambda^4
$$

which is hopefully the same answer as the "companion matrix" identification trick.