Linear Algebra Characteristic Polynomials Let $p(t) = t^n+a_{n-1}t^{n-1}+a_{n-2}t^{n-2} + \cdots + a_1t+a_0$.  Show that the characteristic polynomial of the matrix A below
\begin{bmatrix}
0 & 0 & \cdots & & & -a_0\\
1 & 0 &        & & & -a_1\\
0 & 1 & \ddots &       & & \vdots\\
\vdots  &   & \ddots       &\ddots & & \vdots\\
\vdots & &           & \ddots      & 0 & -a_{n-2}\\
0      & 0 & \cdots          & \cdots      & 1 & -a_{n-1}         
\end{bmatrix}
 is the polynomial $p$.
Please help, currently I got that the $\det(A)=-a_0$ 
 A: Define
$$
     e_{1} =\left[\begin{array}{c}1 \\ 0 \\ 0 \\ \vdots \\ 0\end{array}\right]
     e_{2} =\left[\begin{array}{c}0 \\ 1 \\ 0 \\ \vdots \\ 0\end{array}\right],
     \cdots,
     e_{n} =\left[\begin{array}{c}0 \\ 0 \\ 0 \\ \vdots \\ 1\end{array}\right].
$$
By the definition of your matrix $A$,
$$
\begin{align}
      Ae_{1} & = e_{2} \\
      A^{2}e_{1} & = e_{3} \\
          \vdots \\
     A^{n-1}e_{1} & = e_{n} \\
     A^{n}e_{1} & = -a_{0}e_{1}-a_{1}e_{2}-\cdots-a_{n-1}e_{n}
\end{align}
$$
It follows that
$$
       (A^{n}+a_{n-1}A^{n-1}+\cdots+a_{1}A+a_{0}I)e_{1}=0.
$$
Define $p(\lambda)=\lambda^{n}+a_{n-1}\lambda^{n-1}+\cdots+a_{1}\lambda+a_{0}$. Then $p(A)e_{1}=0$ and no lower order monomial $q$ can have this property. Because $p(A)e_{1}=0$, then $p(A)=0$ because
$$
          p(A)e_{k} = p(A)A^{k-1}e_{1}=A^{k-1}p(A)e_{1} = 0,\;\;\; k=2,3,\cdots,n.
$$
The minimal polynomial for $A$ must be of order $n$ because it cannot be of higher order, and there is no lower order monomial $q$ for which $q(A)e_{1}=0$. So $p$ is the minimal polynomial for $A$.
