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 $\det(A)=-a_0$
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 $A$.