For $a_0,\ldots,a_{n-1}\in\C$ prove that the characteristic polynomial $c_A(z)=\det(A-z)$ of the matrix $$ A=\left(\begin{array}{ccccc} 0&0&\cdots&0&-a_0\\ 1&0&\cdots&0&-a_1\\ 0&1&\cdots&0&-a_2\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&1&-a_{n-1} \end{array}\right)\in\Ma(n,\C) $$ is given by $(-1)^n(z^n+a_{n-1}z^{n-1}~\cdots+a_1z+a_0)$. $A$ is called the companion matrix of the monic polynomial $p(z)=z^n+a_{n-1}z^{n-1}~\cdots+a_1z+a_0$. 2. Check that for all $k=0,\ldots,n-1$: $A^ke_1=e_{k+1}$ and conclude that the minimal polynomial of $A$ is $c_A$. 3. $A$ is diagonalizable iff $c_A$ has only simple roots.

1. By Laplace's expansion and induction on $n$ we get for $c_n(z)=\det(A-z)$ and $c_{n-1}(z)=z^{n-1}+a_{n-1}z^{n-2}+\cdots+a_1$: \begin{eqnarray*} c_n(z) &=&-zc_{n-1}(z)+(-1)^{n-1}(-a_0)\\ &=&(-1)^n(z(z^{n-1}+a_{n-1}z^{n-2}+\cdots+a_1)+a_0)\\ &=&(-1)^n(z^n+a_{n-1}z^{n-1}+\cdots+a_1z+a_0)~. \end{eqnarray*} 2. We obviously have $Ae_1=e_2,\ldots,Ae_{n-1}=e_n$ and thus for all $k=0,\ldots,n-1$: $A^ke_1=e_{k+1}$. Suppose for some $k < n$: $m(A)=A^k+c_{k-1}A^{k-1}+\cdots+c_0=0$, then $m(A)e_1=e_{k+1}+b_{k-1}e_k+\cdots+b_0e_1\neq0$.
3. $A$ is diagonalizable iff the minimal polynom $m$ has only simple roots and since $m=c_A$, $A$ is diagonalizable iff $c_A$ has only simple roots.