I am self studying Chapter -7 of Linear Algebra from Hoffman Kunze and I have a question in 1st section in last corollary whose image I am adding .
Image of Theorem 1:
I have a question in 1st method. How does theorem 1 can be applied with Cayley Hamilton Theorem to deduce the corollary?
Let $p(x)=c_0+c_1x+\dots+c_{k-1}x^{k-1}+x^k$ and $A=\begin{pmatrix}0&0&0&\cdots&0&-c_0\\ 1&0&0&\cdots&0&-c_1\\ 0&1&0&\cdots&0&-c_2\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&0&-c_{k-2}\\ 0&0&0&\cdots&1&-c_{k-1} \end{pmatrix}$. Consider $V=F^k$ with standard (ordered) basis $e=[e_1,e_2,\dots,e_k]$, and the linear operator $U:V\to V$ given by $U(v)=Av$. Since vector $e_i$ has coordinates $(0,\dots,1,\dots,0)$ in $e$ ($1$ is on the $i$-th position), it is easy to see: \begin{eqnarray} U(e_1) &=& e_2\\ U^2(e_1)&=&U(e_2) = e_3\\ U^3(e_1)&=&U(e_3) = e_4\\ &\vdots&\\ U^{k-1}(e_1)&=&U(e_{k-1}) = e_k\\ U^k(e_1)&=&U(e_k) = -c_0e_1-c_1e_2-c_2e_3-\dots-c_{k-2}e_{k-1}-c_{k-1}e_k. \end{eqnarray}
Now, note two things. First, $p(U)(e_1)=0$, which is clear from the previous calculations, so $p\in M(e_1,U)$. Second, $Z(e_1,U)=V$, which is also clear from the previous calculations (all basis vectors belong to $Z(e_1,U)$). By Theorem 1(1), $U$-annihilator of $e_1$ is of degree $\dim(Z(e_1,U))=\dim(V)=k$, so it must be $p$. By Theorem 1(3), $p$ is the minimal polynomial of $U$ (hence of $A$ as well). By Cayley-Hamilton, $p$ is also the characteristic polynomial of $A$ (as minimal polynomial divides characteristic, and characteristic polynomial is of order $k$).