I am studying Jordan forms. I see that Hoffman & Kunze define the companion matrix of a polynomial $a_0+a_1x+\dots+a_{k-1}x^{k-1}+x^k$ as $$ \begin{pmatrix} 0 & 0 & 0 & \dots &0 & -a_0\\ 1 & 0 & 0 & \dots & 0 & -a_1\\ 0 & 1 & 0 & \dots & 0 & -a_2 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & \dots & 1 & -a_{k-1}\end{pmatrix}$$ I know that the reason for doing so is that Hoffman & Kunze define Jordan block to be a lower triangular matrix. To arrive at this definition, they make some arguments as follows: Suppose $T$ is a transformation with cyclic vector $v$ and $p_v=a_0+a_1x+\dots+a_{k-1}x^{k-1}+x^k$, the $T$ - annihilator of $v$ (and also the minimal polynomial of $T$). Then the vectors $v, T(v), \dots, T^{k-1}(v)$ would form a basis $\mathcal{B}$ for $V$. If we let $v_i = T^{i-1}(v)$, $i=1,\dots, k$, then we can write $\mathcal{B} = \{v_1, v_2, \dots, v_k\}$. Notice that $$T(v_i) = v_{i+1}, 1\leq i \leq k-1$$ and $$T(v_k) = T^k(v) = -a_0v_1- a_1v_2-\dots-a_{k-1}v_k.$$ This automatically leads to the matrix as defined above.
Now, I prefer to study the Jordan block in upper triangular form (although it shouldn't matter). For this, I numbered the elements differently. Letting $T^{k-1}(v)=v_1$ and defining $v_i = T^{n-i}(v)$, $2\leq i \leq k$, I observed that $$T(v_1) = T^k(v) = -a_0v-a_1T(v)-\dots-a_{k-1}T^{k-1}(v) = -a_{k-1}v_1-a_{k-2}v_2-\dots-a_1v_{k-1}-a_0v_k$$ and $$T(v_i)=v_{i-1}, 2\leq i \leq k.$$ Thus in the new ordered basis $\{v_1, \dots, v_k\}$, the matrix of $T$ looks like $$\begin{pmatrix} -a_{n-1} & 1 & 0 & \dots & 0 & 0\\ -a_{n-2} & 0 & 1 & \dots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ -a_1 & 0 & 0 & 0 & 0 & 1\\ -a_0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}$$ Now, this companion matrix aids me in obtaining an upper triangular Jordan form.
Now, I do not see any reference to this companion matrix form anywhere. I just need a clarification if I am thinking correctly. Would it be problematic if I go with this definition of companion matrix?