Consider $n-1$ Companion matrices $C^{(i)}$ over $\mathbb{R}$, for $1\leq i \leq n-1$, which are defined by the following form: $$ C^{(i)}=\left( \begin{array}{cccccc} 0 &1 &0 &\cdots &\cdots &0 \\ 0 &0 &1 &\ddots &\ddots &\vdots \\ \vdots &\ddots &\ddots &\ddots &\ddots &\vdots \\ \vdots &\ddots &\ddots &\ddots &\ddots &0 \\ 0 &\cdots &\cdots &0&0 &1 \\ u_1^{(i)} &u_2^{(i)} &u_3^{(i)} &\cdots &u_{n-1}^{(i)} &u_{n}^{(i)} \end{array} \right)_{n\times n} $$ where $u_k^{(i)}$ for $1\leq i \leq n-1$ and $1\leq k \leq n$, are positive integer numbers.
Now consider the matrix $B=\prod_{i=1}^{n-1}\, C^{(i)}$, the product of matrices $C^{(i)}$.
My question: Let $B=(b_{i,j})$. Then how to prove that $b_{1,j}=0$ for $1\leq j \leq n-1$.
for example for $n=3$ we have $$ B = \begin{pmatrix} 0 & 0 & b_{1,3}\\ b_{2,1} & b_{2,2} & b_{2,3}\\ b_{3,1} & b_{3,2}& b_{3,3} \end{pmatrix}. $$
My try: Consider digraphs of $C^{(i)}$ matrices . The elements $u_k^{(i)}$'s are called weight in these digraphs. Therefore, we can assume that they have the same value which means $u_k^{(1)}=u_k^{(2)}=\cdots=u_k^{(n-1)}$ for $1\leq k \leq n$.
Hence it is enough to prove if $E=(C^{(1)})^{n-1}=(e_{i,j})$ then why $e_{1,j}=0$ for $1\leq j \leq n-1$. Consider digraph of $C^{(i)}$ then we can see that there is no walk of length $n-1$ from the vertex $v_1$ to $v_j$ for $1\leq j \leq n-1$ which means $e_{1,j}=0$ for $1\leq j \leq n-1$.
Is this proof correct? If there is other proof, I appreciate you to post it.
Thanks for any suggestions.
The proof is sorta correct, but you should justify better why you can suppose all the weights as the same.
An other proof goes through noticing that $$ e_i^T C^{(k)} = e_{i+1}^T$$ for every $k$ and every $i<n$, since $e_i^T C^{(k)} $ is the $i$-th row of $C^{(k)}$. Now you have $$ b_{1j} = e_1^T B e_j = e_1^T C^{(1)} C^{(2)} \dots C^{(n-1)} e_j = e_2^T C^{(2)} C^{(3)} \dots C^{(n-1)} e_j =$$$$ e_3^T C^{(3)} C^{(4)} \dots C^{(n-1)} e_j = \dots = e_{n-1}^T C^{(n-1)} e_j $$ and the element $(n-1,j)$ of $C^{(n-1)}$ is zero whenever $1\le j<n$.