Show the determinant of a matrix Let $R$ be a commutative ring and let $x_o...x_{n-1} \in \mathbb{R}$. Show for the matrix:
$$M = 
\begin{pmatrix}
0 &  &  & x_0 \\
1 & 0 &  & x_1\\
 & 1 & 0& ... \\
 &  &... & x_{n-1}
\end{pmatrix}
$$
and $v \in \mathbb{R}$ that $\det(vI_n - M) = v^n - \sum\limits^{n-1}_{i=0}x_iv^i$.
I get the general idea of determining the determinant but right now I just can't figure out any good way. Maybe using the Leibniz formula?
 A: First of all a remark: I advised you to expand the determinant using the first column, but I was distracted. I meant to write 'the first row', apologies for that. Anyway, here is the solution (perhaps try it yourself using the first column).
Let us prove this statement by induction.
Suppose $n = 1$. We have the matrix $[x_0]$ in this case and we easily find that $$\det(vI - x_0) = v-x_0.$$
Now suppose that the statement holds for matrices of dimension $n-1$. Consider a matrix of dimension $n$:
\begin{equation} A_{n} = 
\begin{pmatrix}
0 & 0& 0 & \ldots & 0& x_0\\
1 &0& 0& \ldots &0 & x_1\\
0 &1&0& \ldots & 0& x_2\\
\vdots & \vdots & \vdots & &\vdots & \vdots\\
0 & 0 & 0 & \ldots & 1 & x_{n-1}
\end{pmatrix}.
\end{equation}
Therefore the matrix $vI - A_{n}$ is given by 
\begin{equation} vI - A_{n} = 
\begin{pmatrix}
v& 0& 0 & \ldots & 0& -x_0\\
-1 &v& 0& \ldots &0 & -x_1\\
0 &-1&v& \ldots & 0& -x_2\\
\vdots & \vdots & \vdots & &\vdots & \vdots\\
0 & 0 & 0 & \ldots & -1 & v - x_{n-1}
\end{pmatrix}.
\end{equation}
We use the expansion in minors using the first $\textbf{row}$ to determine the determinant of $vI - A_n$. We find that this determinant is equal to 
\begin{equation}
v\cdot(-1)^{1+1}
\det
\begin{pmatrix}
v& 0& \ldots &0 & -x_1\\
-1&v& \ldots & 0& -x_2\\
\vdots & \vdots & \vdots & &\vdots \\
 0 & 0 & \ldots & -1 & v - x_{n-1}
\end{pmatrix}
-x_0\cdot (-1)^{n+1} \det
\begin{pmatrix}
-1 & v & 0 &\ldots & 0\\
0 & -1 & v &\ldots & 0\\
\vdots & \vdots &\vdots& & \vdots\\
0 & 0 & 0&\ldots & -1
\end{pmatrix}.
\end{equation}
Due to the induction hypothesis, the first determinant is equal to 
\begin{equation}
v^{n-1} - \sum_{i = 0}^{n-2}x_{i+1}v^i
\end{equation}
The second determinant is the determinant of a upper triangular matrix and hence equal to the product of the diagonal elements, so we find for this part $(-1)^{n-1}$.
Putting al this info together, we find:
\begin{align}
\det(vI - A_n) &= v\cdot (-1)^2(v^{n-1} - \sum_{i = 0}^{n-2}x_{i+1}v^i) -x_0 \cdot (-1)^{n+1} \cdot (-1)^{n-1}\\
        &= v^n - \sum_{i = 0}^{n-2}x_{i+1}v^{i+1} - x_0 \cdot (-1)^{2n}\\
        &= v^n - \sum_{i = 0}^{n-2}x_{i+1}v^{i+1} - x_0.
\end{align}
We now can rewrite the sum by introducing a new variable: $j = i+1$, we then find that 
\begin{equation}
\det(vI - A_n) = v^n - \sum_{j = 1}^{n-1}x_jv^j - x_0 = v^n - \sum_{j = 0}^{n-1}x_jv^j
\end{equation}
where we could write the last equation since $x_0 = x_0v^0$. 
This proves the induction hypothesis, so we can conclude our proof.
