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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?

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    The $a_i$ are the $x_i$ in the matrix. Use induction and for the induction step: expand the first column of the matrix and use the induction hypotheses. – 2017-02-01
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    oh yes you are right, i will edit my post. And I will try the induction, thanks. – 2017-02-01
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    Ok, I am still struggling a bit. Could you please show the induction basis? – 2017-02-02
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    I'm on mobile now, so i'll type it out when i'm home. – 2017-02-02

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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.