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How to solve following determinant by applying suitable elementary row/column transformations to obtain characteristic polynomial?

\begin{align*} \left\vert \begin{matrix} -\lambda & 0 & 1 & 0 & 0 & \cdots & 0\\ 0 & -\lambda & 0 & 1 & 1 &\cdots & 1\\ 1 & 0 & -\lambda & 1 & 1 &\cdots & 1\\ 0 & 1 & 1 & -\lambda & 0 &\cdots & 0\\ 0 & 1 & 1 & 0 & -\lambda &\cdots & 0\\ \vdots & \vdots & \vdots & \vdots& \vdots &\ddots & \vdots \\ 0 & 1 & 1 & 0 & 0 &\cdots & -\lambda \end{matrix} \right\vert_{n+3} & =0 \\ \end{align*}

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Add $\frac1\lambda$ of all except the first three rows to the second and third rows to get rid of the columns of $1$s. That leaves $(-\lambda)^n$ from the diagonal times

$$ \begin{vmatrix} -\lambda&0&1\\ 0&-\lambda+\frac n\lambda&\frac n\lambda\\ 1&\frac n\lambda&-\lambda+\frac n\lambda \end{vmatrix} =-\lambda\left(-\lambda+\frac n\lambda\right)^2+\lambda-\frac n\lambda+\frac{n^2}\lambda=-\lambda^3+2n\lambda+\lambda-\frac n\lambda\;, $$

so the overall determinant is

$$ (-\lambda)^{n+3}-(2n+1)(-\lambda)^{n+1}+n(-\lambda)^{n-1}\;. $$

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@joriki's proof considered the case $\lambda\ne 0$. Now, for the case that $\lambda=0$, we have \begin{align} \left\vert \begin{matrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{matrix} \right\vert=0\quad\mbox{and}\quad \left\vert \begin{matrix} 0 & 0 & 1 &0\\ 0 & 0 & 0 &1\\ 1 & 0 & 0 &1\\ 0 & 1& 1 & 0 \end{matrix} \right\vert= \left\vert \begin{matrix} 0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1& 0 \end{matrix} \right\vert=1, \end{align} and for $n\geq 2$, \begin{align} \left\vert \begin{matrix} 0 & 0 & 1 & 0 & 0 & \cdots & 0\\ 0 & 0 & 0 & 1 & 1 &\cdots & 1\\ 1 & 0 & 0 & 1 & 1 &\cdots & 1\\ 0 & 1 & 1 & 0 & 0 &\cdots & 0\\ 0 & 1 & 1 & 0 & 0 &\cdots & 0\\ \vdots & \vdots & \vdots & \vdots& \vdots &\ddots & \vdots \\ 0 & 1 & 1 & 0 & 0 &\cdots & 0 \end{matrix} \right\vert_{n+3}=0 \end{align} because the last $n$ columns are identical.

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  • $\begingroup$ I should have perhaps explicated this: My proof doesn't consider only the case $\lambda\ne0$. The determinant is a characteristic polynomial; the fact that I divided by $\lambda$ in an intermediate step doesn't prevent the final result from being the characteristic polynomial valid for all $\lambda$, including $\lambda=0$. Of course in the case $n=0$ the last term should be omitted; this, too, I should have perhaps explicated. $\endgroup$ – joriki Apr 24 '16 at 7:21

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