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I have to show that the determinant of

\begin{pmatrix} \lambda I_p & -I_p & & 0\\ & \ddots &\ddots \\ & & \ddots & \ddots \\ 0 & & & \lambda I_p & -I_p \\ A_0 & A_1 & \cdots & A_{n-2} & \lambda I_p +A_{n-1} \end{pmatrix} equals \begin{equation} \det( A_0+ A_1 \lambda +...+ A_{n-1}{\lambda}^{n-1}+ I_p {\lambda}^n) \end{equation} where each $A_i$ is square of dimension $p$. The case where $p=1$ is straightforward but I am stuck on higher dimensions of $p$. How should I proceed?

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Note that the inverse of the leading principal $(n-1)p\times(n-1)p$ submatrix is given by $$ \begin{pmatrix} \lambda I_p & -I_p\\ & \ddots &\ddots \\ & & \ddots & -I_p \\ & & & \lambda I_p \end{pmatrix}^{-1} =\begin{pmatrix} \lambda^{-1} I_p & \lambda^{-2}I_p & \cdots & \lambda^{1-n}\, I_p\\ & \ddots &\ddots &\vdots\\ & & \ddots & \lambda^{-2}I_p \\ & & & \lambda^{-1} I_p \end{pmatrix} $$ Therefore, using Schur complement, your determinant evaluates to \begin{align*} &\det\begin{pmatrix} \lambda I_p & -I_p\\ & \ddots &\ddots \\ & & \ddots & -I_p \\ & & & \lambda I_p \end{pmatrix} \times \\ &\det\left[(\lambda I_p +A_{n-1}) - (A_0, A_1, \ldots, A_{n-2}) \begin{pmatrix} \lambda I_p & -I_p\\ & \ddots &\ddots \\ & & \ddots & -I_p \\ & & & \lambda I_p \end{pmatrix}^{-1} \begin{pmatrix}0\\ \vdots\\ 0\\-I_p\end{pmatrix}\right]\\ =&\lambda^{n-1} \det\left[(\lambda I_p +A_{n-1}) + (A_0, A_1, \ldots, A_{n-2}) \begin{pmatrix}\lambda^{1-n}\,I_p\\ \vdots\\ \lambda^{-2}\,I_p\\ \lambda^{-1}I_p\end{pmatrix}\right]\\ =&\det( A_0+ A_1 \lambda +...+ A_{n-1}{\lambda}^{n-1}+ I_p {\lambda}^n). \end{align*}

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Thank you very much! –  amun Dec 20 '12 at 20:51
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