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For $a_1,\ldots,a_n \in \mathbb{R}$ I got the following $n \times n$ Matrix

$$ B=\begin{pmatrix}0 & 0 & \cdots & 0 & a_n \\ a_{1} & 0 & \cdots & 0 & 0\\ 0 & a_{2} & \cdots & 0 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & a_{n-1} & 0 \end{pmatrix}, $$ which can be considered as a special case of the Leslie-Matrix.

Now I want to proof, that $B^n = \left(\prod_{i=1}^n a_i \right) \cdot I_n$, where $I_n$ is the Identity Matrix.

Any hints or suggestions?

Ronaldinho
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  • The positions of the $a_k$ are identical to the positions of the $1$s in a generating element of the cyclic group of order $n$. – mathreadler Oct 29 '17 at 21:29

1 Answers1

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Note $$Bv = \begin{bmatrix} a_n v_n \\ a_1 v_1 \\ a_2 v_2 \\ \vdots \\ a_{n-1} v_{n-1}\end{bmatrix}.$$ Can you explain the operation in plain words, and then describe what $B^n v$ is?

angryavian
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