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Let $A$ be the $2\times2$ matrix, $A= \begin{bmatrix} 0 & -b \\ b & 0 \end{bmatrix}$, then $A$ is of the form $A=b \begin{bmatrix} 0 & -I_1 \\ I_1 & 0 \end{bmatrix}$. In particular, $SAS^{-1}=\begin{bmatrix} 0 & -I_1 \\ I_1 & 0 \end{bmatrix}$ where $S= I_2$

I want to show that if $A$ is the $(2k)\times(2k)$ block diagonal matrix,

$\begin{bmatrix} 0 & -b & \cdots & \cdots & \cdots & \cdots & 0 \\ b & 0 & \cdots & \ddots & \ddots & \ddots & \vdots\\ 0 & \cdots & 0 & -b \\ 0 & \cdots & b & 0\\ 0 & \cdots & \cdots & \cdots & \ddots\\ 0 & \cdots & \cdots & \cdots & \cdots & 0 & -b\\ 0 & \cdots & \cdots & \cdots & \cdots & b & 0\\ \end{bmatrix}$

Then by a permutation matrix $S$, $SAS^{-1}=b\begin{bmatrix} 0 & -I_k \\ I_k & 0 \end{bmatrix}$. I tried induction but did not get very far. And by brute force I have not seen any patterns.

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    The question is confused. $A$ is first $2\times2$ then $2k\times2k$. Also you are conjugating by the identity matrix $S$ in the first paragraph, which is curious. Please clarify.2011-12-14

2 Answers 2