The eigenvalues of the following block matrix $$\begin{pmatrix} 0 & 0 & 0 &0 &0 &1 \\ 0 &0&0 &0 &1&0 \\ 0 &0&0 &1 &0&0 \\ 1&0&0&0&0&0\\0&1&0&0&0&0 \\ 0&0&1&0&0&0 \end{pmatrix}$$ are
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2026-03-30 12:31:34.1774873894
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Find the eigenvalues of the block matrix
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Hint : This is a permutation matrix $P$. Note that the components of a vector does not change except its order. The definition of an eigenvector is $Pv=\lambda v$. Now look at the components of $v$, what does it say about $\lambda$?
Let $\mathrm R_3$ be the $3 \times 3$ exchange matrix
$$\mathrm R_3 := \begin{bmatrix}0 & 0 & 1\\0 & 1 & 0\\1 & 0 & 0\end{bmatrix}$$
whose characteristic polynomial is
$$p (s) := (s-1)^2 (s+1)$$
Hence, the characteristic polynomial of the given $6 \times 6$ permutation matrix is
$$\det \begin{bmatrix} s \,\mathrm I_3 & -\mathrm R_3\\ -\mathrm I_3 & s \,\mathrm I_3\end{bmatrix} = \det \left( s^2 \mathrm I_3 - \mathrm R_3 \right) = p (s^2)$$