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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

  1. $ \ 1 \ $

  2. $-1$

  3. $ \ i $

  4. $-i$

Community
  • 1
MAS
  • 10,638

2 Answers2

2

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

$$\det \begin{bmatrix} s \,\mathrm I_3 & -\mathrm R_3\\ -\mathrm I_3 & s \,\mathrm I_3\end{bmatrix} = \det \left( s^2 \mathrm I_3 - \mathrm R_3 \right) = p (s^2)$$

Rodrigo de Azevedo
  • 1
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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 $\lambda$?

Math The Novice
  • 908
  • Sorry, I suddenly realize that my argument only shows that $|\lambda|=1$, which is of no help to your current problem. I think @Rodrigo de Azevedo answer is the correct (and maybe the most efficient way) to do this problem. He study the block matrix eigenvalue. – Math The Novice Jun 01 '18 at 09:11