What are the eigenvalues of this $6 \times 6$ matrix? 
What are the eigenvalues of the following matrix?
$$A=\left(\begin{matrix} 
  0 & 0 & 0 & 1 & 0&0\\
  0 & 0 & 0 & 0 & 1&0\\
  0 & 0& 0 & 0 & 0&1\\
  1 & 0 & 0 & 0 & 0&0\\
  0 & 1 & 0 & 0 & 0&0\\0&0&1&0&0&0
\end{matrix}\right)$$

My attempt: I know how  to find the eigenvalues of a $2 \times 2$ matrix and of a $3 \times 3$ matrix. But here I am very confused, as I don't know how to find the eigenvalues of a $6 \times 6$ matrix.
Is there any easy method or some tricky method?
 A: Hint:
You  can use the fact that for a block matrix we have ( see here):
$$
\det \begin{pmatrix}
A&B\\
C&D
\end{pmatrix}= \det (A-BD^{-1}C)\det D
$$
In this case $B$ and $C$ are the identity matrices and $A,D$ are diagonal, so the result is simple
A: In this case, the simplest way to analyze it is to simply see what it does to vectors.
$A \left(\begin{matrix} 
 a\\
 b\\
 c\\
 d\\
e\\f
\end{matrix}\right) =  \left(\begin{matrix} 
 d\\
 e\\
 f\\
 a\\
b\\c
\end{matrix}\right)$
So for an eigenvector, we have
$A \left(\begin{matrix} 
 a\\
 b\\
 c\\
 d\\
e\\f
\end{matrix}\right) =  \left(\begin{matrix} 
 d\\
 e\\
 f\\
 a\\
b\\c
\end{matrix}\right) =\lambda \left(\begin{matrix} 
 a\\
 b\\
 c\\
 d\\
e\\f
\end{matrix}\right)$
So $d=\lambda a $ and $a = \lambda d$, hence $a = \lambda^2 a$, so $\lambda = \pm 1$.  
You can also observe that if you apply $A$ to a vector, and apply $A$ to the result, you get the original vector. So $A^2=I$, and the minimal polynomial of A is $t^2-1$, which has roots $\pm1$.
A: A permutation of the rows and columns (a change of basis which preserves eigenvalues) transforms your matrix into
$$
A=\left(\begin{matrix} 
  0 & 1 & 0 & 0 & 0&0\\
  1 & 0 & 0 & 0 & 0&0\\
  0 & 0 & 0 & 1 & 0&0\\
  0 & 0 & 1 & 0 & 0&0\\
  0 & 0 & 0 & 0 & 0&1
\\0&0&0&0&1&0
\end{matrix}\right)
$$
which is the direct sum of three copies of $B=\left(\begin{matrix} 
  0 & 1\\
1 & 0
\end{matrix}\right)$. Hence its eigenvalues are those of $B$, i.e., $1$ and $-1$, both with multiplicity 3 (algebraic and geometric).
A: HINT
Let observe that 
$$A(e_1+e_4)=1\cdot (e_1+e_4)$$
$$A(e_1-e_4)=-1\cdot (e_1-e_4)$$
and similarly for others combinations.
A: It's not too hard to find the determinant of $A-\lambda I$ by row operations.  In block form,
$$
    A - \lambda I = \begin{pmatrix} -\lambda I & I \\ I & - \lambda I \end{pmatrix}
$$
By swapping the first and third, second and fourth, and third and sixth rows of $A$, we see that its determinant is $-1$.  This tells us zero is not an eigenvalue.  Therefore, we may take the first, second, and third rows of $A-\lambda I$, multiply them by $\frac{1}{\lambda}$, and add them to the fourth, fifth, and sixth rows.  In block form, we have
$$
    \begin{vmatrix} -\lambda I & I \\ I & - \lambda I \end{vmatrix}
   =\begin{vmatrix} -\lambda I & I \\ 0 & (\lambda^{-1} - \lambda) I \end{vmatrix}
$$
The matrix on the right is upper-triangular, so the determinant is the product of the diagonal entries:
$$
    \begin{vmatrix} -\lambda I & I \\ 0 & (\lambda^{-1} - \lambda) I \end{vmatrix}
   =(-\lambda)^3(\lambda^{-1}-\lambda)^3 = (\lambda^2-1)^3
$$
