What are the eigenvalues of this $6 \times 6$ matrix?

Question

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?

Answer 1

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 $\pm1$.

Answer 2

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.

Answer 3

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

Answer 4

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).

Answer 5

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: $$ \begin{vmatrix} -\lambda I & I \\ 0 & (\lambda^{-1} - \lambda) I \end{vmatrix} =(-\lambda)^3(\lambda^{-1}-\lambda)^3 = (\lambda^2-1)^3 $$