To Find Eigenvalues

Find the eigenvalues of the $6\times 6$ matrix

$$\left[\begin{matrix} 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ \end{matrix}\right]$$

The options are $1, -1, i, -i$

It is a real symmetric matrix and the eigenvalues of a real symmetric matrix are real. Hence $i$ and $-i$ can't be its eigenvalues. Then what else we can say?
Is there any easy way to find it?

Solution 1:

Note that , $$A=\left[\begin{matrix}O&I\\I&O\end{matrix}\right].$$where, $I_{3\times 3}$ and $O_{3\times 3}$ are Identity matrix and Zero matrix respectively.

Now, $AA^T=A^2=I_{6\times 6}\implies A$ is orthogonal. So, eigen values are either $1$ or $-1$. Again, $det(A)=-1$.

If all eigen values are $1$ then $det(A)=1$ , contradiction. Again if all eigen values are $-1$ then $det(A)=1$ ,again a contradiction.

Hence , $1$ and $-1$ both are eigen values of $A$.

Solution 2:

Consider the following permutation, $$\sigma=\left(\begin{matrix}1 & 2 & 3 & 4 & 5 &6\\4&5&6&1&2&3\end{matrix}\right)$$where , the matrix $A$ is defined to be the one whose $i$-th column is the $\sigma(i)$-th column of the Identity matrix. Then order of the permutation $\sigma$ is $2$. So, $A^2=I$. So, $x^2-1$ is an anihilating polynomial of $A$. As, $A\not =\pm I$ so, $x^2-1$ is the minimal polynomial of $A$. So, $\pm 1$ are the eigen values of $A$.

Solution 3:

The matrix has the following effect on the standard basis vectors:

It interchanges $e_1$ with $e_4$, $e_2$ with $e_5$, and $e_3$ with $e_6$. So the three linearly independent (check it!) vectors $e_1+e_4, e_2+e_5, e_3+e_6$ are fixed and so they are eigenvectors with eigenvalue $+1$.

Also check that $e_1-e_4, e_2-e_5, e_3-e_6$ are sent to their negatives, hence are eigenvectors of eigenvalue $-1$.

As these 6 vectors form a basis, $+1$ and $-1$ are the only eigenvalues both of multiplicity $3$.

Solution 4:

Let's call the matrix $A$, and write $A$ as

$A = \begin{bmatrix} 0_3 & I_3 \\ I_3 & 0_3 \end{bmatrix}, \tag{1}$

where $0_3$ and $I_3$ are the $3 \times 3$ zero and identity matrices, respectively. It is easy to see from block matrix multiplication that

$A^2 = I_6, \tag{2}$

which in turn implies that any eigenvalues $\lambda$ satisfy the equation

$\lambda^2 = 1 \tag{3}$

since

$Av = \lambda v \tag{4}$

for some $v \ne 0$; thus

$v = I_6v = A^2 v = \lambda A v = \lambda^2 v \tag{5}$

$(\lambda^2 - 1)v = 0, \tag{6}$

forcing (3) since $v$ doesn't vanish. Thus the only eigenvalues $A$ may have are $\lambda = \pm 1$. In fact, both possibilities occur: for any column three-vector $x \ne 0$, the six-vector

$w = \begin{pmatrix} x \\ x \end{pmatrix} \tag{7}$

satisfies

$Aw = w, \tag{8}$

whereas

$y = \begin{pmatrix} x \\ -x \end{pmatrix} \tag{9}$

solves

$Ay = -y; \tag{10}$

thus both possibilities $\lambda = \pm 1$ occur; the eigenvalues of $A$ are precisely $\pm 1$.

Well, that seems to me like a pretty easy way to do it; we didn't have to evaluate any $6 \times 6$ determinants or do a lot of arithmetic.

Finally, the above easily generalizes to show that the eigenvalues of the $2n \times 2n$ matrix

$\begin{bmatrix} 0_n & I_n \\ I_n & 0_n \end{bmatrix} \tag{11}$

are also exactly $1, -1$.

Solution 5:

Here as you say it's a real symmetric matrix so all the eigen values are real. Now we know that the sum of eigen values is equal to the trace of the matrix. Here trace of the matrix is equal to $0$, so if we take only $1$ or $-1$ as the eigen value, the trace becomes non-zero, so both $1$ and $-1$ are the eigen values.