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?

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    $\begingroup$ You know what, sum of eigenvalues is equal to the trace(matrix), also product of eigenvalues is equal to det(matrix). $\endgroup$ – Ajay May 6 '15 at 5:58
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    $\begingroup$ To add to what RattusRattus said: The trace is quite obviously 0. As Tani has already found, i and -i can't be eigenvalues. That leaves only 1 and -1, the sum of which must equal 0. Thus the eigenvalues are 1, 1, 1, -1, -1, and -1. $\endgroup$ – Yay295 May 6 '15 at 8:06

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 $A$.

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    $\begingroup$ different and simple idea. $\endgroup$ – Tani May 6 '15 at 13:30
  • $\begingroup$ excellent idea +1 $\endgroup$ – Learnmore May 9 '15 at 3:28
  • $\begingroup$ If you wouldn't mind, could you explain how you deduce that $x^2-1$ is an annihilating polynomial? I haven't had advanced linear yet, but I understand all of your argument except that part. $\endgroup$ – Alex Mathers May 9 '15 at 3:59
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    $\begingroup$ @mathers101 To say that $p(x)$ is annihilating polynomial just means that $p(A)=0$. Here, we have $A^2-I=0$. $\endgroup$ – Potato May 9 '15 at 4:22
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    $\begingroup$ Did you not understand what moderators have told you about bumping your posts with insignificant edits? $\endgroup$ – Jyrki Lahtonen Jun 7 '15 at 16:11

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 $A$.


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 $3$.


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

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


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

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

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


$Aw = w, \tag{8}$


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


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

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 $1, -1$.


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.


We can also change basis to get $$I_3 \otimes C_2 = \left(\begin{array}{ccc}C_2&0&0\\0&C_2&0\\0&0&C_2\end{array}\right)$$ where $C_2 = \left(\begin{array}{cc}0&1\\1&0\end{array}\right)$ is a representation for the generating element of the cyclic 2-group. For this small matrix it is easy to find value 1,-1 and vectors $\left(\begin{array}{c}1\\1\end{array}\right)$ and $\left(\begin{array}{r}1\\-1\end{array}\right)$. Please note that the eigenvectors are different from my other answer above, but this is OK (why?)

  • $\begingroup$ It is because there may be many matrix representations of a group but only one which is irreducible. The FFT one with circulant matrices above is not irreducible, but this one is. $\endgroup$ – mathreadler May 6 '15 at 7:19
  • $\begingroup$ This is the easiest method to get all eigenvalues and their multiplicities at once. Moreover it would work (with different size cycle matrices but not with the tensor product) for any permutation matrix (once you know that the eigenvalues of a $k$-cycle matrix are precisely the $k$-th roots of unity). Moreover, of all the properties that one immediately reads off of that matrix, being a permutation matrix is one for the most obvious ones. $\endgroup$ – Marc van Leeuwen May 7 '15 at 10:03

The matrix is circulant. Therefore $SDS^{-1}$ is it's eigendecomposition, where $S$ is the fast fourier transform of $I_6$ and $D$ is the diagonal matrix with elements from the FFT of the first row, which happens to be $diag([1,-1,1,-1,1,-1])$. So the eigenvalues are $-1$ and $1$.


Because sum of elements of each row is equal to one so one is an eigen value of the matrix. Again if we put $x=-1$ in $|A-xI|$ we get two row are equal so $-1$ is also an eigen value of $A$.

  • $\begingroup$ yes its right.. $\endgroup$ – neelkanth May 6 '15 at 7:44
  • $\begingroup$ its simplest way to check eigen values... $\endgroup$ – neelkanth May 6 '15 at 7:45
  • $\begingroup$ How you say that sum of each row is $1$ implies $1$ is an eigen value ? Please explain.. $\endgroup$ – Empty May 6 '15 at 8:16
  • $\begingroup$ Yes corresponding vector will be (1,1,...1) now you can check easily... $\endgroup$ – user237521 May 6 '15 at 8:18
  • $\begingroup$ Actually if the sum of entries is the same constant for all rows (think of magic squares) that constant will be an eigenvalue with $(1,1,\ldots1)^T$ as eigenvector. $\endgroup$ – P Vanchinathan May 8 '15 at 7:29

The given matrix is a non-zero orthogonal symmetric matrix with trace zero, so its eigenvalues are $1$ and $-1$.


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