Determine all eigenvalues of the matrix $$A=\begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}$$ and then determine a base for each eigenspace.

It's easy to compute $\chi_A(z)=z^3-1$ so my roots (and therefore eigenvalues) are $z_1=1, z_2=\cos(2\pi/3)+i \sin(2\pi/3)$ and $z_3=\cos(4\pi/3)+i \sin(4\pi/3)$.

Usually I would determine the eigenspaces by $E_\lambda=\ker(A-I_n\lambda)$, but having the solution to this problem shows that the result should be

$$E_1=\left\langle\begin{bmatrix}1\\1\\1\end{bmatrix}\right\rangle,\qquad\qquad E_{z_2}=\left\langle\begin{bmatrix}z_3\\z_2\\1\end{bmatrix}\right\rangle,\qquad\qquad E_{z_3}=\left\langle\begin{bmatrix}1\\z_2\\z_3\end{bmatrix}\right\rangle.$$

The first one is obvious, but I don't see where the trick is to quickly compute the other two eigenspaces / eigenvectors! Any help would be appreciated!

  • 2
    $\begingroup$ As an aside, it's traditional when working with roots of unity to let $\omega$ or $\zeta$ represent a primitive one (e.g. $z_2$ in your case, or $z_3$ would be just as good), and write things in terms of $\omega$. So your eigenvalues are $1, \omega, \omega^2$, and you can simplify with $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$. It's not a big thing, but it does make things a bit easier to write, and to see patterns. $\endgroup$ – Hurkyl Sep 4 '12 at 20:30
  • $\begingroup$ I got used to write $z$ and $\lambda$ for the variable of my characteristic polynomials so I just didn't change anything to be at least consistent with my notation, thanks for the hint though :) $\endgroup$ – Christian Ivicevic Sep 4 '12 at 20:39

In this particular case, note that $A$ is a permutation matrix corresponding to a 3-cycle. Also, multiplying each of the roots of unity by $z_2$ is the same as permuting the roots in a 3-cycle. So putting those root as the coordinates of vector ought to seem like a good plan for getting something with $z_2$ as an eigenvalue.

In general, once you know an eigenvalue $\lambda$, then corresponding eigenspace is the kernel of (the linear transformation associated with) $A-\lambda I$. So if all attempts to be clever fail, you can always compute that by using Gaussian elimination on $A-\lambda I$.

Example of Gaussian elmination for the eigenvalue $\zeta$. We start with $A-\zeta I$: $$ \begin{pmatrix} -\zeta & 0 & 1 \\ 1 & -\zeta & 0 \\ 0 & 1 & -\zeta \end{pmatrix} $$ Permute the rows to bring some natural pivots to the front and reduce the work: $$ \begin{pmatrix} 1 & -\zeta & 0 \\ 0 & 1 & -\zeta \\ -\zeta & 0 & 1 \end{pmatrix} $$ Add $\zeta$ times the top row to the bottom row: $$ \begin{pmatrix} 1 & -\zeta & 0 \\ 0 & 1 & -\zeta \\ 0 & -\zeta^2 & 1 \end{pmatrix} $$ Add $\zeta^2$ times the middle row to the bottom row: $$ \begin{pmatrix} 1 & -\zeta & 0 \\ 0 & 1 & -\zeta \\ 0 & 0 & 0 \end{pmatrix} $$ So it has rank 2 as expected. Add $\zeta$ times the middle row to the top row, to clear out above the pivot: $$ \begin{pmatrix} 1 & 0 & -\zeta^2 \\ 0 & 1 & -\zeta \\ 0 & 0 & 0 \end{pmatrix} $$ And now we can read off a basis vector for the kernel, namely $(\zeta^2,\zeta,1)$ -- exactly your $E_{z_2}$.

  • $\begingroup$ I am trying to solve it with the general approach as we both mentioned but I get horrible results - is there any way I could persuade you to show me how to use the kernel approach in this particular case? $\endgroup$ – Christian Ivicevic Sep 4 '12 at 20:38
  • $\begingroup$ @ChristianIvicevic: You probably get horrible results if you write the roots explicitly as $\cos(2\pi/3)+i \sin(2\pi/3)$ as in your question and try to use trigonometric identities along the way. The trick to computing with roots of unity is to not split them into real and imaginary parts, write them as $1$, $\zeta$, $\zeta^2$, and simplify using the relations $\zeta^3=1$ and $1+\zeta+\zeta^2=0$ whenever possible. Try that. $\endgroup$ – Henning Makholm Sep 4 '12 at 20:46
  • $\begingroup$ I tried it now a few times and fail though I do use your suggested relations. My current matrix looks like $((1,0,0)^T,(1+\zeta^2,1,0)^T,(0,1+\zeta^2,-2\zeta-\zeta^2)^T)$. I assume that the rank should be 2 and therefore at least my last line is wrong :/ $\endgroup$ – Christian Ivicevic Sep 4 '12 at 21:04
  • $\begingroup$ Worked example added. $\endgroup$ – Henning Makholm Sep 4 '12 at 21:21
  • $\begingroup$ Thanks for your efforts - just because of this I will accept your answer as it is now complete with both descriptions! $\endgroup$ – Christian Ivicevic Sep 4 '12 at 21:29

The matrix you're given is nothing but the permutation matrix corresponding to the permutation $(123)$. Hence, it performs the same permutation on the rows of the matrix $E_{z_2},E_{z_3}$.

Now, just note that multiplying $E_{z_2}$ by $z_2$ and $E_{z_3}$ by $z_3$ also amounts to the same permutation $(123)$.


As far as finding eigenvectors, that's easy once you know the eigenvalue. If 1 is the eigenvalue, then you are looking for solutions to

$$\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} a \\ b \\ c \end{bmatrix}.$$

The left hand side can be multiplied out easily to

$$\begin{bmatrix} c \\ a \\ b \end{bmatrix} = \begin{bmatrix} a \\ b \\ c \end{bmatrix}.$$

Therefore, we must have $c = a$, $a = b$, and $b = c$, or putting it all together, $a = b = c$. Therefore, all eigenvectors are of the form

$$\begin{bmatrix} a \\ a \\ a \end{bmatrix}$$

That is one eigenvector. Do a similar calculation for the other two eigenvalues. It will be more complicated because it will involve the more complicated eigenvalues, but it won't be very complicated because it will just involve $\begin{bmatrix} c \\ a \\ b \end{bmatrix} = k \begin{bmatrix} a \\ b \\ c \end{bmatrix}$, which gives $c = ka$, $a = kb$, and $b = kc$. Pick $a = 1$ and get $c$ and $b$.


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