given is the $6 \times 6$-matrix $A$:

$A = \begin{pmatrix} 0 & 1 & 0 & -1 & 0 & 0 \\ 0 &0&1&1&-1&0\\ -1&0&0&0&-1&-1 \\ 1 & 0&0&0&1&0 \\ 0&1&0&0&0&1 \\ 0&0&1&1&0&0 \end{pmatrix}$

With only the information that

  • $A$ has exactly two different eigenvalues
  • one eigenvalue is $t_1 = i$

I have to determine the Jordan-matrix.

How can I do this with only the information of $t_1 = i$ ?

  • $\begingroup$ The entries are all real, so if there are only 2 eigenvalues, the other must be the complex conjugate of $i$, with a multiplicity of 3. $\endgroup$ – DanielV May 5 '14 at 9:09
  • $\begingroup$ Hi - why is this so? Could you please send me some information about this fact, since this really helps a lot! :) $\endgroup$ – Vazrael May 5 '14 at 9:15
  • $\begingroup$ @DanielV Why must the other eigenvalue be the complex conjugate of the eigenvalue $i$, noting that the entries are real? $\endgroup$ – Mussé Redi May 5 '14 at 9:16
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    $\begingroup$ If $Av = \lambda v,$ taking complex conjugates of everything gives $A \overline{v} = \overline{\lambda} \overline{v}$ as $A$ is real. $\endgroup$ – Geoff Robinson May 5 '14 at 9:34
  • $\begingroup$ Fine question. en.wikipedia.org/wiki/Complex_conjugate_root_theorem and the eigen values are the roots of the en.wikipedia.org/wiki/Characteristic_polynomial which as we can see must have real coefficients. $\endgroup$ – DanielV May 5 '14 at 9:34

Hints. This is a real matrix. So, nonreal eigenvalues must occur in pairs of conjugates. In general, to answer a question like yours, merely knowing the above fact is not enough. We also need to know the (seldom taught) fact that Jordan blocks for nonreal eigenvalues also occur in pairs of conjugates. But fortunately, for your particular $A$, we don't need this latter fact.

Return to the hints. So, the other distinct eigenvalue of $A$ must be $-i$. Therefore, the minimal polynomial of $A$ must take the form of $[(x-i)(x+i)]^k=(x^2+1)^k$ for some $k\in\{1,2,3\}$. What is the smallest $k$ that makes $(A^2+I)^k$ equal to zero?

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    $\begingroup$ In this case, since $A$ has trace zero, we know that $i$ and $-i$ must occur with equal multiplicity. $\endgroup$ – Geoff Robinson May 5 '14 at 9:36
  • $\begingroup$ I thank you very much for this necessary information. By knowing that also $-i$ must be an eigenvalue (and both occur with algebraic multiplicity of 3) the solution is obvious. Again, thank you very much! $\endgroup$ – Vazrael May 5 '14 at 9:38

The characteristic polynomial is given by $(t^2 + 1)^3$. Hence the six eigenvalues are $\pm i$ with multiplicities. Because of $tr (A)=0$ both $i$ and $-i$ have the same multiplicity.

  • $\begingroup$ I think you a made a calculation error. The characteristic polynomial is given by $(t^2+1)^3 = t^6+3t^4+3t^2+1$. $\endgroup$ – Mussé Redi May 5 '14 at 9:23
  • $\begingroup$ The characteristic polynomial is not allowed to be considered, I can only argue with what DanielV said. $\endgroup$ – Vazrael May 5 '14 at 9:26
  • $\begingroup$ @MusséRedi: Thank you, I had a typo in the matrix. $\endgroup$ – Dietrich Burde May 5 '14 at 9:50

Supplement to the above answers.

Let $A$ be a real matrix, then the complex eigenvalues will always occur in complex conjugate pairs.


If an eigenvalue is complex, its eigenvectors will in general be vectors with complex entries. Now, let $\lambda$ be a complex eigenvalue of $A$, with a non-zero eigenvector $v$, so $Av = \lambda v$. Taking complex conjugates of both sides then yields $$A\bar{v} = \bar{A}\bar{v} = \bar{\lambda}\bar{v},$$ where the first equality follows because $A$ is a real matrix.

So $\bar{\lambda}$ is also necessarily an eigenvalue of the matrix $A$.


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