What's the Jordan canonical form of this matrix? 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 $t_1 = i$ ?
 A: 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?
A: 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.
A: Supplement to the above answers.
Let $A$ be a real matrix, then the complex eigenvalues will always occur in complex conjugate pairs.
Proof:
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 $A$. 
