Finding the basis for the Jordan Form with complex eigenvalues The matrix $A = \left[\begin{array}{ccc}2&-1&3\\1&2&1\\0&0&1\end{array}\right]$ has the eigenvalues $\lambda_1 = -1$, $\lambda_2 = 2+i$ and $\lambda_3 = 2-i$ and
the corresponding eigenvectors are $v_1 = \left[\begin{array}{ccc}1\\0\\-1\end{array}\right] $, $v_2 = \left[\begin{array}{ccc}1\\i\\0\end{array}\right] $ and $v_3 = \left[\begin{array}{ccc}1\\-i\\0\end{array}\right] $
How can I find a real matrix $M$ such that:
$$M^{-1}A M = \left[\begin{array}{ccc}1&0&0\\0&2&1\\0&-1&2\end{array}\right] ?$$
 A: $v_1 = \pmatrix{-2\\1\\1},v_2 = \pmatrix{1\\i\\0},v_3 = \pmatrix{1\\-i\\0}$
Now, unfortunately I have forgotten the whys of it but, if you can locate the complex entries such that the complex parts are equal and across the main daigonal from one annother....i.e.
$P = \pmatrix{1&i&-2\\i&-1&-1\\0&0&1}$  then you can do this...
$M = \pmatrix{1&1&-2\\1&-1&-1\\0&0&1}$
$\pmatrix{2&-1&3\\1&2&1\\0&0&1}=M\pmatrix{2&1&0\\-1&2&0\\0&0&1}M^{-1}$
A: See here:

The canonical Jordan decomposition of a matrix that acts on  a vector space over the field $K$ exists only if the  the characteristic polynomial of the matrix splits into linear factors over the field $K$. 

Tis means that this form exists for all matrix if the field $K$ is algebraically closed,  as $\mathbb{C}$, but the existence is not guaranteed if the filed is $\mathbb{R}$. In your case you have two not real eigenvalues (as noted in the comment the  real eigenvalue is $\lambda=1$) so we cannot have a Jordan decomposition with respect to a basis in $\mathbb{R}^3$. 
