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I have the matrix $A = \begin{bmatrix} 5 & 0 & 1 & 0 & 0 & -6 \\ 3 & -1 & 3 & 1 & 0 & -6 \\ 6 & -6 & 5 & 0 & 1 & -6 \\ 7 & -7 & 4 & -2 & 4 & -7 \\ 6 & -6 & 6 & -6 & 5 & -6 \\ 2 & 1 & 0 & 0 & 0 & 0 \end{bmatrix}$

This can be brought in the following Jordan form, i.e. $A = TJT^{-1}$.

$J = \begin{bmatrix} 2-3j & 1 & 0 & 0 & 0 & 0 \\ 0 & 2-3j & 1 & 0 & 0 & 0 \\ 0 & 0 & 2-3j & 0 & 0 & 0 \\ 0 & 0 & 0 & 2+3j & 1 & 0 \\ 0 & 0 & 0 & 0 & 2+3j & 1 \\ 0 & 0 & 0 & 0 & 0 & 2+3j \end{bmatrix}$

$T = \begin{bmatrix} 2j & 2j & 1+j & -2j & -2j & 1-j \\ 1+j & 2j & j & 1-j & -2j & -j \\ 0 & 2j & 2j & 0 & -2j & -2j \\ 0 & 1+j & 2j & 0 & 1-j & -2j \\ 0 & 0 & 2j & 0 & 0 & -2j \\ -1+j & -1+j & j & -1-j & -1-j & -j \end{bmatrix} $

Now I have to bring A into its real Jordan form. This is easy: $J^{R} = \begin{bmatrix} 2 & 3 & 1 & 0 & 0 & 0 \\ -3 & 2 & 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 3 & 1 & 0 \\ 0 & 0 & -3 & 2 & 0 & 1 \\ 0 & 0 & 0 & 0 & 2 & 3 \\ 0 & 0 & 0 & 0 & -3 & 2 \end{bmatrix} $

Now I have to compute $V$ such that $VJ^RV^{-1} = A$. My question now is how do I compute this $V$? For real valued jordan forms this is easy. I just have to compute the eigenvectors of A, $\{T_i\}$ and then $T = \begin{bmatrix} T_1 | T_2 | \ldots | T_n\end{bmatrix}$.

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Notice that, with every complex pair of eigenvalues $\lambda = a \pm ib$, there exists a complex pair of eigenvectors $u \pm i v$. If you look at the columns of your matrix $T$, you can observe that you can pair up your eigenvectors according to complex conjugates in this precise way.

In real canonical form, each of your real Jordan blocks $\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$ comes directly from the eigenspaces generated by the eigenvectors $u \pm iv$. So your $V$ should look like $V = [v_1 | u_1 | v_2 | u_2 | v_3 | u_3]$

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    Solved it. I calculated V*A*inv(V) instead of V*JR*inv(V). Don't know why I made this error hehe.2012-10-08