# Real jordan form to complex jordan form then compute P matrix.

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 $T = \begin{bmatrix} T_1 | T_2 | \ldots | T_n\end{bmatrix}$.

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 $$V = [v_1 | u_1 | v_2 | u_2 | v_3 | u_3]$$
• Hmmm I understand what you are saying but I'm not sure how to compute it. I just pair up the blocks? We can see that indeed column 1 pairs up with column 4, column 2 with 5 and 3 with 6. Then we get; $u_1 \pm iv_1 = \begin{bmatrix}0 & 1 & 0 & 0 & 0 & -1\end{bmatrix}^T \pm i\begin{bmatrix}2 & 1 & 0 & 0 & 0 & 1\end{bmatrix}^T$? – WG- Oct 5 '12 at 12:46