so my lecturer gave me this task of finding real cannonical form of 4x4. I can easily find Jordan cannonical form of my matrix, In my case I have one repeated real root with $AM=2>GM=1$ and two complex ones so $J=\begin{bmatrix} J_2(\lambda_1) & \\ & J_1(\lambda_2) \\ & & J_1(\lambda_3) \end{bmatrix}$
However he stated real cannonical form and obviously $J$ is complex and not real.
I have found some notes online that given $3x3$ matrix with 1 real eigenvalue we can extend the basis, "split" the complex eigenvalues and we obtain $C=\begin{bmatrix} a & 0 &0 \\ 0 &x & y\\ 0 & -y & x \end{bmatrix}$ the real canonnical form with eigenvalues $\lambda_1=a, \lambda_2=x+iy, \lambda_3=x-iy$
and $P=\begin{bmatrix} v_1 & x &y \\ \end{bmatrix}$ where $v_1$= the eigenvector of $\lambda_1$
$x$ is the real part of eigenvectors $\lambda=x\pm iy $ and $y$ is the imaginery part.
I tried to extend the idea to $4x4$ but I can't seem to work it out. I do hope I explained it clearly enough, the whole idea is very new to me so i am sorry for not being as clear as I desire. The notes I found online are here
I don't really want to give ma matrix out, because i wanted to do it myself, so if anyone has a general way of finding real canonnical form for $4x4$ that would be more than enough.
Hint:
In this case the extension to a $4\times 4$ real matrix is simple. The Jordan block for the real eigenvalue $\lambda_1$ with algebraic multiplicity $2$ and geometric multiplicity $1$ is $$ \begin{bmatrix} \lambda_1&1\\ 0&\lambda_1 \end{bmatrix} $$ Since the characteristic equation has degree $4$ (and $\lambda_1$ is a double root), the two complex eigenvalues have to be conjugate: $\lambda_2=x+iy$ and $\lambda_3= \overline{\lambda_2}=x-iy$ so the corresponding real Jordan block is $$ \begin{bmatrix} x&y\\ -y&x \end{bmatrix} $$
and these are the two diagonal blocks of the real canonical form.