# Find “REAL” cannonical form of 4x4 matrix

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.

• numbertheory.org/courses/MP274/realjord.pdf – Will Jagy Mar 12 '17 at 16:59
• I looked at it, but the notation is way over my head. – Kuurrwwaa Mar 12 '17 at 17:06
• It basically says we can do Jordan form also for real values if we consider the 2x2 complex representation matrix in $R^{2\times 2}$ and allow $I_2$ blocks as replacement of the 1s off diagonal. – mathreadler Mar 12 '17 at 21:35

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 $$\begin{bmatrix} x&y\\ -y&x \end{bmatrix}$$
• I figured it out. the one thing holding me down was wikipedia's answer saying there should be $I_2$ on the superdiagonal. But then i realised that $I_2$ works as number 1 in Jordan form and since my complex roots only appears once, it is not involved. Thank you anyway I shall upvote your answer. – Kuurrwwaa Mar 12 '17 at 22:00