2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ numbertheory.org/courses/MP274/realjord.pdf $\endgroup$ – Will Jagy Mar 12 '17 at 16:59
  • $\begingroup$ I looked at it, but the notation is way over my head. $\endgroup$ – Kuurrwwaa Mar 12 '17 at 17:06
  • $\begingroup$ 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. $\endgroup$ – mathreadler Mar 12 '17 at 21:35
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Kuurrwwaa Mar 12 '17 at 22:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.