Weyr form of a matrix

Question

I have a real Jordan form $$ C = \begin{bmatrix} 3 & 1 & 1& 0 & 0\\ -1& 3& 0 & 1& 0\\ 0& 0& 3& 1 &0 \\ 0 & 0& -1& 3& 0\\ 0 & 0& 0& 0 & 4 \end{bmatrix} $$

Could you please tell me how to find the Weyr form of $C$?

I confuse because this is not a normal form of Jordan block.

Answer 1

Because there is only one Jordan block for each eigenvalue (i.e. each eigenvalue has geometric multiplicity $1$), the Weyr and complex Jordan canonical forms coincide. In particular, the complex Jordan canonical form of your matrix will be $$ J = \pmatrix{ 3+i&1\\ &3+i\\ &&3-i&1\\ &&&3-i\\ &&&&4} $$ Note that this matrix is already in Weyr canonical form.