A real $8\times 8$ matrix $A$ has $2-i$ and $3+4i$ among its eigenvalues, and their algebraic multiplicity is 2. Write down the possible generalized (real) Jordan matrices for $A$.

How can I use the complex roots condition? I know that for real matrix the complex eigenvalues comes in pairs, right? And the "algebraic multiplicity is 2" means the characteristic equation has double that roots. This is all I can get from this problem.


You are exactly correct about the information that you can extract. It turns out, that's everything you need.

We can deduce that the Jordan form of $A$ has two $2 \times 2$ blocks on the diagonal, each corresponding to $2 \pm i$, and another two $2 \times 2$ blocks on the diagonal, each corresponding to $3 \pm 4i$.

All together, the Jordan form has the diagonal $$ \pmatrix{ \pmatrix{2 & -1\\1&2}\\ &\pmatrix{2 & -1\\1&2}\\ &&\pmatrix{3 & -4\\4&3}\\ &&&\pmatrix{3 & -4\\4&3} } $$ We therefore have $4$ distinct real Jordan forms, corresponding to whether we put both copies of a given $2 \times 2$ matrix into the same block. So, for example, one Jordan form would be given by $$ \pmatrix{ \pmatrix{2 & -1\\1&2}&I\\ &\pmatrix{2 & -1\\1&2}\\ &&\pmatrix{3 & -4\\4&3}&0\\ &&&\pmatrix{3 & -4\\4&3} } $$

  • $\begingroup$ Thank you! I see. So you change the $\lambda$ to a 2 by 2 matrix(with that complex eigenvalue). This is new to me, but it sounds reasonable, since we restrict it in real number. So the only change is the $I$ and $0$ matrices in your example(and there are 4 type of JCF), right? $\endgroup$ – breezeintopl Apr 3 '15 at 15:57
  • $\begingroup$ Right. The $I$ is instead of a $1$. And yes, $4$ types. $\endgroup$ – Omnomnomnom Apr 3 '15 at 16:08

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