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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.

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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 $$ \pmatrix{ \pmatrix{2 & -1\\1&2}&I\\ &\pmatrix{2 & -1\\1&2}\\ &&\pmatrix{3 & -4\\4&3}&0\\ &&&\pmatrix{3 & -4\\4&3} } $$

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  • $\begingroup$ 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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