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Let $A\in M_n(\mathbb{R})$. I know that $A$ has a real Jordan form which is obtained from the complex Jordan form by using the usual Jordan blocks for the real eigenvalues and by associating to the complex conjugate eigenvalues $\lambda=a+bi$ and $\overline{\lambda}=a-bi$ block diagonal matrices formed by $2\times 2$ blocks of the form $\begin{pmatrix} a_i & b_i \\ -b_i & a_i \end{pmatrix}$.

I understand this and I am able to write the real Jordan form from the complex Jordan form by proceeding like this, but what I don't get is what kind of uniqueness property the real Jordan form has. I mean, what if instead of the block $\begin{pmatrix} a_i & b_i \\ -b_i & a_i \end{pmatrix}$ I use the block $\begin{pmatrix} a_i & -b_i \\ b_i & a_i \end{pmatrix}$, i.e. what if I use $\overline{\lambda}$ instead of $\lambda$? Is this still a valid real Jordan form? If this is the case, I think that the real Jordan form is unique up to a permutation of the elements on the secondary diagonal of these $2\times 2$ blocks, but I am not sure.

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It depends. Let $R=\pmatrix{a&-b\\ b&a}$. If the real Jordan form of $A$ contains a Jordan block of the form $$ \pmatrix{R&I\\ &R&I\\ &&\ddots&\ddots\\ &&&R&I\\ &&&&R}, $$ you may take the transposes of all copies of $R$ within the same Jordan block (and leave other copies of the same $R$ in other Jordan blocks unchanged) and still obtain a valid real Jordan form of $A$. However, you cannot take the transposes of only some $R$ in a Jordan block and leave others in the same block unchanged. E.g. let $$ J_1=\left[\begin{array}{c|cc}R\\ \hline&R&I\\ &&R\end{array}\right],\quad J_2=\left[\begin{array}{c|cc}R\\ \hline&R^T&I\\ &&R^T\end{array}\right] \quad\text{and}\quad M=\left[\begin{array}{c|cc}R\\ \hline&R&I\\ &&\color{red}{R^T}\end{array}\right]. $$ Then $J_1$ is always similar to $J_2$ and both are valid real Jordan forms, but they are not always similar to $M$. In particular, when $R=\pmatrix{0&-1\\ 1&0}$, the minimal polynomial of $M$ is $x^2+1$, but $J_1^2,J_2^2\ne-I$. The matrix $M$ is not considered a real Jordan form.

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  • $\begingroup$ I see, so basically you have to be consistent with taking the transpose of $R$ within the same Jordan block, thank you! $\endgroup$ Feb 22 at 19:53

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