Is the real Jordan form unique?

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.

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