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Let $A$ be an $m \times m$ real matrix, and let \begin{equation} A=C^{-1} J C, \end{equation} \begin{equation} A=\tilde{C}^{-1} J \tilde{C}, \end{equation} be two Jordan decompositions of $A$, where $J$ is the same Jordan canonical form. Consider a partition $\{ I_1, I_2 \}$ of the set of eigenvalues of A such that if an eigenvalue $\lambda$ belongs to $I_i$ then also the conjugate of $\lambda$ belongs to $I_i$. Suppose that \begin{equation} J= \left( \begin{array}{ccc} J_1 & O \\ O & J_2 \end{array} \right), \end{equation} where $J_i$ is the Jordan block of dimension $n_i \times n_i$ corresponding to the eigenvalues in $I_i$, $i=1,2$. Now let $B=C^{-1}$ and $\tilde{B}=\tilde{C}^{-1}$. Partition $C$ as follows \begin{equation} C=\left( \begin{array}{ccc} C_{11} & C_{12} \\ C_{21} & C_{22} \end{array} \right), \end{equation} where $C_{ij}$ has dimension $n_{i} \times n_{j}$, for $i,j=1,2$, and do the same for $\tilde{C}$, $B$ and $\tilde{B}$. Suppose that $B_{11}$ and $\tilde{B}_{11}$ are invertible, so that also $C_{22}$ and $\tilde{C}_{22}$ are invertible. Then I guess the following statements are true.

(I) The matrices $B_{11} J_1 B_{11}^{-1}$, $B_{21} B_{11}^{-1}$, and $C_{22}^{-1} J_2 C_{22}$ are real matrices, and

(II) $B_{11} J_1 B_{11}^{-1} = \tilde{B}_{11} J_1 \tilde{B}_{11}^{-1}$, $B_{21} B_{11}^{-1} = \tilde{B}_{21} \tilde{B}_{11}^{-1}$, $C_{22}^{-1} J_2 C_{22} = \tilde{C}_{22}^{-1} J_2 \tilde{C}_{22}$.

What do you think about?

Thank you very much for your help.

Best Regards, Maurizio Barbato

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The answers to my two questions are two definitevly Yes! Let us start from (II). We know that the columns vectors in the matrix \begin{equation} \left( \begin{array}{ccc} B_{11} \\ B_{21} \end{array} \right) \end{equation}

are a basis of the direct sum of the generalized eigenspaces corresponding to $I_{1}$, and the same holds for the matrix \begin{equation} \left( \begin{array}{ccc} \tilde{B}_{11} \\ \tilde{B}_{21} \end{array} \right). \end{equation}

So there exists an invertible $n_1 \times n_1$ matrix $P$ such that \begin{equation} \left( \begin{array}{ccc} B_{11} \\ B_{21} \end{array} \right) = \left( \begin{array}{ccc} \tilde{B}_{11} \\ \tilde{B}_{21} \end{array} \right) P. \end{equation}

From this we obtain \begin{equation} B_{11} J_1 B_{11}^{-1} = \tilde{B}_{11} J_1 \tilde{B}_{11}^{-1}, \end{equation}

and

\begin{equation} B_{21} B_{11}^{-1} = \tilde{B}_{21} \tilde{B}_{11}^{-1}. \end{equation}

Analogously one can obtain the other equality in (II).

Now, to prove (I), it is enough to consider the case in which one chooses as basis of the direct sum of the generlized eigenspaces relative to $I_{i}$ a special one, in which to the vector $v$ relative to the eigenvalue $\lambda$ in $I_{i}$ corresponds the conjugate of $u$ for the conjugate of $\lambda$. One then can check directly from the definition of matrix product that $B_{21} B_{11}^{-1}$ is a real matrix. Moreover let $F$ be the restriction of the linear map $F_{A}$ associated to $A$ to the direct sum of the generalized eigenspaces relative to the eigenvalues in $I_{1}$, and let $\mathbf{f}$ be the basis of this space made up of the columns of

\begin{equation} \left( \begin{array}{ccc} B_{11} \\ B_{21} \end{array} \right) \end{equation}

Then $J_1$ is the matrix which represents $F$ in this basis. Now interpret $B_{11}$ as the matrix of a change of basis from $\mathbf{f}$ to a new basis $\mathbf{g}$. NThe columns of $B_{11}$ are interpreted now as the coordinates of the vectors in $\mathbf{f}$ in this new basis, and since in this new basis the conjugate of a vector $v$ in $\mathbf{f}$ has as coordinates the conjugate of the vector of the coordinates of $v$, it is easily deduced that $g$ is a real basis, so that we end up that \begin{equation} B_{11} J_1 B_{11}^{-1} \end{equation}

represents $F$ in a real basis, so it has to be a real matrix. One can analogously proceed for the other matrix in (I).

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  • $\begingroup$ Let us note here that the proof shows also that $B_{11}$ is invertible if and only if $\tilde{B}_{11}$ is. Moreover, the proof also applies to the more general case in which $A=\tilde{C}^{-1} \tilde{J} \tilde{C}$, where $\tilde{J}$ decomposes into $\left( \begin{array}{ccc} \tilde{J}_1 & O \\ O & \tilde{J}_2 \end{array} \right)$, where $\tilde{J}_i$ contains all the blocks corresponding to the eigenvalues in $I_i$. $\endgroup$ – Maurizio Barbato Mar 9 '15 at 16:28
  • $\begingroup$ Now I realize that there is a more simple way to prove statement (i). First note that if $ A = \tilde{C}^{-1} \tilde{J} \tilde{C}$ is a $ \mathbf{Jordan}$ $\mathbf{real}$ $\mathbf{canonical}$ form of $A$, with $\tilde{J} =\left( \begin{array}{ccc} \tilde{J}_{1} & O \\ O & \tilde{J}_{2} \end{array} \right)$, where $\tilde{J}_{i}$ contains the real Jordan blocks corresponding to the eigenvalues in $I_i$, then statement (ii) still holds and the proof is identical. Since now $\tilde{C}$ and $\tilde{J}$ are real matrices, statement (i) immediately follows. $\endgroup$ – Maurizio Barbato Apr 12 '15 at 11:28

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