Return to Answer

• Step 1: Find the eigen decomposition of $A \sigma $. The output is (X,d) where d is diagonal and X is the change of basis matrix.

• Step 2: Compute $X \sigma X^T $ which will be of the form $\textrm{Diag}_1(n) \oplus \textrm{Diag}_2(n) $ , compute $C=(\textrm{Diag}_1)^{-1/2} \oplus i (\textrm{Diag}_2)^{-1/2} $. The solution is then $\tilde{X}=X C$ and the diagonal is $d \sigma$

• Step 1: Find the eigen decomposition of $A \sigma $. The output is (X,d) where d is diagonal and X is the change of basis matrix.

• Step 2: Compute $X \sigma X^T $ which will be of the form $\textrm{Diag}_1(n) \oplus \textrm{Diag}_2(n) $ , compute $C=(\textrm{Diag}_1)^{-1/2} \oplus i (\textrm{Diag}_2)^{-1/2} $. The solution is then $\tilde{X}=X C$ and the diagonal matrix is $d \sigma$

For the second point, we will assume a hypothetical natural extension of results from a paper that is cited in the last discussion below. That discussion concerns results from an analogue scenario of sympletic matrices. Assuming that the paper can be extended in a natural way, there exists (I will explain why in the last section) an $\tilde{X}$ that verifies the eigen factorization mentioned before

For the second point, we will assume a hypothetical natural extension of results from a paper that is cited in the last discussion below. That discussion concerns results from an analogue scenario of sympletic matrices. Assuming that the paper can be extended in a natural way, there exists (I explain why in the last section, the reader can go there now if they wish to do so) an $\tilde{X}$ that verifies the eigen factorization mentioned before

Diag$(1,1,...,-1,-1,..,-1)=\{ \{ 1_{n \times n},0_{n \times n}\},\{0_{n \times n},1_{n \times n}\}\}$

For the second point, we will assume a hypothetical natural extension of results from a paper that is cited in the last discussion below. That discussion concerns results from an analogue scenario of sympletic matrices. Assuming that the paper can be extended in a natural way, there exists an $\tilde{X}$ that verifies the eigen factorization mentioned before

In the following we will consider that the answer is yes and we will check after.

d // DiagonalMatrixQ
X . A . Transpose[X] - d . σ // Norm
Inverse[X] - σ . Transpose[X] . σ // Norm


(* True

 4.85014*10^-15

 5.88943*10^-15 *)

If I take the hypothesis that the same applies here then basically it suffices to show that:

Diag$(1,1,...,-1,-1,..,-1)=\{ \{ 1_{n \times n},0_{n \times n}\},\{0_{n \times n},-1_{n \times n}\}\}$

For the second point, we will assume a hypothetical natural extension of results from a paper that is cited in the last discussion below. That discussion concerns results from an analogue scenario of sympletic matrices. Assuming that the paper can be extended in a natural way, there exists (I will explain why in the last section) an $\tilde{X}$ that verifies the eigen factorization mentioned before

d // DiagonalMatrixQ
X . A . Transpose[X] - d . σ // Norm (* check that
the factorization works *)
Inverse[X] - σ . Transpose[X] . σ // Norm (* check that 
X is an element of O(n,n) *)


(* True

 4.85014*10^-15

 5.88943*10^-15 *)

If I take the hypothesis that the same applies here then basically from the eigen decomposition:

$$ X A \sigma X^{-1}=B\sigma$$

we see that $A \sigma$ and $ B\sigma $ are similar that is there is a change of basis that relates the two. Extending the results from the papers cited to this case, if $ A \sigma $ and $ B \sigma$ are similar and are auto adjoint (a generalization of being symmetric) with respect to the pseudo scalar product whose matrix is $\sigma$=Diag$\left(1,1,...-1,-1...-1\right)$, that is they verify the equation :

$$ \sigma Y^{T} \sigma= Y $$

where the left hand side is the adjoint with respect to $\sigma$, then there exists a change of matrix basis $\tilde{X}$ that is an element of $O(n,n)$. Basically, if both matrices respect the $O(n,n)$ symmetry and are similar then their is a change of basis matrix that respects the symmetry and that is the analogue of a unitary matrix.

It suffices then to show that: