Appendix

Proof 35 In [22], Rim proposed an elementary proof that real symplectic matrices have determinant one; following the same procedure, we prove that the symplectic matrices have determinant μn. From the definition

det <sup>M</sup>TJM <sup>¼</sup> det <sup>M</sup><sup>T</sup> detð Þ<sup>J</sup> detð Þ¼ <sup>M</sup> detð Þ¼ <sup>μ</sup><sup>J</sup> <sup>μ</sup><sup>2</sup><sup>n</sup> ! detð Þ¼� <sup>M</sup> <sup>μ</sup>n, therefore it is necessary to prove that det½ �¼� <sup>M</sup> <sup>μ</sup><sup>2</sup><sup>n</sup> is false. Considering the matrix

<sup>S</sup> <sup>¼</sup> <sup>M</sup>TM <sup>þ</sup> <sup>μ</sup>I2<sup>n</sup> since MTM <sup>≥</sup><sup>0</sup> and <sup>μ</sup>∈ð � <sup>0</sup>; <sup>1</sup> , the matrix S has real and greater than <sup>μ</sup> eigenvalues:

$$\det(\mathbf{S}) = \det(\mathbf{M}^T \mathbf{M} + \mu I\_{2n}) \ge \mu. \tag{53}$$

Now from the definition M�<sup>T</sup> <sup>¼</sup> <sup>μ</sup>�<sup>1</sup>JMJ�<sup>1</sup> and rewriting S,

$$\mathbf{S} = \mathbf{M}^T \mathbf{M} + \mu \mathbf{I}\_{2n} = \mathbf{M}^T \left( \mathbf{M} + \mu \mathbf{M}^{-T} \right) = \mathbf{M}^T \left( \mathbf{M} + \mathbf{J} \mathbf{M} \mathbf{J}^{-1} \right),$$

denotes the subblocks of M as follows:

$$M = \begin{bmatrix} M\_{11} & M\_{12} \\ M\_{21} & M\_{21} \end{bmatrix}$$

with <sup>M</sup>11, M12, M21, M<sup>21</sup> <sup>∈</sup> <sup>R</sup>n�n; thus <sup>M</sup> <sup>þ</sup> JMJ�<sup>1</sup> <sup>¼</sup> <sup>M</sup><sup>11</sup> <sup>þ</sup> <sup>M</sup><sup>22</sup> <sup>M</sup><sup>12</sup> � <sup>M</sup><sup>21</sup> <sup>M</sup><sup>21</sup> � <sup>M</sup><sup>12</sup> <sup>M</sup><sup>11</sup> <sup>þ</sup> <sup>M</sup><sup>22</sup> � � if A ¼ M<sup>11</sup> þ M<sup>22</sup> and B ¼ M<sup>12</sup> � M21; then

$$M + jM\!\!\!/\!^{-1} = \begin{bmatrix} A & B \\ -B & A \end{bmatrix}.\tag{54}$$

We rewrite (54) with the unitary transformation T:

$$T = \frac{1}{\sqrt{2}} \begin{bmatrix} I\_n & I\_n \\ iI\_n & -iI\_n \end{bmatrix} \Leftrightarrow T^{-1} = \frac{1}{\sqrt{2}} \begin{bmatrix} I\_n & -iI\_n \\ I\_n & iI\_n \end{bmatrix}$$

$$[M + jM]^{-1} = \begin{bmatrix} A & B \\ -B & A \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} I\_n & I\_n \\ iI\_n & -iI\_n \end{bmatrix} \begin{bmatrix} A + iB & 0 \\ 0 & A - iB \end{bmatrix} \frac{1}{\sqrt{2}} \begin{bmatrix} I\_n & -iI\_n \\ I\_n & iI\_n \end{bmatrix}$$

therefore

$$\begin{aligned} \det(S) &= \det\left(\mathcal{M}^T (\mathcal{M} + j\mathcal{M}\mathcal{J}^{-1})\right) = \det(\mathcal{M}^T) \det(\mathcal{M} + j\mathcal{M}\mathcal{J}^{-1}) \\ &= \det(\mathcal{M}) \det(A + iB) \det(A - iB) = \det(\mathcal{M}) \det(A + iB) \det(\overline{A + iB}) \end{aligned}$$

since A, B are real, the complex conjugation commute with the determinant, and then

$$\det(\mathbf{S}) = \det(M)\det(A + iB)\overline{\det(A - iB)} = \det(M)|\det(A + iB)|^2$$

and from Eq. (53)

$$0 \le \mu \le \det(\mathbb{S}) = \det(M) \left| \det(A + iB) \right|^2$$

then j j detð Þ <sup>A</sup> <sup>þ</sup> iB <sup>2</sup> . <sup>0</sup> and detð Þ <sup>M</sup> . <sup>0</sup>; therefore detð Þ¼ <sup>M</sup> <sup>μ</sup><sup>n</sup>. ■

Dynamical Systems Theory
