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suppose $A \in M_{2n}(\mathbb{R})$. and$$J=\begin{pmatrix} 0 & E_n\\ -E_n&0 \end{pmatrix}$$ where $E_n$ represents identity matrix.

if $A$ satisfies $$AJA^T=J$$

How to figure out $$\det(A)=1$$

My approach:

I have tried to separate $A$ into four submartix:$$A=\begin{pmatrix}A_1&A_2 \\A_3&A_4 \end{pmatrix}$$ and I must add a assumption that $A_1$ is invertible. by elementary transfromation:$$\begin{pmatrix}A_1&A_2 \\ A_3&A_4\end{pmatrix}\rightarrow \begin{pmatrix}A_1&A_2 \\ 0&A_4-A_3A_1^{-1}A_2\end{pmatrix}$$

we have: $$\det(A)=\det(A_1)\det(A_4-A_3A_1^{-1}A_2)$$ from$$\begin{pmatrix}A_1&A_2 \\ A_3&A_4\end{pmatrix}\begin{pmatrix}0&E_n \\ -E_n&0\end{pmatrix}\begin{pmatrix}A_1&A_2 \\ A_3&A_4\end{pmatrix}^T=\begin{pmatrix}0&E_n \\ -E_n&0\end{pmatrix}$$ we get two equalities:$$A_1A_2^T=A_2A_1^T$$ and $$A_1A_4^T-A_2A_3^T=E_n$$

then $$\det(A)=\det(A_1(A_4-A_3A_1^{-1}A_2)^T)=\det(A_1A_4^T-A_1A_2^T(A_1^T)^{-1}A_3^T)=\det(A_1A_4^T-A_2A_1^T(A_1^T)^{-1}A_3^T)=\det(E_n)=1$$

but I have no idea to deal with this problem when $A_1$ is not invertible...


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Another remark: You may assume that $A_1$ is invertible since it can be approximated by invertible matrices. – Yang Zhou Jan 3 '13 at 15:52
up vote 15 down vote accepted

First, taking the determinant of the condition $$ \det AJA^T = \det J \implies \det A^TA = 1 $$ using that $\det J \neq 0$. This immediately implies $$ \det A = \pm 1$$ if $A$ is real valued. The quickest way, if you know it, to show that the determinant is positive is via the Pfaffian of the expression $A J A^T = J$.

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Does the "Pfaffian argument" coincide with my answer? – Yang Zhou Jan 2 '13 at 3:01
@Andrew: basically, yes. Thanks for writing it up. – Willie Wong Jan 3 '13 at 9:01
Is there a way to connect the fact that skew-symmetric matrices have positive determinant to the fact that they are the Lie algebra of $SO(n)$? – Doug Oct 7 '13 at 16:23
@DanDouglas: the $3\times 3$ matrix $\begin{pmatrix} 0 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & 1 & 0 \end{pmatrix}$ is skew, and has determinant 0. The condition that $AJA^T = J$ is stronger than skew symmetry. – Willie Wong Oct 8 '13 at 8:34
Gotcha. The Pfaffian can be zero. My bad. Thanks! – Doug Oct 18 '13 at 20:54

The determinant is a continuous function, and the set of symplectic matrices with invertible $A_1$ is dense in the set of all symplectic matrices. So if you've proven that it equals 1 for all invertible $A_1$, then it equals 1 for all $A_1$.

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Hi @Ted, do you have a reference for this claim? I have a simple proof that the determinant of a symplectic matrix equals 1 for all invertible $A_1$, but I have no idea how to show that the set of symplectic matrices with invertible $A_1$ is dense in the set of all symplectic matrices. – William Jul 13 at 17:43
@William I can't seem to reconstruct the argument now and maybe it wasn't as easy as I thought when I thought when I wrote this. One method may be to start with the fact that the block matrix (1 B ; 0 1) (where B is symmetric) and its transpose are both symplectic matrices. So by left and right multiplication, for any $A_1$ we can also get all $A_1 + B A_3$ and $A_1 + A_2 B$ in the upper left corner. – Ted Jul 22 at 4:02
Maybe I should ask this as a separate question so if you do remember the argument later you can get more recognition for it? It leads to a very nice and clean argument to prove the overall result, since the proof that the determinant is 1 when $A_1$ has non-zero determinant is very simple. – William Jul 22 at 4:04

Let me first restate your question in a somewhat more abstract way. Let $V$ be a finite dimensional real vector space. A sympletic form is a 2-form $\omega\in \Lambda^2(V^\vee)$ which is non-degenerate in the sense that $\omega(x,y)=0$ for all $y\in V$ implies that $x=0$. $V$ together with such a specified $\omega$ nondegerate 2-form is called a symplectic space. It can be shown that $V$ must be of even dimension, say, $2n$.

A linear operator $T:V\to V$ is said to be a symplectic transformations if $\omega(x,y)=\omega(Tx,Ty)$ for all $x,y\in V$. This is the same as saying $T^*\omega=\omega$. What you want to show is that $T$ is orientation preserving. Now I claim that $\omega^n\neq 0$. This can be shown by choosing a basis $\{a_i,b_j|i,j=1,\ldots,n\}$ such that $\omega(a_i,b_j)=\delta_{ij}$ and $\omega(a_i,a_j)=\omega(b_i,b_j)=0$, for all $i,j=1,\ldots,n $. Then $\omega=\sum_ia_i^\vee\wedge b_i^\vee$, where $\{a_i^\vee,b_j^\vee\}$ is the dual basis. We can compute $\omega^n=n!a_1^\vee\wedge b_1^\vee\wedge\dots\wedge a_n^\vee \wedge b_n^\vee$, which is clearly nonzero.

Now let me digress to say a word about determinants. Let $W$ be an n-dimensional vector space and $f:W\to W$ be linear. Then we have induced maps $f_*:\Lambda^n(W)\to \Lambda^n(W)$. Since $\Lambda^n(W)$ is 1-dimensional, $f_*$ is multiplication by a number. This is just the determinant of $f$. And the dual map $f^*:\Lambda^n(W^\vee)\to \Lambda^n(W^\vee)$ is also multiplication by the determinant of $f$.

Since $T^*(\omega^n)=\omega^n$, we can see from the above argument that $\det(T)=1$. The key point here is that the sympletic form $\omega$ give a canonical orientation of the space, via the top from $\omega^n$.

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There is an easy proof for real and complex case which does not require the use of Pfaffians. This proof first appeared in a Chinese text. Please see for the reference.

I reproduce the proof for the real case here. The approach extends to complex symplectic matrices.

Taking the determinant on both sides of $A^T J A = J$, $$\det(A^T J A) = \det(A^T) \det(J) \det(A) = \det(J).$$ So we immediately have that $\det(A) = \pm 1$.

Then let us consider the matrix $A^TA + I.$ Since $A^TA$ is symmetric positive definite,
its eigenvalues are real and greater than $1$.Therefore its determinant, being the product of its eigenvalues, has $\det(A^TA +I) > 1$.

Now as $\det(A) \ne 0$, $A$ is invertible. Using this we may write $$ A^TA + I = A^T( A + A^{-T}) = A^T(A + JAJ^{-1}).$$ Denote the four $N \times N$ subblocks of $A$ as follows, $$ A = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix}, \text{ where } A_{11},A_{12},A_{21},A_{22} \in \mathbb{R}^{N \times N}. $$ Then we compute $$ A + JAJ^{-1} = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix} + \begin{bmatrix} O & I_N \\ -I_N & O \end{bmatrix} \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix} \begin{bmatrix} O & - I_N \\ I_N & O \end{bmatrix} = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix} + \begin{bmatrix} A_{22} & -A_{21} \\ -A_{12} & A_{11} \end{bmatrix} = \begin{bmatrix} A_{11}+ A_{22} & A_{12} - A_{21} \\ - A_{12}+ A_{21} & A_{11} + A_{22} \end{bmatrix}.$$ Writing the blocks as $C := A_{11} + A_{22}$ and $D:= A_{12} - A_{21}$, we make use of a unitary transform $$ A + JAJ^{-1} = \begin{bmatrix} C & D \\ -D & C \end{bmatrix} = \frac{1}{\sqrt{2}}\begin{bmatrix} I_N & I_N \\ iI_N & -iI_N \end{bmatrix} \begin{bmatrix} C + i D & O \\ O & C - i D \end{bmatrix} \frac{1}{\sqrt{2}} \begin{bmatrix} I_N & -iI_N \\ I_N & iI_N \end{bmatrix}. $$ We plug this factorization into our identity. Note that $C,D$ are both real. This allows the complex conjugation to commute with the determinant (as it is a polynomial of its entries) $$0 < 1 < \det(A^TA + I) = \det(A^T(A + JAJ^{-1})) \\ = \det(A) \det(C + i D) \det(C - iD) \\ = \det(A) \det(C + i D) \det\left(\overline{C + iD}\right)\\ = \det(A) \det(C + iD) \overline{\det(C + iD)} = \det(A) \left\lvert \det(C + iD)\right\rvert^2. $$ Clearly, none of the two determinants on the RHS can be zero, so we may conclude $\left\lvert \det(C + iD) \right\rvert^2 > 0$. Dividing this through on both sides, we have $\det(A) > 0$, and thus $\det(A) = 1$.

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Every symplectic matrix is the product of two symplectic matrices with lower-left corner invertible. See: M. de Gosson, s Symplectic Geometry and Quantum Mechanics, Birkhäuser, Basel, series "Operator Theory: Advances and Applications" (subseries: "Advances in Partial Differential Equations"), Vol. 166 (2006)

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