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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. –  Andrew Jan 3 '13 at 15:52
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3 Answers

up vote 13 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? –  Andrew 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 $SO(n)$? –  Dan Douglas 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! –  Dan Douglas Oct 18 '13 at 20:54
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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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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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