Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

Thanks

share|improve this question
    
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
add comment

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 $A J A^T = J$.

share|improve this answer
    
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
1  
@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
add comment

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 $A_1$.

share|improve this answer
add comment

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 $\omega^n$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.