Why is the determinant of a symplectic matrix 1?

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.

• Another remark: You may assume that $A_1$ is invertible since it can be approximated by invertible matrices. Jan 3, 2013 at 15:52
• The symplectic group is generated by Chevalley generators, each of determinant 1. Thus, every element of the symplectic group has determinant 1. A proof similar to row-column operations is given here arxiv.org/pdf/1504.03794.pdf Jul 27, 2019 at 3:33

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 $A J A^T = J$.

• Does the "Pfaffian argument" coincide with my answer? Jan 2, 2013 at 3:01
• @Andrew: basically, yes. Thanks for writing it up. Jan 3, 2013 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, 2013 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. Oct 8, 2013 at 8:34
• Gotcha. The Pfaffian can be zero. My bad. Thanks!
– Doug
Oct 18, 2013 at 20:54

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 $\omega^n$.

• Correction: A symplectic for is an alternating two-form ... Jul 23, 2017 at 3:02
• I wonder whether there exists a basis-free proof of $\omega^n\ne0$ from the non-degeneracy of $\omega$. Mar 10, 2022 at 3:49

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 http://arxiv.org/abs/1505.04240 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 $\det(A) = 1$.

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 $A_1$.

• 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. Jul 13, 2016 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, 2016 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. Jul 22, 2016 at 4:04
1. The most natural way to show [independent of the field $$\mathbb{F}$$ of characteristics $$\neq 2$$] that the symplectic group $$Sp(2n,\mathbb{F})~:=~\{M\in {\rm Mat}_{2n\times 2n}(\mathbb{F})\mid MJM^T=J\} \tag{1}$$ consists of matrices $$M$$ with unit determinant is to use the following Pfaffian property \begin{align} {\rm Pf}(J)~\stackrel{(1)}{=}~{\rm Pf}(MJM^T) =~&{\rm Det}(M)~{\rm Pf}(J) \cr \Downarrow~&\cr {\rm Det}(M)~=~&+1,\end{align}\tag{2} as Willie Wong hints in his answer.

2. An elementary proof [independent of the field $$\mathbb{F}$$ of characteristics $$=0$$] of the Pfaffian property (2) is e.g. given in my Math.SE answer here.

• Since your proof of the Pfaffian property in this other answer that you refer to does not hold in arbitrary characteristic (already because the formula of the Pfaffian you are using there involves a division by $n!$), I would like to give a proper reference to this Pfaffian property in arbitrary characteristic: see for example E. Artin, Geometric Algebra, Theorem 3.28 of the first edition (1957). Feb 14, 2019 at 18:43
• Good point. Thanks. Feb 14, 2019 at 18:53
• @user293657: There is a way of proving the Pfaffian formula using Grassmann-Berezin integration (much along the lines of math.stackexchange.com/a/2705240/11127 ) that works over any commutative ring. You will need to use Propositions A.10 and A.13 of Caracciolo/Sokal/Sportiello, arXiv:1105.6270v2. (See Sections A.3-A.5 of that paper for the definition of Grassmann integrals and $e^{\frac12 \chi^T A \chi}$ over an arbitrary commutative ring.) Feb 14, 2019 at 19:18
• @darij This paper is an amazing reference, thank you for giving it. Feb 14, 2019 at 20:36

Your approach can be remedied to work over any field $$\mathbb F$$. As you already know, if $$A=\pmatrix{X&Y\\ Z&W} \text{ where } X,Y,Z,W\in M_n(\mathbb F),$$ the equation $$A^TJA=J$$ means precisely that $$X^TZ,\,Y^TW$$ are symmetric and $$X^TW-Z^TY=I$$. In particular, every matrix in the form of $$M^T\oplus M^{-1}$$ is a solution. Let $$Z=PDQ$$ where $$P$$ and $$Q$$ are some invertible matrices and $$D$$ is a diagonal matrix. Then $$P^T\oplus P^{-1}$$ and $$Q^{-1}\oplus Q^T$$ are members of $$Sp(2n,\mathbb F)$$. It follows that $$A\in Sp(2n,\mathbb F)$$ if and only if $$B=\pmatrix{P^T\\ &P^{-1}}A\pmatrix{Q^{-1}\\ &Q^T}=\pmatrix{\ast&\ast\\ D&\ast} \in Sp(2n,\mathbb F)$$ and $$\det(A)=1$$ if and only if $$\det(B)=1$$.

In other words, we may assume without loss of generality that the sub-block $$Z$$ is a diagonal matrix. Let $$t$$ be an indeterminate. Then $$A+tI$$ is invertible over $$\mathbb F(t)$$. Therefore, using Schur complement and the conditions that $$X^TZ$$ is symmetric and $$X^TW-Z^TY=I$$, we obtain \begin{aligned} &\det\pmatrix{X+tI&Y\\ Z&W}\\ &=\det(X+tI)\det\left(W-Z(X+tI)^{-1}Y\right)\\ &=\det(X^T+tI)\det\left(W-Z(X+tI)^{-1}Y\right)\\ &=\det\left((X^T+tI)W-(X^T+tI)Z(X+tI)^{-1}Y\right)\\ &=\det\left(X^TW+tW-(X^TZ+tZ)(X+tI)^{-1}Y\right)\\ &=\det\left(X^TW+tW-(X^TZ+tZ^T)(X+tI)^{-1}Y\right)\ \text{(because Z is diagonal)}\\ &=\det\left(X^TW+tW-(Z^TX+tZ^T)(X+tI)^{-1}Y\right)\ \text{(because X^TZ is symmetric)}\\ &=\det\left(X^TW+tW-Z^TY\right)\\ &=\det(I+tW) \ \text{(because X^TW-Z^TY=I)}. \end{aligned} Put $$t=0$$, we get $$\det(A)=\det(I)=1$$.

Edit (2021-05-07). I have just learnt that a proof very similar in spirit to mine was given in Büngera and Rump, Yet more elementary proofs that the determinant of a symplectic matrix is 1, Linear Algebra and Its Applications, 515 (2017) 87-95. See proof II of their lemma 1 on p.92.

• thanks very much. $(A\cos\theta+J\sin\theta)J(A\cos\theta+J\sin\theta)^T=J+(A-A^T)\cos\theta \sin\theta$. I can't verify the above identity. Oct 9, 2013 at 7:47

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)

For $$A \in M_{2n}(\mathbb R)$$, a very simple proof comes via Polar Decomposition. If $$A \in G$$, the symplectic group, then $$\det\big(A\big)\in \big\{-1,1\big\}$$ by taking determinants of the stabilizer equation, i.e. of $$AJA^T = J$$. And by taking inverses of the stabilizer equation $$A \in G \implies A^T \in G\implies A^TA \in G$$.

Using Polar Decomposition, write $$A= U\big(A^TA\big)^\frac{1}{2}$$. Now $$A^TA$$ is diagonalizable with all positive eigenvalues, so we may use a polynomial argument to show $$A^TA \in G\implies (A^TA)^\frac{1}{2} \in G$$. The proof is identical to the 'shorter proof' here Polar Decomposition of $O(p,q)$ , except we write $$J$$ instead of $$I_{p,q}$$.

Thus we have $$A \in G \implies (A^TA)^\frac{1}{2} \in G\implies U\in G$$, i.e.
$$UJU^{-1}= UJU^T =J\implies \det\big(U\big)\geq 0\implies \det\Big(A\Big)\geq 0\implies \det\Big(A\Big)=1$$. The first implication is justified because any matrix with real components commuting with $$J$$ must have determinant $$\geq 0$$, see Nonnegativity of the determinant of a commuting matrix. I suggest my analytic proof, i.e. that via simultaneous triangularization (in fact simultaneous diagonalization since $$U$$ and $$J$$ are normal) we can conclude that $$\big(U+\frac{1}{k}J\big)$$ has all non-real eigenvalues (coming in conjugate pairs) for $$k$$ large enough, which gives the result by taking limits. The fact that $$\dim \ker \big(U+I\big)$$, a J-invariant subspace, cannot be odd dimensional will give the desired result as well.