A $2n\times 2n$ matrix $A$ is called symplectic if $A^T J A = J$, where $J$ is a fixed invertible, skew symmetric matrix. Generally, $J$ is taken to be the block matrix $J = \begin{pmatrix} 0 & I_n \\ -I_n & 0\end{pmatrix}$.

Is the notion of symplectic matrix independent of the choice of $J$?

  • $\begingroup$ So do you mean, if you choose another $J$, are the resulting matrix groups the same? or isomorphic? or something else? $\endgroup$ Mar 29 '17 at 14:47
  • $\begingroup$ Yes, ideally the class of symplectic matrices would be the same. This is what I am asking. $\endgroup$
    – Ron
    Mar 29 '17 at 15:04
  • $\begingroup$ Hi Rohan, welcome to Math.SE! You have taken good care to properly format and present your question. +1 I can see that you wanted to reply to a comment; you can directly address someone by using @ mentions like so: @Rohan. Usually if you type this at the very beginning of a comment, the comment box will provide you with an autocomplete list which you can navigate through with the up and down arrow keys if there's more than one possible completion, and press tab to select your choice. I think you will enjoy becoming a part of this community and find it very productive and helpful! $\endgroup$ Mar 29 '17 at 15:22
  • $\begingroup$ @ThisIsNotAnId Thanks for the tip. $\endgroup$
    – Ron
    Mar 29 '17 at 15:35

As mentioned above, the groups are all isomorphic, however they are not all the same subset of $M_{2n}(\mathbb R)$.

Example. From MathWorld we have the following nice examples of symplectic matrices:

$$I_4, X = \begin{pmatrix}1&0&0&1\\0&1&1&0\\0&0&1&0\\ 0&0&0&1\end{pmatrix}, Y = \begin{pmatrix}0&1&0&1\\1&0&1&0\\0&0&0&1\\0&0&1&0\end{pmatrix}$$

Let's now try a different skew-symmetric nondegenerate matrix like $$J = \begin{pmatrix}0&0&2&0\\0&0&0&1\\-2&0&0&0\\0&-1&0&0\end{pmatrix}$$

A computation shows $X^TJX \neq J$. You can copy this following into WolframAlpha to check it:


Please note that one should not be that surprised by this! The same thing happens for instance with orthogonal groups. If one changes to a basis that is not orthonormal, the orthogonal matrices expressed in the new basis will no longer belong be in the standard subset $O_n \subset M_n(\mathbb R)$.

There is a little disparity between Wikipedia's definitions of symplectic matrix and orthogonal matrix in this sense. Orthogonal groups are of the form $A^TIA = I$ where $I$ is the identity matrix, but for a general nondegenerate symmetric bilinear form one replaces $I$ with any invertible symmetric matrix $S$. The group $\{A \in GL_n(\mathbb R) | A^TSA = S\} \cong O_n$ is certainly not equal to $O_n$ for all the same reasons as above.


Let $W^{2n}$ be a real vector space of dimension $2n$ and let $\omega$ be an alternating bilinear form which is non-degenerate, then according to the linear symplectic Darboux's theorem, there exists $(e_1,\ldots,e_n,f_1,\ldots,f_n)$ a basis of $W$ such that: $$\omega=\sum_{k=1}^n{e_k}^*\wedge{f_k}^*,$$ said differently the matrix of $\omega$ in this basis is equal to $J$, indeed for all $(i,j)\in\{1,\ldots,n\}^2$, one has: $$\omega(e_i,e_j)=\omega(f_i,f_j)=0,\omega(e_i,f_j)=\delta_{i,j}.$$ Whence, the group of linear endomorphisms of $W$ that preserves $\omega$: $$\{\ell\in\operatorname{End}(W);\forall(v,w)\in W^2,\omega(\ell(v),\ell(w))=\omega(v,w)\}$$ is isomorphic to the usual set of symplectic matrices (taking their matrices in the Darboux basis), which is now an intrinsic definition for $\operatorname{Sp}(2n)$.


  • A bilinear form $\omega$ is non-degenerate if and only if for all $v\in W$, $\omega(v,\cdot)\colon W\to W^*$ is a linear isomorphism.

  • The proof of the linear symplectic Darboux's theorem is done by induction on the half dimension of $W$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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