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Recently I come into symplectic vector space and its properties in my linear algebra class.

However, this interesting thing is so different from the usual inner-product spaces I've met before, and I cannot have an intuitive impression of it.

I believe there must be many interesting examples about symplectic spaces, and they may help me understand better~

Coming to the answers: of course the more the better~ thanks~

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Do you know about the natural symplectic structure on $V \oplus V^{\ast}$ for $V$ a vector space? Up to isomorphism all examples look like this... – Qiaochu Yuan May 24 '12 at 12:09
Not exactly, could you explain it? Thanks~ – rhenskyyy May 24 '12 at 12:19
An even more basic example would be $k^{2n}$ with the canonical symplectic form given by the block matrix $\begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix}$. – Zhen Lin May 24 '12 at 12:22

Just like for inner product spaces (I'm assuming here the inner product is positive definite), all symplectic vector spaces are really the same (i.e. they are isomorphic). In the case of an inner product space you can always find an orthonormal basis. In the case of a symplectic vector space you can always find a ``symplectic basis," which is a basis $\{a_1,b_1, \ldots, a_n, b_n\}$ such that $\omega(a_i, b_j) = \delta_{ij}, \omega(a_i,a_j) = 0 = \omega(b_i,b_j)$ where $\omega$ is the symplectic form. To see this, you first pick some $0 \ne a_1 \in V$. Then by the non-degenerate condition you can find a $b_1$ such that $\omega(a_1,b_1) = 1$. Then you consider the symplectic-orthogonal complement of $\text{span}\{a_1,b_1\}$ and repeat the process. Note in particular that this implies any symplectic vector space is even dimensional.

In the example of $V\oplus V^*$ Qiaochu gave in the comments, the symplectic form is $$ \omega((v,\phi),(w,\psi)) = \psi(v) - \phi(w). $$ Then if $v_1,\ldots,v_n$ is a basis for $V$ and $\phi_1,\ldots,\phi_n$ the dual basis for $V^*$ (i.e. $\phi_i(v_j) = \delta_{ij}$), the set $\{v_1,\phi_1,\ldots,v_n,\phi_n\}$ is a symplectic basis for $V\oplus V^*$.

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