= Four-dimensional symplectic basis change
On $K^4$, use the <standard basis> $\mathcal U=(u_1,u_2,u_3,u_4)$ and the <alternating bilinear form> with <matrix> $J$ above for $n=2$. A second <symplectic basis> is
$$
e_1=u_1,\quad e_2=u_2,\quad
f_1=u_1+u_2+u_3,\quad f_2=u_1+u_4.
$$
For instance, $\omega(f_1,f_2)=1-1=0$, while $\omega(e_1,f_1)=\omega(e_2,f_2)=1$ and the cross-pairings vanish. The <change of basis> <matrix>, whose columns are these new <basis vectors> in $\mathcal U$, is
$$
S=\begin{pmatrix}
1&0&1&1\\
0&1&1&0\\
0&0&1&0\\
0&0&0&1
\end{pmatrix}
=\begin{pmatrix}I_2&C\\0&I_2\end{pmatrix},\qquad
C=\begin{pmatrix}1&1\\1&0\end{pmatrix}.
$$
The <symmetric matrix> $C$ makes $S^TJS=J$, so $S$ is a <symplectic matrix>. Explicitly, multiplying the blocks gives a lower-right block $C^T-C=0$; the other blocks agree with $J$. For a fixed <vector>, <change of basis> means $[v]_{\mathcal U}=S[v]_{\mathcal B}$, hence
$$
[v]_{\mathcal U}=(3,2,1,1)^T\quad\Longleftrightarrow\quad
[v]_{\mathcal B}=(1,1,1,1)^T,\qquad
[w]_{\mathcal U}=(1,0,0,1)^T\quad\Longleftrightarrow\quad
[w]_{\mathcal B}=(0,0,0,1)^T.
$$
Their <symplectic form> gives $3\cdot0+2\cdot1-1\cdot1-1\cdot0=1$ in $\mathcal U$, and $1\cdot0+1\cdot1-1\cdot0-1\cdot0=1$ in $\mathcal B$. The separate coordinate-plane contributions change, but their sum does not.
The same <matrix> also represents an active <symplectic linear map>: keeping one <basis> fixed, it sends $(q,p)$ to $(q+Cp,p)$. The following diagram shows this active interpretation over the <real numbers>. Its preserved total pairing is precisely the test used for linear <canonical transformations>.
\Image[/symplectic-shear-pairing-preservation.png]
{title=A symplectic shear redistributes projected areas while preserving their sum}
{description=Before the map the two signed areas are zero and one; afterwards they are minus one and two. The block shear q maps to q plus Cp with symmetric C, so the total symplectic pairing remains one.}
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