Complete the proof by showing that, given any non-zero vector  , there is always a non-zero vector  such that  is symplectic.

To prove that given any non-zero vector \(\begin{bmatrix} a \\ b \end{bmatrix}\), there is always a non-zero vector \(\begin{bmatrix} a' \\ b' \end{bmatrix}\) such that \(\begin{bmatrix} a&a' \\ b&b' \end{bmatrix}\) is symplectic, we need to ensure that the determinant of the matrix is 1. A symplectic matrix is defined by having the property that \(AA^T - I = 0\), where \(A\) is the symplectic matrix and \(I\) is the identity matrix.

Let's consider the matrix \(\begin{bmatrix} a&a' \\ b&b' \end{bmatrix}\). To ensure it is symplectic, we need:

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