Completing the Symplectic Basis

When determining the Stinespring dilation for a Gaussian channel, we require to complete the symplectic matrix from some given set of vectors. Here is a formal problems statement.

Problem. Suppose $n$ is a positive integer. We are given a set of vectors $(s_j)_{j = 1}^{2n}$ in $\mathbb R^{6n}$ such that $\langle s_j, \Omega_{3n} s_k\rangle = (\Omega_{3n})_{j, k}$ for all $k, j \in [2n]$ . Here $\Omega_{3n}$ is the symplectic form of order $6n$ . Prove that there exist a completion of this symplectic basis set, i.e. $(s_{k})_{k = 2n + 1}^{6n}$ such that $\langle s_j, \Omega_{3n} s_k\rangle = (\Omega_{3n})_{j, k}$ for all $j, k \in [6n]$ .

Solution. We show that a choice of vectors $(s_{2n +1}, s_{2n + 2})$ exist. Given $(s_j)_{j = 1}^n$ , choose any pair of orthogonal vectors $v_1, v_2$ both of norm $1$ that is in the orthogonal compliment of $span\{(\Omega_{3n} s_j)_{j = 1}^{2n}\}$ . At least two such vectors are guaranteed to exist by a dimensionality argument (Gram-Schmidt). Set

$s_{2n + 1} = v_1, \ \ s_{2n + 2} = \frac{v_2}{\langle v_1, \Omega_{3n} v_2\rangle^{-1}}$

where the denominator of the fraction can be chosen nonzero. Such a choice is possible, and we leave it as an exercise that such choice exist. It is straightforward to verify

$\langle s_j, \Omega s_{2n + 1}\rangle = \langle s_j, \Omega s_{2n + 1}\rangle = 0, \ \forall j \in [2n]$

and

$\langle s_{2n +1}, s_{2n}\rangle \ = \ 1.$

We can repeat the inductive argument to choose rest of the vectors.

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