and $B\psi=i x \psi$. A complete set of eigenfunctions of $A$ is $$\phi_k(x)=
\begin{cases}
 \sqrt{2} \cos(k\pi x)& \mbox{ when } k \mbox{ is odd,}\\
 \sqrt{2} \sin(k\pi x)& \mbox{ when } k \mbox{ is even,}
\end{cases}
\qquad
k \in {\mathbb{N}},$$ associated with the eigenvalues $i\lambda_k=  - i\frac{k^2 \pi^2}{2}, k \in {\mathbb{N}}$. Notice that $$\langle \phi_{j}, B \phi_{k} \rangle \neq 0$$ if and only if $j+k$ is odd. In particular $\mathcal{V}_{n} \subset\mathfrak{su}(n) $ for every $n$.

We prove by induction on $n$ that $
\mathrm{Lie} \mathcal{V}_{n} = \mathfrak{su}(n)$, and hence that the [Lie–Galerkin Stalking Condition]{} is fulfilled. Notice that the matrices $\mathcal{E}_{2k-1}(B^{(N)})$, for $k \leq N$, have only zero elements in the positions $(j,l)$ for $j\leq N$ and $l\geq N+1$, since $$l^{2}-j^{2} \geq (N+1)^{2} - N^{2} = 2N+1 > 2k-1.$$ Hence $\mathcal{E}_{2k-1}(B^{(N)}) \in \mathcal{V}_{N}$ for $k=1,\ldots,N$. We prove the claim by showing that $$\label{disamina}
\mathrm{Lie}\left( \{\mathcal{E}_{2k-1}(B^{(n)})\mid k=1,\ldots,n\} \right) = \mathfrak{su}(n).$$ For $n=2$, $\mathcal{E}_{3}(B^{(2)}) = 
 \begin{pmatrix}
 0 & b_{12}\\
 -\bar b_{12} & 0
 \end{pmatrix}
$ generates $\mathfrak{su}(2)$ because $b_{12}\neq0$.

Now assume that $$\mathrm{Lie}\left( \{\mathcal{E}_{2k-1}(B^{(n-1)})\mid k=1,\ldots,n-1\} \right) = \mathfrak{su}(n-1),$$ and let us prove . The matrices $\mathcal{E}_{2k-1}(B^{(n)})$ are in $ \mathfrak{su}(n)$ for every $k=1,\ldots,n-1$ and generate the subalgebra of matrices in $ \mathfrak{su}(n)$ with 