I wish to have a proof for the following result:

Let $U_n$ be an $n\times n$ upper shift matrix, and $L_n = U_n^T$ be a lower shift matrix. For example, $$ U_5 = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}, \quad L_5 = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{pmatrix}. $$ Then the Lie algebra generated by $U_n$ and $L_n$ over $\mathbb{C}$ is isomorphic to $\mathfrak{so}(n, \mathbb{C})$ when $n$ is odd, and is isomorphic to $\mathfrak{sp}(n, \mathbb{C})$ when $n$ is even.

I tested this numerically with `GAP`

over the field $\mathbb{Q}$: the Lie algebra generated by $U_{2n+1}$ and $L_{2n+1}$ over $\mathbb{Q}$ is of type $B_n$, while the Lie algebra generated by $U_{2n}$ and $L_{2n}$ over $\mathbb{Q}$ is of type $C_n$.

isomorphicto $\mathfrak{so}(n,\mathbb{C})$ or $\mathfrak{sp}(n,\mathbb{C})$. $\endgroup$5more comments