I wish to have a proof for the following result:


Let $U\_n$ be an $n\times n$ upper [shift matrix](https://en.wikipedia.org/wiki/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 $C\_n$.



Prove: Lie algebra generated by two $n\times n$ shift matrices is $\mathfrak{so}(n,\mathbb{C})$ ($n$ odd) or $\mathfrak{sp}(n,\mathbb{C})$ ($n$ even)
Let ${\mathfrak g} $ be the Lie algebra generated by $U\_n$ and $L\_n$. It's easy to check that $U\_n$ and $L\_n$ preserve the bilinear form determined by the matrix with $1,-1,1,\ldots$ down the antidiagonal and zero elsewhere. So ${\mathfrak g}$ is contained in (a Lie algebra isomorphic to) $\mathfrak{so} \_n$ for odd $n$, $\mathfrak{sp} \_n$ for even $n$. Let's assume $n$ is odd; the argument is the same for even $n$. Clearly ${\mathfrak g}=\mathfrak{so} \_3$ if $n=3$. In the general case, $[U\_n, L\_n]=H\_n$ is a diagonal matrix with just two non-zero entries $\pm 1$ and $U\_n-[H\_n, U\_n]$, resp. $L\_n+[H\_n, L\_n]$ equals $U\_{n-2}$, resp. $L\_{n-2}$ in the "middle" matrix subalgebra. By induction, ${\mathfrak g}$ contains $\mathfrak{so} \_{n-2}$. Now it's easy to see that $[H\_n, U\_n]$, resp. $[H\_n, L\_n]$ is a multiple of the "missing" positive, resp. negative root element, so in fact ${\mathfrak g}$ equals $\mathfrak{so} \_n$.


