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)

Question

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 $C_n$.

Answer 1

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 $\mathfrak{so} _n$.