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 $U_n-$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 $\mathfrak{so} _n$.
