A completely different answer:

Given $n$-by-$n$ $K$ we can easily cook an $H$ such that $K=H*H'$ (so your $L=I$).

Let $m=n+1$-choose-$2$, associate the columns of $H$ with singletons and pairs of the original rows.

Populate a row $r$ of $H$ so that, in particular, $r$ has a 0 in any column not associated to singleton $\{r\}$ or a pair containing $r$.

Then pick values for the other entries of $H$, first to get the right off-diagonal entries of $K$ (the doubleton columns), and lastly to get the right diagonal entries (the singleton columns).

Since $H$ comes so cheap giving $K$, receiving such an $H$ tied-up-with-string can't genuinely simplify the decomposition of $K$.