@pete The previous attempt yielded the same result.

For those interested in a more citeable resource. The result in this answer has been derived in Lemma 1 of doi.org/10.1016/0304-4076(89)90059-6.

@pete man are you difficult to convince :) . It does check out numerically as well, obviously.

@pete In fact you can write $Q = \frac{1}{2} (DL)^\top$ using Lemma 4.4 in epubs.siam.org/doi/abs/10.1137/0601049.

@pete If $D$ is the duplication matrix and $L$ is an elimination matrix (which is not unique, see math.stackexchange.com/a/4081768/185047) then we can just take the Moore-Penrose inverse of $D$ as $L$, $L=(DD')^{-1}D'$, in which case it should work as I said.

It must be the Kronecker delta: en.wikipedia.org/wiki/Kronecker_delta

Quick question: What is $\delta_{kl}$ here?