The title says it (almost) all.
I try to minimise $\Vert V^T V - T \Vert_2$ over $V \in \mathbb{R}^{m,n}$ for a given matrix $T \in \mathbb{R}^{n,n}$. $n$ is bigger than $m$. The L1-norm would probably also be ok.
Additionally I remove the diagonal of $V^T V$ and $T$. But if that isn't possible, I can probably work around it.
There are some further constraints, for example the entries of $V$ are supposed to be between 0 and 1.0., but those seem to pop out of a solution anyway.
I tried minimising by gradient descent, but though the results make sense, they aren't close enough to the global minimum. So now I'm looking for a more sophisticated solver. The equation is so simple that it seems like there should be something, but I'm not sure how to search for it.
There is also the option of making $m$ equal to $n$, maybe that allows for some exact solution of $V^T V = T$?
I also asked this question on CrossValidated, where nothing much seems to be happening. Maybe Maths is a better fit. I delete the other one if double questions are frowned upon.
Define the symmetric matrix $\,S=\tfrac{1}{2}(T+T^T)$
Let's also define the difference in dimensions as $\,\,k=(n-m)$
Since $S$ is real symmetric its eigen-decomposition consists of an orthogonal and diagonal matrix $$S = QDQ^T$$
Since we want a reduced rank factorization, identify the $(k)$ eigenvalues with the smallest magnitudes. Eliminate the columns of $Q$ which correspond to these eigenvalues, as well as the corresponding rows/columns of $D$. This leaves a factorization in terms of the reduced matrices $$S \approx Q_rD_rQ_r^T = V^TV$$ Now identify the desired matrix as $$V = D_r^{1/2}Q_r^T$$