After my question whose answer turned out to be false, I re-examined the course of my proof, which is actually seperate from the one in my question, and found out that there's another condition, at the start of the proof, that may make the proof work.
Again, long story short, the question I'll be asking here is the following: given two real singular matrices $A, B$ s.t. $AB^T$ is symmetric, is the purely real pair $E, Z$ guaranteed to exist s.t. $BE = A + Z$ with $E$ being symmetric and $Z$ diagonal? (Note that, were $B$ not singular, $B^{-1}A$ would have been symmetric and the answer would be trivially yes)
EDIT: Both matrices are assumed to be not identically zero.
